13.7: EXTREME VALUES AND SADDLE POINTS
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 function when the 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 be a function of two variables that is differentiable on an open set containing the point . The point is called a critical point of a function of two variables if one of the two following conditions holds:
1.
2. Either does not exist.
EXAMPLE : FINDING CRITICAL POINTS
Find the critical points of each of the following functions:
a.
b.
Solution:
a. First, we calculate
z = f(x, y) ( , )x0 y0 ( , )x0 y0
f ( , ) = ( , ) = 0 fx x0 y0 fy x0 y0
( , ) or ( , ) fx x0 y0 fy x0 y0
13.7.1
f(x, y) = 4 − 9 + 24y + 36x + 36√−−−−−−−−−−−−−−−−−−−−−y2 x2 − g(x, y) =x2+ 2xy − 4 + 4x − 6y + 4y2
(x, y) and (x, y) :
fx fy
(x, y)
fx = (−18x + 36)(4 − 9 + 24y + 36x + 361
2 y2 x2 )−1/2
= −9x + 18
4 − 9 + 24y + 36x + 36y2 x2
−−−−−−−−−−−−−−−−−−−−−−
√
. (x, y)
fy = (8y + 24)(4 − 9 + 24y + 36x + 361
2 y2 x2 )−1/2
= 4y + 12
4 − 9 + 24y + 36x + 36y2 x2
−−−−−−−−−−−−−−−−−−−−−−
√
Next, we set each of these expressions equal to zero:
Then, multiply each equation by its common denominator:
Therefore, and so is a critical point of .
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:
This equation represents a hyperbola. We should also note that the domain of consists of points satisfying the inequality
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:
Dividing both sides by puts the equation in standard form:
Notice that point is the center of the hyperbola.
Thus, the critical points of the function are and all points on the hyperbola, .
−9x + 18
4 − 9 + 24y + 36x + 36y2 x2
−−−−−−−−−−−−−−−−−−−−−−
√
4y + 12
4 − 9 + 24y + 36x + 36y2 x2
−−−−−−−−−−−−−−−−−−−−−−
√
= 0
= 0.
−9x + 18 4y + 12
= 0
= 0.
x = 2 y = −3, (2, −3) f
4 − 9 + 24y + 36x + 36 = 0.y2 x2 (13.7.1)
f
4 − 9 + 24y + 36x + 36 ≥ 0.y2 x2 (13.7.2)
4 − 9 + 24y + 36x + 36y2 x2 4 − 9 + 24y + 36xy2 x2 4 + 24y − 9 + 36xy2 x2 4( + 6y) − 9( − 4x)y2 x2 4( + 6y + 9) − 9( − 4x + 4)y2 x2 4(y + 3 − 9(x − 2)2 )2
= 0
= −36
= −36
= −36
= −36 − 36 + 36
= −36.
−36
4(y + 3)2 −
−36
9(x − 2)2
−36 (x − 2)2 −
4
(y + 3)2 9
= 1
= 1.
(2, −3)
f (2, −3) (x − 2)2 − = 1
4
(y + 3)2 9
b. First, we calculate and :
Next, we set each of these expressions equal to zero, which gives a system of equations in and :
Subtracting the second equation from the first gives , so . Substituting this into the first equation gives , so . Therefore is a critical point of . There are no points in that make either partial derivative not exist.
Figure shows the behavior of the surface at the critical point.
Figure : The function has a critical point at .
EXERCISE :
Find the critical point of the function HintAnswer
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.
(x, y)
gx gy(x, y)
(x, y) gx
(x, y) gy
= 2x + 2y + 4
= 2x − 8y − 6.
x y
2x + 2y + 4 2x − 8y − 6
= 0
= 0.
10y + 10 = 0 y = −1 2x + 2(−1) + 4 = 0 x = −1
(−1, −1) g R2
13.7.1
13.7.1 g(x, y) (−1, −1, 6)
13.7.1
f(x, y) =x3+ 2xy − 2x − 4y.
