Learning Objectives

- Use partial derivatives to locate critical points for a function of two variables.
- Apply a second derivative test to identify a critical point as a local maximum, local minimum, or saddle point for a function of two variables.
- Examine critical points and boundary points to find absolute maximum and minimum values for a function of two variables.

One of the most useful applications for derivatives of a function of one variable is the determination of maximum and/or minimum values. This application is also important for functions of two or more variables, but as we have seen in earlier sections of this chapter, the introduction of more independent variables leads to more possible outcomes for the calculations. The main ideas of finding critical points and using derivative tests are still valid, but new wrinkles appear when assessing the results.

## Critical Points

For functions of a single variable, we defined critical points as the values of the variable at which the function's derivative equals zero or does not exist. For functions of two or more variables, the concept is essentially the same, except for the fact that we are now working with partial derivatives.

Definition: Critical Points

Let (z=f(x,y)) be a function of two variables that is differentiable on an open set containing the point ((x_0,y_0)). The point ((x_0,y_0)) is called a *critical point* of a function of two variables (f) if one of the two following conditions holds:

- (f_x(x_0,y_0)=f_y(x_0,y_0)=0)
- Either (f_x(x_0,y_0) ; ext{or} ; f_y(x_0,y_0)) does not exist.

Example (PageIndex{1}): Finding Critical Points

Find the critical points of each of the following functions:

- (f(x,y)=sqrt{4y^2−9x^2+24y+36x+36})
- (g(x,y)=x^2+2xy−4y^2+4x−6y+4)

**Solution**

**a**. First, we calculate (f_x(x,y) ; ext{and} ; f_y(x,y):)

[egin{align*} f_x(x,y) &=dfrac{1}{2}(−18x+36)(4y^2−9x^2+24y+36x+36)^{−1/2} [4pt] &=dfrac{−9x+18}{sqrt{4y^2−9x^2+24y+36x+36}} end{align*}]

[egin{align*} f_y(x,y) &=dfrac{1}{2}(8y+24)(4y^2−9x^2+24y+36x+36)^{−1/2} [4pt] &=dfrac{4y+12}{sqrt{4y^2−9x^2+24y+36x+36}} end{align*}.]

Next, we set each of these expressions equal to zero:

[egin{align*} dfrac{−9x+18}{sqrt{4y^2−9x^2+24y+36x+36}} &=0 [4pt] dfrac{4y+12}{sqrt{4y^2−9x^2+24y+36x+36}} &=0. end{align*}]

Then, multiply each equation by its common denominator:

[egin{align*} −9x+18 &=0 [4pt] 4y+12 &=0. end{align*}]

Therefore, (x=2) and (y=−3,) so ((2,−3)) is a critical point of (f).

We must also check for the possibility that the denominator of each partial derivative can equal zero, thus causing the partial derivative not to exist. Since the denominator is the same in each partial derivative, we need only do this once:

[4y^2−9x^2+24y+36x+36=0. label{critical1}]

Equation ef{critical1} represents a hyperbola. We should also note that the domain of (f) consists of points satisfying the inequality

[4y^2−9x^2+24y+36x+36≥0.]

Therefore, any points on the hyperbola are not only critical points, they are also on the boundary of the domain. To put the hyperbola in standard form, we use the method of completing the square:

[egin{align*} 4y^2−9x^2+24y+36x+36 &=0 [4pt] 4y^2−9x^2+24y+36x &=−36 [4pt] 4y^2+24y−9x^2+36x &=−36 [4pt] 4(y^2+6y)−9(x^2−4x) &=−36 [4pt] 4(y^2+6y+9)−9(x^2−4x+4) &=−36−36+36 [4pt] 4(y+3)^2−9(x−2)^2 &=−36.end{align*}]

Dividing both sides by (−36) puts the equation in standard form:

[egin{align*} dfrac{4(y+3)^2}{−36}−dfrac{9(x−2)^2}{−36} &=1 [4pt] dfrac{(x−2)^2}{4}−dfrac{(y+3)^2}{9} &=1. end{align*}]

Notice that point ((2,−3)) is the center of the hyperbola.

Thus, the critical points of the function (f) are ( (2, -3) ) and all points on the hyperbola, (dfrac{(x−2)^2}{4}−dfrac{(y+3)^2}{9}=1).

**b**. First, we calculate (g_x(x,y)) and (g_y(x,y)):

[egin{align*} g_x(x,y) &=2x+2y+4 [4pt] g_y(x,y) &=2x−8y−6. end{align*}]

Next, we set each of these expressions equal to zero, which gives a system of equations in (x) and (y):

[egin{align*} 2x+2y+4 &=0 [4pt] 2x−8y−6 &=0. end{align*}]

Subtracting the second equation from the first gives (10y+10=0), so (y=−1). Substituting this into the first equation gives (2x+2(−1)+4=0), so (x=−1).

Therefore ((−1,−1)) is a critical point of (g). There are no points in (mathbb{R}^2) that make either partial derivative not exist.

Figure (PageIndex{1}) shows the behavior of the surface at the critical point.

Exercise (PageIndex{1})

Find the critical point of the function (f(x,y)=x^3+2xy−2x−4y.)

**Hint**Calculate (f_x(x,y)) and (f_y(x,y)), then set them equal to zero.

**Answer**The only critical point of (f) is ((2,−5)).

The main purpose for determining critical points is to locate relative maxima and minima, as in single-variable calculus. When working with a function of one variable, the definition of a local extremum involves finding an interval around the critical point such that the function value is either greater than or less than all the other function values in that interval. When working with a function of two or more variables, we work with an open disk around the point.

Definition: Global and Local Extrema

Let (z=f(x,y)) be a function of two variables that is defined and continuous on an open set containing the point ((x_0,y_0).) Then (f) has a **local maximum** at ((x_0,y_0)) if

[f(x_0,y_0)≥f(x,y)]

for all points ((x,y)) within some disk centered at ((x_0,y_0)). The number (f(x_0,y_0)) is called a local maximum value. If the preceding inequality holds for every point ((x,y)) in the domain of (f), then (f) has a** global maximum** (also called an absolute maximum) at ((x_0,y_0).)

The function (f) has a local minimum at ((x_0,y_0)) if

[f(x_0,y_0)≤f(x,y)]

for all points ((x,y)) within some disk centered at ((x_0,y_0)). The number (f(x_0,y_0)) is called a local minimum value. If the preceding inequality holds for every point ((x,y)) in the domain of (f), then (f) has a **global minimum** (also called an absolute minimum) at ((x_0,y_0)).

If (f(x_0,y_0)) is either a local maximum or local minimum value, then it is called a **local extremum** (see the following figure).

In Calculus 1, we showed that extrema of functions of one variable occur at critical points. The same is true for functions of more than one variable, as stated in the following theorem.

Fermat’s Theorem for Functions of Two Variables

Let (z=f(x,y)) be a function of two variables that is defined and continuous on an open set containing the point ((x_0,y_0)). Suppose (f_x) and (f_y) each exist at ((x_0,y_0)). If f has a local extremum at ((x_0,y_0)), then ((x_0,y_0)) is a critical point of (f).

## Second Derivative Test

Consider the function (f(x)=x^3.) This function has a critical point at (x=0), since (f'(0)=3(0)^2=0). However, (f) does not have an extreme value at (x=0). Therefore, the existence of a critical value at (x=x_0) does not guarantee a local extremum at (x=x_0). The same is true for a function of two or more variables. One way this can happen is at a **saddle point**. An example of a saddle point appears in the following figure.