DEFINITION: GLOBAL AND LOCAL EXTREMA
Let be a function of two variables that is defined and continuous on an open set containing the point Then has a local maximum at if
for all points within some disk centered at . The number is called a local maximum value. If the preceding inequality holds for every point in the domain of , then has a global maximum (also called an absolute maximum) at
The function has a local minimum at if
for all points within some disk centered at . The number is called a local minimum value. If the preceding inequality holds for every point in the domain of , then has a global minimum (also called an absolute minimum) at .
If is either a local maximum or local minimum value, then it is called a local extremum (see the following figure).
Figure : The graph of has a maximum value when . It attains its minimum value at the boundary of its domain, which is the circle
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
z = f(x, y) ( , ).x0 y0 f ( ,x0 y0
f( , ) ≥ f(x, y)x0 y0 (13.7.3)
(x, y) ( , )x0 y0 f( , )x0 y0
(x, y) f f ( , ).x0 y0
f ( , )x0 y0
f( , ) ≤ f(x, y)x0 y0 (13.7.4)
(x, y) ( , )x0 y0 f( , )x0 y0
(x, y) f f ( , )x0 y0
f( , )x0 y0
13.7.2 z = 16 −√−−−−−−−−−x2−y−2 (x, y) = (0, 0) + = 16.
x2 y2
Let be a function of two variables that is defined and continuous on an open set containing the point . Suppose and each exists at . If f has a local extremum at , then is a critical point of .
SECOND DERIVATIVE TEST
Consider the function This function has a critical point at , since . However, does not have an extreme value at . Therefore, the existence of a critical value at does not guarantee a local extremum at . 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 : Graph of the function . 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 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 is (a parabola opening upward), but the vertical trace corresponding to
is (a parabola opening downward). Therefore, it is both a global maximum for one trace and a global minimum for another.
z = x2− y2
z = f(x, y) ( , )x0 y0 fx fy ( , )x0 y0
( , )x0 y0 ( , )x0 y0 f
f(x) = .x3 x = 0 f′(0) = 3(0 = 0)2 f x = 0
x = x0 x = x0
13.7.3 z =x2−y2
(x, y) =x2−y2 y = 0 z = x2 x = 0 z = −y2
DEFINITION: SADDLE POINT
Given the function the point is a saddle point if both and , but does not have a local extremum at
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 that replaces in the second derivative test for a function of one variable.
SECOND DERIVATIVE TEST
Let be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point .
Suppose and Define the quantity
Then:
i. If and , then f has a local minimum at . ii. If and , then f has a local maximum at . iii. If , then has a saddle point at .
iv. If , then the test is inconclusive.
See Figure .
Figure : The second derivative test can often determine whether a function of two variables has a local minima (a), a local maxima (b), or a saddle point (c).
z = f(x, y), ( , , f( , ))x0 y0 x0 y0 f0( , ) = 0x0 y0 fy( , ) = 0x0 y0 f ( , ).x0 y0
D ( )
f′′ x0
z = f(x, y) ( , )x0 y0
( , ) = 0
fx x0 y0 fy( , ) = 0.x0 y0
D =fxx( , ) ( , ) − (x0 y0 fyy x0 y0 fxy( , ) .x0 y0 )2 (13.7.5)
D > 0 fxx( , ) > 0x0 y0 ( , )x0 y0 D > 0 fxx( , ) < 0x0 y0 ( , )x0 y0
D < 0 f ( , )x0 y0
D = 0 13.7.4
13.7.4
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 be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point To apply the second derivative test to find local extrema, use the following steps:
1. Determine the critical points of the function where Discard any points where at least one of the partial derivatives does not exist.
2. Calculate the discriminant for each critical point of .
3. 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 : 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:
a.
b.
Solution:
a. Step 1 of the problem-solving strategy involves finding the critical points of . To do this, we first calculate and , then set each of them equal to zero:
Setting them equal to zero yields the system of equations
The solution to this system is and . Therefore is a critical point of .