**Figure (PageIndex{3}): **Graph of the function (z=x^2−y^2). This graph has a saddle point at the origin.

In this graph, the origin is a saddle point. This is because the first partial derivatives of f((x,y)=x^2−y^2) are both equal to zero at this point, but it is neither a maximum nor a minimum for the function. Furthermore the vertical trace corresponding to (y=0) is (z=x^2) (a parabola opening upward), but the vertical trace corresponding to (x=0) is (z=−y^2) (a parabola opening downward). Therefore, it is both a global maximum for one trace and a global minimum for another.

Definition: Saddle Point

Given the function (z=f(x,y),) the point (ig(x_0,y_0,f(x_0,y_0)ig)) is a saddle point if both (f_x(x_0,y_0)=0) and (f_y(x_0,y_0)=0), but (f) does not have a local extremum at ((x_0,y_0).)

The second derivative test for a function of one variable provides a method for determining whether an extremum occurs at a critical point of a function. When extending this result to a function of two variables, an issue arises related to the fact that there are, in fact, four different second-order partial derivatives, although equality of mixed partials reduces this to three. The second derivative test for a function of two variables, stated in the following theorem, uses a **discriminant **(D) that replaces (f''(x_0)) in the second derivative test for a function of one variable.

Second Derivative Test

Let (z=f(x,y)) be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point ((x_0,y_0)). Suppose (f_x(x_0,y_0)=0) and (f_y(x_0,y_0)=0.) Define the quantity

[D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)−ig(f_{xy}(x_0,y_0)ig)^2.]

Then:

- If (D>0) and (f_{xx}(x_0,y_0)>0), then f has a local minimum at ((x_0,y_0)).
- If (D>0) and (f_{xx}(x_0,y_0)<0), then f has a local maximum at ((x_0,y_0)).
- If (D<0), then (f) has a saddle point at ((x_0,y_0)).
- If (D=0), then the test is inconclusive.

See Figure (PageIndex{4}).

To apply the second derivative test, it is necessary that we first find the critical points of the function. There are several steps involved in the entire procedure, which are outlined in a problem-solving strategy.

Problem-Solving Strategy: Using the Second Derivative Test for Functions of Two Variables

Let (z=f(x,y)) be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point ((x_0,y_0).) To apply the second derivative test to find local extrema, use the following steps:

- Determine the critical points ((x_0,y_0)) of the function (f) where (f_x(x_0,y_0)=f_y(x_0,y_0)=0.) Discard any points where at least one of the partial derivatives does not exist.
- Calculate the discriminant (D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)−ig(f_{xy}(x_0,y_0)ig)^2) for each critical point of (f).
- Apply the four cases of the test to determine whether each critical point is a local maximum, local minimum, or saddle point, or whether the theorem is inconclusive.

Example (PageIndex{2}): Using the Second Derivative Test

Find the critical points for each of the following functions, and use the second derivative test to find the local extrema:

- (f(x,y)=4x^2+9y^2+8x−36y+24)
- (g(x,y)=dfrac{1}{3}x^3+y^2+2xy−6x−3y+4)

**Solution**

a. **Step 1** of the problem-solving strategy involves finding the critical points of (f). To do this, we first calculate (f_x(x,y)) and (f_y(x,y)), then set each of them equal to zero:

[egin{align*} f_x(x,y) &=8x+8 [4pt] f_y(x,y) &=18y−36. end{align*}]

Setting them equal to zero yields the system of equations

[egin{align*} 8x+8 &=0 [4pt] 18y−36 &=0. end{align*}]

The solution to this system is (x=−1) and (y=2). Therefore ((−1,2)) is a critical point of (f).

**Step 2** of the problem-solving strategy involves calculating (D.) To do this, we first calculate the second partial derivatives of (f:)

[egin{align*} f_{xx}(x,y) &=8 [4pt] f_{xy}(x,y) &=0 [4pt] f_{yy}(x,y) &=18. end{align*}]

Therefore, (D=f_{xx}(−1,2)f_{yy}(−1,2)−ig(f_{xy}(−1,2)ig)^2=(8)(18)−(0)^2=144.)

**Step 3** states to apply the four cases of the test to classify the function's behavior at this critical point.

Since (D>0) and (f_{xx}(−1,2)>0,) this corresponds to case 1. Therefore, (f) has a local minimum at ((−1,2)) as shown in the following figure.

**Figure (PageIndex{5}): **The function (f(x,y)) has a local minimum at ((−1,2,−16).) Note the scale on the (y)-axis in this plot is in thousands.

b. For **step 1**, we first calculate (g_x(x,y)) and (g_y(x,y)), then set each of them equal to zero:

[egin{align*} g_x(x,y) &=x^2+2y−6 [4pt] g_y(x,y) &=2y+2x−3. end{align*}]

Setting them equal to zero yields the system of equations

[egin{align*} x^2+2y−6 &=0 [4pt] 2y+2x−3 &=0. end{align*}]

To solve this system, first solve the second equation for (y). This gives (y=dfrac{3−2x}{2}). Substituting this into the first equation gives

[egin{align*} x^2+3−2x−6 &=0 [4pt] x^2−2x−3 &=0 [4pt] (x−3)(x+1) &=0. end{align*}]

Therefore, (x=−1) or (x=3). Substituting these values into the equation (y=dfrac{3−2x}{2}) yields the critical points (left(−1,frac{5}{2} ight)) and (left(3,−frac{3}{2} ight)).

**Step 2 **involves calculating the second partial derivatives of (g):

[egin{align*} g_{xx}(x,y) &=2x [4pt] g_{xy}(x,y) &=2[4pt] g_{yy}(x,y) &=2. end{align*}]

Then, we find a general formula for (D):

[egin{align*} D(x_0, y_0) &=g_{xx}(x_0,y_0)g_{yy}(x_0,y_0)−ig(g_{xy}(x_0,y_0)ig)^2 [4pt] &=(2x_0)(2)−2^2[4pt] &=4x_0−4.end{align*}]

Next, we substitute each critical point into this formula:

[egin{align*} Dleft(−1, frac{5}{2} ight) &=(2(−1))(2)−(2)^2=−4−4=−8 [4pt] Dleft(3,− frac{3}{2} ight) &=(2(3))(2)−(2)^2=12−4=8. end{align*}]

In step 3, we note that, applying Note to point (left(−1,frac{5}{2} ight)) leads to case (3), which means that (left(−1,frac{5}{2} ight)) is a saddle point. Applying the theorem to point (left(3,−frac{3}{2} ight)) leads to case (1), which means that (left(3,−frac{3}{2} ight)) corresponds to a local minimum as shown in the following figure.

Exercise (PageIndex{2})

Use the second derivative test to find the local extrema of the function

[ f(x,y)=x^3+2xy−6x−4y^2. onumber]

**Hint**Follow the problem-solving strategy for applying the second derivative test.

**Answer**(left(frac{4}{3},frac{1}{3} ight)) is a saddle point, (left(−frac{3}{2},−frac{3}{8} ight)) is a local maximum.

## Absolute Maxima and Minima

When finding global extrema of functions of one variable on a closed interval, we start by checking the critical values over that interval and then evaluate the function at the endpoints of the interval. When working with a function of two variables, the closed interval is replaced by a closed, bounded set. A set is bounded if all the points in that set can be contained within a ball (or disk) of finite radius. First, we need to find the critical points inside the set and calculate the corresponding critical values. Then, it is necessary to find the maximum and minimum value of the function on the boundary of the set. When we have all these values, the largest function value corresponds to the global maximum and the smallest function value corresponds to the absolute minimum. First, however, we need to be assured that such values exist. The following theorem does this.