Step 2 of the problem-solving strategy involves calculating To do this, we first calculate the second partial derivatives of
Therefore,
z = f(x, y) ( , ).x0 y0
( , )x0 y0 f fx( , ) = ( , ) = 0.x0 y0 fy x0 y0
D =fxx( , ) ( , ) − (x0 y0 fyy x0 y0 fxy( , )x0 y0 )2 f
13.7.2
f(x, y) = 4 + 9 + 8x − 36y + 24x2 y2 g(x, y) = 1 + + 2xy − 6x − 3y + 4
3x3 y2
f fx(x, y) fy(x, y)
(x, y) fx
(x, y) fy
= 8x + 8
= 18y − 36.
8x + 8 18y − 36= 0
= 0.
x = −1 y = 2 (−1, 2) f
D. f :
(x, y) fxx
(x, y) fxy
(x, y) fyy
= 8
= 0
= 18.
D =fxx(−1, 2) (−1, 2) − (fyy fxy(−1, 2) = (8)(18) − (0 = 144.)2 )2
Step 3 states to check Note. Since and this corresponds to case 1. Therefore, has a local minimum at as shown in the following figure.
Figure : The function has a local minimum at Note the scale on the -axis in this plot is in thousands.
b. For step 1, we first calculate and , then set each of them equal to zero:
Setting them equal to zero yields the system of equations
To solve this system, first solve the second equation for . This gives . Substituting this into the first equation gives z = 4x2+ 9y2+ 8x − 36y + 24
D > 0 fxx(−1, 2) > 0, f (−1, 2)
13.7.5 f(x, y) (−1, 2, −16). y
(x, y)
gx gy(x, y)
(x, y) gx
(x, y) gy
=x2+ 2y − 6
= 2y + 2x − 3.
+ 2y − 6 x2
2y + 2x − 3= 0
= 0.
y y = 3 − 2x
2
Therefore, or . Substituting these values into the equation yields the critical points and . Step 2 involves calculating the second partial derivatives of :
Then, we find a general formula for :
Next, we substitute each critical point into this formula:
In step 3, we note that, applying Note to point leads to case , which means that is a saddle point. Applying the theorem to point leads to case , which means that corresponds to a local minimum as shown in the following figure.
+ 3 − 2x − 6 x2
− 2x − 3 x2
(x − 3)(x + 1)
= 0
= 0
= 0.
x = −1 x = 3 y = 3 − 2x2 (−1, )52 (3, − )32
g
(x, y) gxx
(x, y) gxy
(x, y) gyy
= 2x
= 2
= 2.
D
D( , )x0 y0 =gxx( , ) ( , ) − (x0 y0 gyy x0 y0 gxy( , )x0 y0 )2
= (2 )(2) −x0 22
= 4 − 4.x0
D (−1, )52 D (3, − )32
= (2(−1))(2) − (2 = −4 − 4 = −8)2
= (2(3))(2) − (2 = 12 − 4 = 8.)2
(−1, )52 3 (−1, )52 (3, − )32
1 (3, − )32
Figure : The function has a local minimum and a saddle point.
EXERCISE
Use the second derivative to find the local extrema of the function
Hint Answer
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
13.7.6 g(x, y)
13.7.2
f(x, y) =x3+ 2xy − 6x − 4 .y2
A continuous function on a closed and bounded set in the plane attains an absolute maximum value at some point of and an absolute minimum value at some point of .
Now that we know any continuous function 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 is a differentiable function of two variables defined on a closed, bounded set . Then will attain the absolute maximum value and the absolute minimum value, which are, respectively, the largest and smallest values found among the following:
1. The values of at the critical points of in . 2. The values of on the boundary of .
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 . But an interior point of that’s an absolute extremum is also a local extremum; hence, is a critical point of by Fermat’s theorem. Therefore the only possible values for the global extrema of on are the extreme values of on the interior or boundary of .