Extreme Value Theorem

A continuous function (f(x,y)) on a closed and bounded set (D) in the plane attains an absolute maximum value at some point of (D) and an absolute minimum value at some point of (D).

Now that we know any continuous function (f) defined on a closed, bounded set attains its extreme values, we need to know how to find them.

Finding Extreme Values of a Function of Two Variables

Assume (z=f(x,y)) is a differentiable function of two variables defined on a closed, bounded set (D). Then (f) will attain the absolute maximum value and the absolute minimum value, which are, respectively, the largest and smallest values found among the following:

- The values of (f) at the critical points of (f) in (D).
- The values of (f) on the boundary of (D).

The proof of this theorem is a direct consequence of the extreme value theorem and Fermat’s theorem. In particular, if either extremum is not located on the boundary of (D), then it is located at an interior point of (D). But an interior point ((x_0,y_0)) of (D) that’s an absolute extremum is also a local extremum; hence, ((x_0,y_0)) is a critical point of (f) by Fermat’s theorem. Therefore the only possible values for the global extrema of (f) on (D) are the extreme values of (f) on the interior or boundary of (D).

Problem-Solving Strategy: Finding Absolute Maximum and Minimum Values

Let (z=f(x,y)) be a continuous function of two variables defined on a closed, bounded set (D), and assume (f) is differentiable on (D). To find the absolute maximum and minimum values of (f) on (D), do the following:

- Determine the critical points of (f) in (D).
- Calculate (f) at each of these critical points.
- Determine the maximum and minimum values of (f) on the boundary of its domain.
- The maximum and minimum values of (f) will occur at one of the values obtained in steps (2) and (3).

Finding the maximum and minimum values of (f) on the boundary of (D) can be challenging. If the boundary is a rectangle or set of straight lines, then it is possible to parameterize the line segments and determine the maxima on each of these segments, as seen in Example (PageIndex{3}). The same approach can be used for other shapes such as circles and ellipses.

If the boundary of the set (D) is a more complicated curve defined by a function (g(x,y)=c) for some constant (c), and the first-order partial derivatives of (g) exist, then the method of Lagrange multipliers can prove useful for determining the extrema of (f) on the boundary which is introduced in Lagrange Multipliers.

Example (PageIndex{3}): Finding Absolute Extrema

Use the problem-solving strategy for finding absolute extrema of a function to determine the absolute extrema of each of the following functions:

- (f(x,y)=x^2−2xy+4y^2−4x−2y+24) on the domain defined by (0≤x≤4) and (0≤y≤2)
- (g(x,y)=x^2+y^2+4x−6y) on the domain defined by (x^2+y^2≤16)

**Solution**

**a**. Using the problem-solving strategy, step (1) involves finding the critical points of (f) on its domain. Therefore, we first calculate (f_x(x,y)) and (f_y(x,y)), then set them each equal to zero:

[egin{align*} f_x(x,y) &=2x−2y−4 [4pt] f_y(x,y) &=−2x+8y−2. end{align*}]

Setting them equal to zero yields the system of equations

[egin{align*} 2x−2y−4 &=0[4pt] −2x+8y−2 &=0. end{align*}]

The solution to this system is (x=3) and (y=1). Therefore ((3,1)) is a critical point of (f). Calculating (f(3,1)) gives (f(3,1)=17.)

The next step involves finding the extrema of (f) on the boundary of its domain. The boundary of its domain consists of four line segments as shown in the following graph:

(L_1) is the line segment connecting ((0,0)) and ((4,0)), and it can be parameterized by the equations (x(t)=t,y(t)=0) for (0≤t≤4). Define (g(t)=fig(x(t),y(t)ig)). This gives (g(t)=t^2−4t+24). Differentiating (g) leads to (g′(t)=2t−4.) Therefore, (g) has a critical value at (t=2), which corresponds to the point ((2,0)). Calculating (f(2,0)) gives the (z)-value (20).

(L_2) is the line segment connecting ((4,0)) and ((4,2)), and it can be parameterized by the equations (x(t)=4,y(t)=t) for (0≤t≤2.) Again, define (g(t)=fig(x(t),y(t)ig).) This gives (g(t)=4t^2−10t+24.) Then, (g′(t)=8t−10). g has a critical value at (t=frac{5}{4}), which corresponds to the point (left(0,frac{5}{4} ight).) Calculating (fleft(0,frac{5}{4} ight)) gives the (z)-value (27.75).

(L_3) is the line segment connecting ((0,2)) and ((4,2)), and it can be parameterized by the equations (x(t)=t,y(t)=2) for (0≤t≤4.) Again, define (g(t)=fig(x(t),y(t)ig).) This gives (g(t)=t^2−8t+36.) The critical value corresponds to the point ((4,2).) So, calculating (f(4,2)) gives the (z)-value (20).

(L_4) is the line segment connecting ((0,0)) and ((0,2)), and it can be parameterized by the equations (x(t)=0,y(t)=t) for (0≤t≤2.) This time, (g(t)=4t^2−2t+24) and the critical value (t=frac{1}{4}) correspond to the point (left(0,frac{1}{4} ight)). Calculating (fleft(0,frac{1}{4} ight)) gives the (z)-value (23.75.)

We also need to find the values of (f(x,y)) at the corners of its domain. These corners are located at ((0,0),(4,0),(4,2)) and ((0,2)):

[egin{align*} f(0,0) &=(0)^2−2(0)(0)+4(0)^2−4(0)−2(0)+24=24 [4pt] f(4,0) &=(4)^2−2(4)(0)+4(0)^2−4(4)−2(0)+24=24 [4pt] f(4,2) &=(4)^2−2(4)(2)+4(2)^2−4(4)−2(2)+24=20[4pt] f(0,2) &=(0)^2−2(0)(2)+4(2)^2−4(0)−2(2)+24=36. end{align*}]

The absolute maximum value is (36), which occurs at ((0,2)), and the global minimum value is (20), which occurs at both ((4,2)) and ((2,0)) as shown in the following figure.

**b**. Using the problem-solving strategy, step (1) involves finding the critical points of (g) on its domain. Therefore, we first calculate (g_x(x,y)) and (g_y(x,y)), then set them each equal to zero:

[egin{align*} g_x(x,y) &=2x+4 [4pt] g_y(x,y) &=2y−6. end{align*}]

Setting them equal to zero yields the system of equations

[egin{align*} 2x+4 &=0 [4pt] 2y−6 &=0. end{align*}]

The solution to this system is (x=−2) and (y=3). Therefore, ((−2,3)) is a critical point of (g). Calculating (g(−2,3),) we get

[g(−2,3)=(−2)^2+3^2+4(−2)−6(3)=4+9−8−18=−13.]

The next step involves finding the extrema of *g* on the boundary of its domain. The boundary of its domain consists of a circle of radius (4) centered at the origin as shown in the following graph.

The boundary of the domain of (g) can be parameterized using the functions (x(t)=4cos t,, y(t)=4sin t) for (0≤t≤2π). Define (h(t)=gig(x(t),y(t)ig):)

[egin{align*} h(t) &=gig(x(t),y(t)ig) [4pt] &=(4cos t)^2+(4sin t)^2+4(4cos t)−6(4sin t) [4pt] &=16cos^2t+16sin^2t+16cos t−24sin t[4pt] &=16+16cos t−24sin t. end{align*}]

Setting (h′(t)=0) leads to

[egin{align*} −16sin t−24cos t &=0 [4pt] −16sin t &=24cos t[4pt] dfrac{−16sin t}{−16cos t} &=dfrac{24cos t}{−16cos t} [4pt] an t &=−dfrac{3}{2}. end{align*}]

This equation has two solutions over the interval (0≤t≤2π). One is (t=π−arctan (frac{3}{2})) and the other is (t=2π−arctan (frac{3}{2})). For the first angle,

[egin{align*} sin t &=sin(π−arctan( frac{3}{2}))=sin (arctan ( frac{3}{2}))=dfrac{3sqrt{13}}{13} [4pt] cos t &=cos (π−arctan ( frac{3}{2}))=−cos (arctan ( frac{3}{2}))=−dfrac{2sqrt{13}}{13}. end{align*}]

Therefore, (x(t)=4cos t =−frac{8sqrt{13}}{13}) and (y(t)=4sin t=frac{12sqrt{13}}{13}), so (left(−frac{8sqrt{13}}{13},frac{12sqrt{13}}{13} ight)) is a critical point on the boundary and

[egin{align*} gleft(− frac{8sqrt{13}}{13}, frac{12sqrt{13}}{13} ight) &=left(− frac{8sqrt{13}}{13} ight)^2+left( frac{12sqrt{13}}{13} ight)^2+4left(− frac{8sqrt{13}}{13} ight)−6left( frac{12sqrt{13}}{13} ight) [4pt] &=frac{144}{13}+frac{64}{13}−frac{32sqrt{13}}{13}−frac{72sqrt{13}}{13} [4pt] &=frac{208−104sqrt{13}}{13}≈−12.844. end{align*}]

For the second angle,

[egin{align*} sin t &=sin (2π−arctan ( frac{3}{2}))=−sin (arctan ( frac{3}{2}))=−dfrac{3sqrt{13}}{13} [4pt] cos t &=cos (2π−arctan ( frac{3}{2}))=cos (arctan ( frac{3}{2}))=dfrac{2sqrt{13}}{13}. end{align*}]

Therefore, (x(t)=4cos t=frac{8sqrt{13}}{13}) and (y(t)=4sin t=−frac{12sqrt{13}}{13}), so (left(frac{8sqrt{13}}{13},−frac{12sqrt{13}}{13} ight)) is a critical point on the boundary and

[egin{align*} gleft( frac{8sqrt{13}}{13},− frac{12sqrt{13}}{13} ight) &=left( frac{8sqrt{13}}{13} ight)^2+left(− frac{12sqrt{13}}{13} ight)^2+4left( frac{8sqrt{13}}{13} ight)−6left(− frac{12sqrt{13}}{13} ight) [4pt] &=dfrac{144}{13}+dfrac{64}{13}+dfrac{32sqrt{13}}{13}+dfrac{72sqrt{13}}{13}[4pt] &=dfrac{208+104sqrt{13}}{13}≈44.844. end{align*}]

The absolute minimum of (g) is (−13,) which is attained at the point ((−2,3)), which is an interior point of (D). The absolute maximum of (g) is approximately equal to 44.844, which is attained at the boundary point (left(frac{8sqrt{13}}{13},−frac{12sqrt{13}}{13} ight)). These are the absolute extrema of (g) on (D) as shown in the following figure.

Exercise (PageIndex{3}):

Use the problem-solving strategy for finding absolute extrema of a function to find the absolute extrema of the function

[f(x,y)=4x^2−2xy+6y^2−8x+2y+3 onumber]

on the domain defined by (0≤x≤2) and (−1≤y≤3.)

**Hint**Calculate (f_x(x,y)) and (f_y(x,y)), and set them equal to zero. Then, calculate (f) for each critical point and find the extrema of (f) on the boundary of (D).

**Answer**The absolute minimum occurs at ((1,0): f(1,0)=−1.)

The absolute maximum occurs at ((0,3): f(0,3)=63.)

Example (PageIndex{4}): Profitable Golf Balls

Pro-(T) company has developed a profit model that depends on the number (x) of golf balls sold per month (measured in thousands), and the number of hours per month of advertising (y), according to the function

[ z=f(x,y)=48x+96y−x^2−2xy−9y^2,]

where (z) is measured in thousands of dollars. The maximum number of golf balls that can be produced and sold is (50,000), and the maximum number of hours of advertising that can be purchased is (25). Find the values of (x) and (y) that maximize profit, and find the maximum profit.

**Solution**

Using the problem-solving strategy, step (1) involves finding the critical points of (f) on its domain. Therefore, we first calculate (f_x(x,y)) and (f_y(x,y),) then set them each equal to zero:

[egin{align*} f_x(x,y) &=48−2x−2y [4pt] f_y(x,y) &=96−2x−18y. end{align*}]

Setting them equal to zero yields the system of equations

[egin{align*} 48−2x−2y &=0 [4pt] 96−2x−18y &=0. end{align*}]

The solution to this system is (x=21) and (y=3). Therefore ((21,3)) is a critical point of (f). Calculating (f(21,3)) gives (f(21,3)=48(21)+96(3)−21^2−2(21)(3)−9(3)^2=648.)

The domain of this function is (0≤x≤50) and (0≤y≤25) as shown in the following graph.

(L_1) is the line segment connecting ((0,0)) and ((50,0),) and it can be parameterized by the equations (x(t)=t,y(t)=0) for (0≤t≤50.) We then define (g(t)=f(x(t),y(t)):)

[egin{align*} g(t) &=f(x(t),y(t)) [4pt] &=f(t,0)[4pt] &=48t+96(0)−y^2−2(t)(0)−9(0)^2[4pt] &=48t−t^2. end{align*}]

Setting (g′(t)=0) yields the critical point (t=24,) which corresponds to the point ((24,0)) in the domain of (f). Calculating (f(24,0)) gives (576.)

(L_2) is the line segment connecting ((50,0)) and ((50,25)), and it can be parameterized by the equations (x(t)=50,y(t)=t) for (0≤t≤25). Once again, we define (g(t)=fig(x(t),y(t)ig):)

[egin{align*} g(t) &=fig(x(t),y(t)ig)[4pt] &=f(50,t)[4pt] &=48(50)+96t−50^2−2(50)t−9t^2 [4pt] &=−9t^2−4t−100. end{align*}]

This function has a critical point at (t =−frac{2}{9}), which corresponds to the point ((50,−29)). This point is not in the domain of (f).

(L_3) is the line segment connecting ((0,25)) and ((50,25)), and it can be parameterized by the equations (x(t)=t,y(t)=25) for (0≤t≤50). We define (g(t)=fig(x(t),y(t)ig)):

[egin{align*} g(t) &=fig(x(t),y(t)ig)[4pt] &=f(t,25) [4pt] &=48t+96(25)−t^2−2t(25)−9(25^2) [4pt] &=−t^2−2t−3225. end{align*}]

This function has a critical point at (t=−1), which corresponds to the point ((−1,25),) which is not in the domain.

(L_4) is the line segment connecting ((0,0)) to ((0,25)), and it can be parameterized by the equations (x(t)=0,y(t)=t) for (0≤t≤25). We define (g(t)=fig(x(t),y(t)ig)):

[egin{align*} g(t) &=fig(x(t),y(t)ig) [4pt] &=f(0,t) [4pt] &=48(0)+96t−(0)^2−2(0)t−9t^2 [4pt] &=96t−9t^2. end{align*}]

This function has a critical point at (t=frac{16}{3}), which corresponds to the point (left(0,frac{16}{3} ight)), which is on the boundary of the domain. Calculating (fleft(0,frac{16}{3} ight)) gives (256).