PROBLEM-SOLVING STRATEGY: FINDING ABSOLUTE MAXIMUM AND MINIMUM VALUES
Let be a continuous function of two variables defined on a closed, bounded set , and assume is differentiable on . To find the absolute maximum and minimum values of on , do the following:
1. Determine the critical points of in . 2. Calculate at each of these critical points.
3. Determine the maximum and minimum values of on the boundary of its domain.
4. The maximum and minimum values of will occur at one of the values obtained in steps and .
Finding the maximum and minimum values of on the boundary of 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. The same approach can be used for other shapes such as circles and ellipses.
If the boundary of the set is a more complicated curve defined by a function for some constant , and the first-order partial derivatives of exist, then the method of Lagrange multipliers can prove useful for determining the extrema of on the boundary which is introduced in Lagrange Multipliers.
EXAMPLE : 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:
a. on the domain defined by and
b. on the domain defined by
f(x, y) D D
D
f
z = f(x, y) D f
f f D
f D
D ( , )x0 y0 D ( , )x0 y0
f f D f D
z = f(x, y) D f D
f D
f D
f
f
f 2 3
f D
D g(x, y) = c c g
f
13.7.3
f(x, y) =x2− 2xy + 4 − 4x − 2y + 24y2 0 ≤ x ≤ 4 0 ≤ y ≤ 2 g(x, y) =x2+y2+ 4x − 6y x2+y2≤ 16
Solution:
a. Using the problem-solving strategy, step involves finding the critical points of on its domain. Therefore, we first calculate and , then set them each equal to zero:
Setting them equal to zero yields the system of equations
The solution to this system is and . Therefore is a critical point of . Calculating gives
The next step involves finding the extrema of on the boundary of its domain. The boundary of its domain consists of four line segments as shown in the following graph:
Figure : Graph of the domain of the function
is the line segment connecting and , and it can be parameterized by the equations for . Define . This gives . Differentiating leads to Therefore, has a critical value at , which corresponds to the point . Calculating gives the -value .
is the line segment connecting and , and it can be parameterized by the equations for Again, define This gives Then, . g has a critical value at , which corresponds to the point
Calculating gives the -value .
is the line segment connecting and , and it can be parameterized by the equations for Again, define This gives The critical value corresponds to the point So, calculating gives the -value .
is the line segment connecting and , and it can be parameterized by the equations for This time, and the critical value correspond to the point . Calculating gives the -value
1 f fx(x, y) fy(x, y)
(x, y) fx
(x, y) fy
= 2x − 2y − 4
= −2x + 8y − 2.
2x − 2y − 4
−2x + 8y − 2
= 0
= 0.
x = 3 y = 1 (3, 1) f f(3, 1) f(3, 1) = 17.
f
13.7.7 f(x, y) =x2− 2xy + 4 − 4x − 2y + 24.y2
L1 (0, 0) (4, 0) x(t) = t, y(t) = 0 0 ≤ t ≤ 4 g(t) = f(x(t), y(t))
g(t) = − 4t + 24t2 g g'(t) = 2t − 4. g t = 2 (2, 0)
f(2, 0) z 20
L2 (4, 0) (4, 2) x(t) = 4, y(t) = t 0 ≤ t ≤ 2.
g(t) = f(x(t), y(t)). g(t) = 4 − 10t + 24.t2 g'(t) = 8t − 10 t = 54 (0, ) .54
f (0, )54 z 27.75
L3 (0, 2) (4, 2) x(t) = t, y(t) = 2 0 ≤ t ≤ 4.
g(t) = f(x(t), y(t)). g(t) = − 8t + 36.t2 (4, 2). f(4, 2) z 20
L4 (0, 0) (0, 2) x(t) = 0, y(t) = t 0 ≤ t ≤ 2.
g(t) = 4 − 2t + 24t2 t = 14 (0, )14 f (0, )14 z 23.75.
We also need to find the values of at the corners of its domain. These corners are located at and :
The absolute maximum value is , which occurs at , and the global minimum value is , which occurs at both and as shown in the following figure.