We also need to find the values of (f(x,y)) at the corners of its domain. These corners are located at ((0,0),(50,0),(50,25)) and ((0,25)):

[egin{align*} f(0,0) &=48(0)+96(0)−(0)^2−2(0)(0)−9(0)^2=0[4pt] f(50,0) &=48(50)+96(0)−(50)^2−2(50)(0)−9(0)^2=−100 [4pt] f(50,25) &=48(50)+96(25)−(50)^2−2(50)(25)−9(25)^2=−5825 [4pt] f(0,25) &=48(0)+96(25)−(0)^2−2(0)(25)−9(25)^2=−3225. end{align*}]

The maximum value is (648), which occurs at ((21,3)). Therefore, a maximum profit of ($648,000) is realized when (21,000) golf balls are sold and (3) hours of advertising are purchased per month as shown in the following figure.

## Key Concepts

- A critical point of the function (f(x,y)) is any point ((x_0,y_0)) where either (f_x(x_0,y_0)=f_y(x_0,y_0)=0), or at least one of (f_x(x_0,y_0)) and (f_y(x_0,y_0)) do not exist.
- A saddle point is a point ((x_0,y_0)) where (f_x(x_0,y_0)=f_y(x_0,y_0)=0), but (f(x_0,y_0)) is neither a maximum nor a minimum at that point.
- To find extrema of functions of two variables, first find the critical points, then calculate the discriminant and apply the second derivative test.

## Key Equations

**Discriminant**

(D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)−(f_{xy}(x_0,y_0))^2)

## Glossary

**critical point of a function of two variables**the point ((x_0,y_0)) is called a critical point of (f(x,y)) if one of the two following conditions holds:

1. (f_x(x_0,y_0)=f_y(x_0,y_0)=0)

2. At least one of (f_x(x_0,y_0)) and (f_y(x_0,y_0)) do not exist

**discriminant**- the discriminant of the function (f(x,y)) is given by the formula (D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)−(f_{xy}(x_0,y_0))^2)

**saddle point**- given the function (z=f(x,y),) the point ((x_0,y_0,f(x_0,y_0))) is a saddle point if both (f_x(x_0,y_0)=0) and (f_y(x_0,y_0)=0), but (f) does not have a local extremum at ((x_0,y_0))

## Contributors and Attributions

Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.

## 14.7: Maximum and Minimum Values - Mathematics

In engineering, there are many optimization problems which can be reduced to finding the maximum or minimum values of a given function. In this section, the basic concept and theorem related to maximum and minimum are discussed.

Maximum and Minimum Value

Maximum value is the absolute maximum value of the function in its domain. In mathematics it is defined as following:

A function f has an absolute maximum at point x_{0} if

f(x_{0}) &ge f(x) for all x in its domain D. The number f(x_{0}) is called the maximum value of f on its domain. For example, the maximum value of the function plotted on the left is f(f) between a and h.

Similarly, minimum value is the absolute minimum value of the function in its domain. In mathematics it is defined as following:

Function f has an absolute minimum at point x_{0} if

f(x_{0}) &le f(x) for all x in its domain D. The number f(x_{0}) is called the minimum value of f on its domain. The minimum value for the function shown on the left is f(c) in [a, h].

The maximum and minimum values of the function are called the extreme values of the function. In the function show on the left, f(c) and f(f) are the extreme values.

The local maximum is defined in mathematics as:

A function f has a local maximum at x_{0} if there is an open interval u that contains x_{0} such that f(x_{0}) &ge f(x) for all x in this interval u. Local maximum is also called relative maximum. In *Example Description Diagram*, f(b), f(d) and f(f) are the local maximum. Notice that f(f) is also absolute maximum and extreme value of the function.

Similarly, local minimum is defined as,

Function f has local minimum at x_{0} if there is an open interval u that contains x_{0} such that f(x_{0}) &le f(x) for all x in this interval u. Local minimum is also called relative minimum. In the previous example, f(c), f(e) and f(g) are the local minimum. Notice that f(c) is also absolute minimum and extreme value of the function.

Some functions have extreme value such as function y = x 2 . Since in (-&infin, +&infin ), f(x) &ge f(0). Therefore, f(0) = 0 is the absolute and local minimum value of this function. It is also the extreme value. However, some functions do not have extreme value such as function y = x 3 . In the function's domain (-&infin, +&infin ), it does not have an absolute maximum or an absolute minimum value. In reality, it has no local maximum and local minimum too. and thus function y = x 3 does not have extreme value. The Extreme Value Theorem has been introduced in the Continuity sectiion. It gives the conditions under which a function has an extreme value.

Extreme Value Fermat's Theorem

The Fermat's Theorem can be used to find the location of an extreme value. The theorem states,

If a function has a local extreme which is the maximum or minimum at x_{0}, and if df(x_{0})/dx exist, then df(x_{0})/dx = 0.

However, using Fermat's Theorem does not guarantee to find all extreme value. For example, although the absolute and local minimum value is 0 for function y = (x 2 ) 0.5 , but this value cannot be found by setting df(x_{0})/dx = 0. The reason is df(0 )/dx does not exist which has been shown in the Differentiability section. Fermat's Theorem suggest that looking for extreme values at point x_{0} where df(x_{0})/dx = 0 or df(x_{0})/dx does not exist. Such points are given a special name - critical point.

A critical point is the point where the derivative of the function is 0 or does not exist. The method of calculating the critical point can be understood by finding the critical point of function f(x) = 6x 1/3 - 5x.

Calculating the derivative of f(x) gives

When df(x)/dx does not exist,

Thus, the critical point is at x = 0.34 and x = 0.

In terms of critical points, Fermat's Theorem can be rephrased as:

If a function has a local extreme at x_{0}, then x_{0} is a critical point of the function.

In order to find the absolute maximum and minimum of a continuous function in its domain [a, b], the following steps need to be taken:

## How to Find the Extreme Points of a Function

At a local minimum, the function changes direction. This is because it is the lowest point in its neighborhood. Therefore the slope of the function goes from negative to positive, since the function was decreasing until it reached the minimum and then it started increasing again. This means that in the local minimum, the slope is equal to zero, and hence the derivative of the function must be equal to zero in the point that is the minimum. The same holds for the local maximum of a function, since there the function goes from increasing to decreasing.

## Uses of the Maximum and Minimum

Beyond giving us some very basic information about a data set, the maximum and minimum show up in the calculations for other summary statistics.

Both of these two numbers are used to calculate the range, which is simply the difference of the maximum and minimum.

The maximum and minimum also make an appearance alongside the first, second, and third quartiles in the composition of values comprising the five number summary for a data set. The minimum is the first number listed as it is the lowest, and the maximum is the last number listed because it is the highest. Due to this connection with the five number summary, the maximum and minimum both appear on a box and whisker diagram.

## How To Find Maximum And Minimum Value Of Quadratic Function Using General Form Of The Function

The maximum and the minimum value of the quadratic function can be determined using the general form of function. This is an algebraic method, and does not involve the use of graphs.

General steps for this method of determination are mentioned below:

- Set up the function in its general form.
- Using the co-efficient of the term, determine the direction of the graph.
- Calculate the x coordinate of the vertex using the formula .
- Find the co-responding value of the y coordinate of the vertex by putting the value of x coordinate in the original function.

Let’s solve some examples with the help of these rules:

Considering we are given with a function:

We see that the given function is not present in it’s general form . Therefore, we’ll first set up the function in it’s general form by combing the x² and x terms in the following manner:

The co-efficient of the x² term is +3.

This means that the parabola of the given function would be **opening upwards** and whenever it open upwards, we **find the minimum value of the quadratic function.**

The minimum value of the quadratic function is the y-coordinate of the vertex. Hence, to find the y coordinate of the vertex we first find the x coordinate.

In order to find the x coordinate of the vertex we put the relevant values of ** a** and

**in the formula of:**

*b** *

By assigning values of the variables we get:

The above evaluation shows that the x coordinate of the vertex is -1. Replacing *x* with -1 in the function to calculate the y-coordinate of the vertex we get:

H*ere f(x*) represents the y-coordinate

f ( – 1 ) = 3 ( – 1 ) 2 + 6 ( – 1 ) + 4

The above shows that the y-coordinate of the vertex is 1.

Hence, the minimum value of the quadratic function f(x)=3x+3x-x²+4x²+4 is 1.

Both coordinates of the vertex are given as (-1 , 1).

*Let’s look at another example:*

The is function is present in it’s general form.

The co-efficient of the x² term is – 7 for the above function.

This means that the *parabola of the given function would be opening downwards*.

In the case of downward opening, we **find the maximum value of the quadratic function.**

To find the y-coordinate of the vertex, we first find the x-coordinate using the formula:

We derive x from the values of the equation below

By assigning values of the variables we get

The above evaluation shows that the x coordinate of the vertex is +1. Replacing with +1 in the function to calculate the y coordinate of the vertex we get:

## MAXIMUM AND MINIMUMVALUES

W E SAY THAT A FUNCTION f ( x ) has a relative maximum value at x = a ,

if f ( a ) is greater than any value immediately preceding or follwing.

We call it a "relative" maximum because other values of the function may in fact be greater.

We say that a function f ( x ) has a relative minimum value at x = b ,

if f ( b ) is less than any value immediately preceding or follwing.

Again, other values of the function may in fact be less. With that understanding, then, we will drop the term relative.

The value of the function, the value of y , at either a maximum or a minimum is called an extreme value.

Now, what characterizes the graph at an extreme value?

The tangent to the curve is horizontal . We see this at the points A and B . The slope of each tangent line -- the derivative when evaluated at a or b -- is 0.

Moreover, at points immediately to the left of a maximum -- at a point C -- the slope of the tangent is positive: f ' ( x ) > 0. While at points immediately to the right -- at a point D -- the slope is negative: f ' ( x ) f ' ( x ) changes sign from + to &minus .

At a minimum, f ' ( x ) changes sign from &minus to + . We can see that at the points E and F .

We can also observe that at a maximum, at A , the graph is concave downward. (Topic 14 of Precalculus.) While at a minimum, at B , it is concave upward.

A value of x at which the function has either a maximum or a minimum is called a critical value . In the figure --

-- the critical values are x = a and x = b .

The critical values determine turning points , at which the tangent is parallel to the x -axis. The critical values -- if any -- will be the solutions to f ' ( x ) = 0.

Example 1. Let f ( x ) = x 2 &minus 6 x + 5.

Are there any critical values -- any turning points? If so, do they determine a maximum or a minimum? And what are the coördinates on the graph of that maximum or minimum?

Solution . f ' ( x ) = 2 x &minus 6 = 0 implies x = 3. (Lesson 9 of Algebra.)

x = 3 is the only critical value. It is the x -coördinate of the turning point. To determine the y -coördinate, evaluate f at that critical value -- evaluate f (3):

f ( x ) | = | x 2 &minus 6 x + 5 |

f (3) | = | 3 2 &minus 6 · 3 + 5 |

= | &minus4. |

The extreme value is &minus4. To see whether it is a maximum or a minimum, in this case we can simply look at the graph.

f ( x ) is a parabola, and we can see that the turning point is a minimum.

By finding the value of x where the derivative is 0, then, we have discovered that the vertex of the parabola is at (3, &minus4).

But we will not always be able to look at the graph. The algebraic condition for a minimum is that f ' ( x ) changes sign from &minus to + . We see this at the points E , B , F above. The value of the slope is increasing.

Now to say that the slope is increasing, is to say that, at a critical value, the second derivative (Lesson 9) -- which is rate of change of the slope -- is positive .

f ( x ) | = | x 2 &minus 6 x + 5. |

f ' ( x ) | = | 2 x &minus 6. |

f '' ( x ) | = | 2. |

f '' evaluated at the critical value 3 -- f'' (3) = 2 -- is positive. This tells us algebraically that the critical value 3 determines a minimum.

We can now state these sufficient conditions for extreme values of a function at a critical value a :

The function has a minimum value at x = a if f ' ( a ) = 0

and f '' ( a ) = a positive number.

The function has a maximum value at x = a if f ' ( a ) = 0

and f '' ( a ) = a negative number.

In the case of the maximum, the slope of the tangent is decreasing -- it is going from positive to negative. We can see that at the points C , A , D .

Example 2. Let f ( x ) = 2 x 3 &minus 9 x 2 + 12 x &minus 3.

Are there any extreme values? First, are there any critical values -- solutions to f ' ( x ) = 0 -- and do they determine a maximum or a minimum? And what are the coördinates on the graph of that maximum or minimum? Where are the turning points?

Solution . f ' ( x ) = 6 x 2 &minus 18 x + 12 | = | 6( x 2 &minus 3 x + 2) |

= | 6( x &minus 1)( x &minus 2) | |

= | 0 |

Those are the critical values. Does each one determine a maximum or does it determine a minimum? To answer, we must evaluate the second derivative at each value .

f ' ( x ) | = | 6 x 2 &minus 18 x + 12. |

f '' ( x ) | = | 12 x &minus 18. |

f '' (1) | = | 12 &minus 18 = &minus6. |

The second derivative is negative. The function therefore has a maximum at x = 1.

To find the y -coördinate -- the extreme value -- at that maximum we evaluate f (1):

f ( x ) | = | 2 x 3 &minus 9 x 2 + 12 x &minus 3 |

f (1) | = | 2 &minus 9 + 12 &minus 3 |

= | 2. |

The maximum occurs at the point (1, 2).

Next, does x = 2 determine a maximum or a minimum?

f '' ( x ) | = | 12 x &minus 18. |

f '' (2) | = | 24 &minus 18 = 6. |

The second derivative is positive. The function therefore has a minimum at x = 2.

To find the y -coördinate -- the extreme value -- at that minimum, we evaluate f (2):

f ( x ) | = | 2 x 3 &minus 9 x 2 + 12 x &minus 3. |

f (2) | = | 16 &minus 36 + 24 &minus 3 |

= | 1. |

The minimum occurs at the point (2, 1).

Here in fact is the graph of f ( x ):

Solutions to f '' ( x ) = 0 indicate a point of inflection at those solutions, not a maximum or minimum. An example is y = x 3 . y'' = 6 x = 0 implies x = 0. But x = 0 is a point of inflection in the graph of y = x 3 , not a maximum or minimum.

Another example is y = sin x . The solutions to y'' = 0 are the multiplies of &pi , which are points of inflection.

Problem 1. Find the coördinates of the vertex of the parabola,

To see the answer, pass your mouse over the colored area.

To cover the answer again, click "Refresh" ("Reload").

Do the problem yourself first!

That implies x = 4. That's the x -coördinate of the vertex. To find the y -coördinate, evaluate y at x = 4:

Problem 2. Examine each function for maxima and minima.

y' = 3 x 2 &minus 6 x = 3 x ( x &minus 2) = 0 implies

The second derivative is negative. That means there is a maximum at x = 0. That maximum value is

The second derivative is positive. That means there is a minimum at x = 2. That minimum value is

y (2) = 2 3 &minus 3 · 2 2 + 2 = 8 &minus 12 + 2 = &minus2.

b) y = &minus2 x 3 &minus 3 x 2 + 12 x + 10.

At x = 1 there is a maximum of y = 17.

At x = &minus2 there is a minimum of y = &minus10.

c) y = 2 x 3 + 3 x 2 + 12 x &minus 4.

Since f ' ( x ) = 0 has no real solutions, there are no extreme values.

d) y = 3 x 4 &minus 4 x 3 &minus 12 x 2 + 2.

At x = 0 there is a maximum of y = 2.

At x = &minus1 there is a minimum of y = &minus3.

At x = 2 there is a minimum of y = &minus30.

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## MAXIMUM AND MINIMUMVALUES

F INDING a maximum or a minimum (Lesson 10) has its application in pure mathematics, where for example we could find the largest rectangle that has a given perimeter. It also has its application to commercial problems, such as finding the least dimensions of a carton that is to contain a given volume.

Example 1. Find the dimensions of the rectangle that, for a given perimeter, will have the largest area.

Solution . Let the base of the rectangle be x , let its height be y , let A be its area, and let P be the given perimeter. Then

Since we are going to maximize A , we would like to have A as a function only of x . And we can do that because in the expression for P we can solve for y :

A | = | x (½ P &minus x ) | ||

= | ½ x P &minus x 2 . | |||

On taking the derivative of A and setting it equal to 0, | ||||

dA dx | = | ½ P &minus 2 x | = | 0, |

x | = | ¼ P . |

The base is one quarter of the perimeter. We can now find the value of y :

The height is also one quarter of the perimeter. That figure is a square The rectangle that has the largest area for a given perimeter

is a square.

(Note: The value we found is a maximum because the second derivative is negative.)

All maximum-minimum problems follow this same procedure:

&bull | Write the function whose maximum or minimum value is to be determined . (In the Example, we wrote A = xy .) |

&bull | The resulting expression will typically contain more than one variable. Use the information given in the problem to express every variable in terms of a single variable. (In the Example, we expressed y in terms of x .) |

&bull | Find the critical value of that single variable by taking the derivative and setting it equal to 0. (In the Example, we took the derivative of A with respect to x .) |

&bull | If necessary, determine the values of the other variables. (In the Example, we evaluated y by substituting the critical value of x .) |

In the following, notice how we follow these steps.

Example 2. A box having a square base and an open top is to contain 108 cubic feet. What should its dimensions be so that the material to make it will be a minimum? That is, what dimensions will cost the least?

Solution . Let x be the side of the square base, and let y be its height. Then

Area of base | = | x 2 . |

Area of four sides | = | 4 xy . |

Let M be the total amount of material. Then | ||

M | = | x 2 + 4 xy . |

Now, how shall we express y in terms of x ?

We have not yet used the fact that the volume must be 108 cubic feet. The volume is equal to

y | = | 108 x 2 | ||||

and therefore in the expression for M , | ||||||

4 xy | = | 4 x · | 108 x 2 | = | ||

M | = | |||||

= | ||||||

This implies, on multiplying through by the denominator x 2 : | ||||||

2 x 3 &minus 432 | = | 0 | ||||

x 3 | = | 216 | ||||

x | = | 6 feet. | ||||

We can now evaluate y : | ||||||

y | = |

These are the dimensions that will cost the least.

Example 3. Find the dimensions of the rectangle with the most area that can be inscribed in a semi-circle of radius r . Show, in fact, that the area of that rectangle is r 2 .

Solution . First, it should be clear that there is a rectangle with the

greatest area, as the figures above show.

Let x be the base of the rectangle, and let y be its height. Then, since r is the radius:

= | r 2 | |

= | r 2 | |

x 2 + 4 y 2 | = | 4 r 2 . |

Therefore, | ||

y | = |

Let A be the area we want to maximize. A = xy . That is,

A | = | |

According to the product rule and the chain rule (Example 1): | ||

dA dx | = |

&minus x 2 + (4 r 2 &minus x 2 ) | = | 0. |

This implies: | ||

x 2 | = | 2 r 2 |

x | = | |

This is the base of the largest rectangle. As for the height y : | ||

y | = | |

y | = | |

y | = |

The area of this largest rectangle, then, is

Problem 1. Find two numbers whose sum is 42 and whose product will be the largest.

(Hint: Call the two numbers x and y . For convenience, call the product something. You will then have two equations in two unknowns. Express the product as function of a single variable, and find its maximum.)

To see the answer, pass your mouse over the colored area.

To cover the answer again, click "Refresh" ("Reload").

Do the problem yourself first!

The two numbers are 21 and 21.

Problem 2. You have a given length of fence. Using the wall of a house as one side of a rectangular fence, how would you place the fence around the other three sides in order to enclose the largest possible area?

Place half the fence parallel to the house.

Problem 3. Find the dimensions of the rectangle of maximum area that can be inscribed in a circle of radius r . Show, in fact, that that area will be 2 r 2 .

That figure is a square. Each side of the square is r .

Problem 4. A can is to be constructed in the form of a right circular cylinder. If it is to contain a given volume V , what dimensions will require the least amount of material?

Show, in fact, that the height h of the can must equal its width, which is twice the radius r .

First, the top and bottom of the can are each a circle. And the area of the lateral surface is equivalent to a rectangle whose dimensions are 2 &pi r × h . Therefore,

The area A of the material | = | 2 &pi r 2 + 2 &pi rh . |

From the formula for V , express h in terms of r.

A therefore has a minimum at

Compare Lesson 29 of Algebra, Problem 2.

The height of the can must equal its width, which is 2 r .

Problem 5. Find the volume V of the largest right circular cone that can be inscribed in a sphere of radius r .

V | = | 1 3 | (area of the base) · (height). |

Let P be the center of the sphere of radius r . Let APB be the height of the cone, and call that height h . Then PB = h &minus r .

Let s be the radius of the cone. Then

We want to maximize V as a function of h alone. Therefore we must express s in terms of h and the constant r .

Now, s and h &minus r are the sides of a right triangle. Therefore,

s 2 + ( h &minus r ) 2 | = | r 2 |

s 2 + h 2 &minus 2 hr + r 2 | = | r 2 |

s 2 | = | 2 hr &minus h 2 . |

V | = | 1 3 | &pi (2 hr &minus h 2 ) h . |

= | &pi 3 | (2 h 2 r &minus h 3 ). | |

dV dh | = | &pi 3 | (4 h r &minus 3 h 2 ). |

V therefore has a maximum at

Upon substituting that value for h in the expression for V above, that maximum volume is:

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## IB Mathematics HL SL-Maxima and Minima

In the previous post, **IB Maths Tutors** discussed how to find the equation of tangents and normal to a curve. There are a few more Applications of Derivatives in IB Mathematics HL SL, ‘Maxima and Minima’ is one of them.

**Maxima and Minima:-**

1. A function f(x) is said to have a maximum at x = a if f(a) is greater than every other value assumed by f(x) in the immediate neighbourhood of x = a. Symbolically

gives maxima for a sufficiently small positive h.

Similarly, a function f(x) is said to have a minimum value at x = b if f(b) is least than every other value assumed by f(x) in the immediate neighbourhood at x = b. Symbolically

If x = b gives minima for a sufficiently small positive h.

**►**The maximum & minimum values of a function are also known as local/relative maxima or local/relative minima as these are the greatest & least values of the function relative to some neighborhood of the point in question.

**►** The term ‘extremum’ or (extremal) or **‘turning value’** is used both for maximum or a minimum value.

**►** A maximum (minimum) value of a function may not be the greatest (least) value in a finite interval.

**►**A function can have several maximum & minimum values & a minimum value may even be greater than a maximum value.

**►**Maximum & minimum values of a continuous function occur alternately & between two consecutive maximum values, there is a minimum value & vice versa.

**2. A Necessary Condition For Maximum & Minimum:-** If f(x) is a maximum or minimum at x = c & if f'(c) exists then f'(c) = 0.

**►**The set of values of x for which f'(x) = 0 are often called stationary points or critical points . The rate of change of function is zero at a stationary point. In IB Mathematics HL SL questions are asked on these points

**►**In case f'(c) does not exist f(c) may be a maximum or a minimum & in this case left hand and right-hand derivatives are of opposite signs.

**►** The greatest **(global maxima)** and the least **(global minima)** values of a function f in an interval [a, b] are f(a) or f(b) or are given by the values of x for which f'(x) = 0.

**►** Critical points are those where f'(x)= 0 if it exists, or it fails to exist either by virtue of a vertical tangent or by virtue of a geometrical sharp corner but not because of discontinuity of function.

**3. Sufficient Condition For Extreme Values:-**

x=c is a point of local maxima where f'(c)=0

similarly, x=c is a point of local maxima where f'(c)=0. Here ‘h’ is

a sufficiently small positive quantity.

**►** If f'(x) does not change sign i.e. has the same sign in a certain complete neighbourhood of c, then f(x) is either strictly increasing or decreasing throughout this neighborhood implying that f(c) is not an extreme value of the given function.

**4.Use Of Second Order Derivative In Ascertaining The Maxima Or Minima:-**

(a) f(c) is a minimum value of the function f, if f'(c) = 0 & f”(c) > 0.

(b) f(c) is a maximum value of the function f, if f'(c) = 0 & f”(c) < 0.

**►**If f'(c) = 0 then the test fails. Revert back to the first order derivative check for ascertaining the maxima or minima.

**►** If y = f make the function (x) is a quantity to be maximum or minimum, find those values of x for which f'(x)=0

**►**Test each value of x for which f'(x) = 0 to determine whether it provides a maximum or minimum or neither. The usual tests are :

(a) If is positive when y is minimum. If is negative when y is

maximum. If when the test fails.

(b) If is positive for then a maximum occurs at

(c)If is zero for then a maximum occurs at

(d)If is negative for then a maximum occurs at

(e) But if dy/dx changes sign from negative to zero to positive as x advances through xo there is a minimum. If dy/dx does not change sign, neither a maximum nor a minimum. Such points are called **Points of Inflection**.

Point of inflexion is a point where the shape of f(x) changes from concave to convex or convex to concave. This concept is usually asked in IB Mathematics HL SL

**►**If the derivative fails to exist at some point, examine this point as possible maximum or minimum.

**►**If the function y = f (x) is defined for only a limited range of values then we should examine x = a & x=b for possible extreme values.

**►**If the derivative fails to exist at some point, we should examine this point as possible maximum or minimum.

**Important Notes -:**

Given a fixed point

A(x_{1}, y_{1}) and a moving point P(x, f (x)) on the curve y = f(x). Then AP will be maximum or minimum if it is normal to the curve at P

**►**If the sum of two positive numbers x and y is constant than their product is maximum if they are equal, i.e. x + y = c , x > 0 , y > 0 , then xy = [(x + y) 2 – (x – y) 2 ]

**►**If the product of two positive numbers is constant then their sum is least if they are equal.i.e. (x+y) 2 =(x-y) 2 + 4xy

**6. Useful Formulae Of Mensuration To Remember**

a. The volume of a cuboid = lbh

b. Surface area of a cuboid = 2 (lb + bh + hl)

c. The volume of a prism = area of the base x height.

d. The lateral surface of a prism = perimeter of the base x height.

e. The total surface of a prism = lateral surface + 2 area of the base (Note that lateral surfaces of a prism are all rectangles) f.

## 14.7: Maximum and Minimum Values - Mathematics

Statistics: double gsl_stats_max (const double data [], size_t stride , size_t n ) This function returns the maximum value in data , a dataset of length n with stride stride . The maximum value is defined as the value of the element @math

If you want instead to find the element with the largest absolute magnitude you will need to apply fabs or abs to your data before calling this function.

Statistics: double gsl_stats_min (const double data [], size_t stride , size_t n ) This function returns the minimum value in data , a dataset of length n with stride stride . The minimum value is defined as the value of the element @math

If you want instead to find the element with the smallest absolute magnitude you will need to apply fabs or abs to your data before calling this function.

Statistics: void gsl_stats_minmax (double * min , double * max , const double data [], size_t stride , size_t n ) This function finds both the minimum and maximum values min , max in data in a single pass.

Statistics: size_t gsl_stats_max_index (const double data [], size_t stride , size_t n ) This function returns the index of the maximum value in data , a dataset of length n with stride stride . The maximum value is defined as the value of the element @math

Statistics: size_t gsl_stats_min_index (const double data [], size_t stride , size_t n ) This function returns the index of the minimum value in data , a dataset of length n with stride stride . The minimum value is defined as the value of the element @math

Statistics: void gsl_stats_minmax_index (size_t * min_index , size_t * max_index , const double data [], size_t stride , size_t n ) This function returns the indexes min_index , max_index of the minimum and maximum values in data in a single pass.

## Maxima and Minima in Calculus

The local maxima are the largest values (maximum) that a function takes in a point within a given neighborhood.

The local minima are the smallest values (minimum), that a function takes in a point within a given neighborhood.

Definition of local maximum and local minimum

A function *f* has a local maximum (or relative maximum) at *c*, if *f*(*c*) &ge *f*(*x*) where *x* is near *c*.

Definition of global maximum or global minimum

A function *f* has a global maximum (or absolute maximum) at *c* if *f*(*c*) &ge *f*(*x*) for all *x* in *D*, where *D* is the domain of *f*. The number *f*(*c*) is called the maximum value of *f* on *D*.

Similarly, *f* has a global minimum (or absolute minimum) at *c* if *f*(*c*) &le *f*(*x*) for all *x* in *D* and the number *f*(*c*) is called the minimum value of *f* on *D.*

The maximum and minimum values of *f* are called the extreme values of *f*

The following diagram illustrates local minimum, global minimum, local maximum, global maximum. Scroll down the page for examples and solutions.

If *f* has a local maximum or minimum at *c*, and if *f &lsquo*(*c*) exists then * f &lsquo*(

*c*) = 0

Definition of critical number

A critical number of a function *f* is a number *c* in the domain of *f* such that either *f* &lsquo(*c*) = 0 of *f* &lsquo(c) does not exists.

**Example:**

Find the critical numbers of

So, the critical numbers are and 0.

### The Extreme Value Theorem

If *f* is continuous on a closed interval [*a, b*], then *f* attains an absolute maximum value *f*(*c*) and an absolute minimum value *f*(*d*) at some number *c* and *d* in [*a*, *b*]

### The Closed Interval Method

These are the steps to find the absolute maximum and minimum values of a continuous function *f* on a closed interval [*a*, *b*]:

* Step 1:* Find the values of

*f*at the critical numbers of

*f*in (

*a*,

*b*).

* Step 2:* Find the values of

*f*at the endpoints of the interval.

* Step 3:* The largest of the values from Steps 1 and 2 is the absolute maximum value and the smallest of these values is the absolute minimum value.

Find the absolute maximum and minimum value of the function

Since *f* is continuous on , we can use the Closed Interval Method.

So, the critical numbers are *x* = 0 and *x* = 2

The values of *f* at these critical numbers are *f*(0) = 1 and *f*(2) = &ndash3

* Step 2:* Find the values of

*f*at the endpoints of the interval.

The values of *f* at the endpoints of the interval are

* Step 3:* The largest of the values from Steps 1 and 2 is the absolute maximum value and the smallest of these values is the absolute minimum value.

Comparing the four numbers, we see that the absolute maximum value is *f*(4) = 17 and the absolute minimum is *f*(2) = &ndash3.

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