Figure : The function has two global minima and one global maximum over its domain.
b. Using the problem-solving strategy, step involves finding the critical points of on its domain. Therefore, we first calculate and , then set them each equal to zero:
Setting them equal to zero yields the system of equations
The solution to this system is and . Therefore, is a critical point of . Calculating we get
f(x, y) (0, 0), (4, 0), (4, 2) (0, 2)
f(0, 0) f(4, 0) f(4, 2) f(0, 2)
= (0 − 2(0)(0) + 4(0 − 4(0) − 2(0) + 24 = 24)2 )2
= (4 − 2(4)(0) + 4(0 − 4(4) − 2(0) + 24 = 24)2 )2
= (4 − 2(4)(2) + 4(2 − 4(4) − 2(2) + 24 = 20)2 )2
= (0 − 2(0)(2) + 4(2 − 4(0) − 2(2) + 24 = 36.)2 )2
36 (0, 2) 20 (4, 2) (2, 0)
13.7.8 f(x, y)
1 g gx(x, y) gy(x, y)
(x, y) gx
(x, y)
gy = 2x + 4
= 2y − 6.
2x + 4 2y − 6= 0
= 0.
x = −2 y = 3 (−2, 3) g g(−2, 3),
g(−2, 3) = (−2 +)2 32 + 4(−2) − 6(3) = 4 + 9 − 8 − 18 = −13. (13.7.6)
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 centered at the origin as shown in the following graph.
Figure : Graph of the domain of the function .
The boundary of the domain of g can be parameterized using the functions for . Define
Setting leads to
4
13.7.9 g(x, y) =x2+y2+ 4x − 6y
x(t) = 4 cos t, y(t) = 4 sin t 0 ≤ t ≤ 2π h(t) = g(x(t), y(t)) : h(t) = g(x(t), y(t))
= (4 cos t + (4 sin t + 4(4 cos t) − 6(4 sin t))2 )2
= 16cos2t + 16sin2t + 16 cos t − 24 sin t
= 16 + 16 cos t − 24 sin t.
h'(t) = 0
−16 sin t − 24 cos t
−16 sin t
−16 sin t
−16 cos t tan t
= 0
= 24 cos t
= 24 cos t−16 cos t
= − .4 3
−16 sin t − 24 cos t
−16 sin t
−16 sin t
−16 cos t tan t
= 0
= 24 cos t
= 24 cos t
−16 cos t
= − .3 2
This equation has two solutions over the interval . One is and the other is . For the first angle,
Therefore, and , so is a critical point on the boundary and
For the second angle,
Therefore, and , so is a critical point on the boundary and
The absolute minimum of g is which is attained at the point , 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 . These are the absolute extrema of g on D as shown in the following figure.
0 ≤ t ≤ 2π t = π − arctan( )32 t = 2π − arctan( )32 sin t
cos t
= sin(π − arctan( )) = sin(arctan( )) =32 32 3 13√−−
13
= cos(π − arctan( )) = − cos(arctan( )) = −32 32 2 13√−−. 13 x(t) = 4 cos t = −8 13√13 y(t) = 4 sin t = 12 13√13 (−8 13√13 , 12 1313√ )
g (−8 13√13 ,12 1313√ )=(−8 13√13 )2+(12 1313√ )2+ 4 (−8 13√13 ) − 6 (√121313)
=144 + − −
13 64 13
32 13√−−
13
72 13√−−
13
= 208 − 104 13√−−≈ −12.844.
13
sin t cos t
= sin(2π − arctan( )) = − sin(arctan( )) = −32 32 3 13√−−
13
= cos(2π − arctan( )) = cos(arctan( )) =32 32 2 13√−−. 13 x(t) = 4 cos t = 8 13√13 y(t) = 4 sin t = −12 1313√ (8 13√13 , −12 1313√ )
g (8 13√13 , −12 1313√ )=(8 13√13 )2+(−12 1313√ )2+ 4 (8 13√13 ) − 6 (−12 1313√ )
= 144 + + +
13 64 13
32 13√−−
13
72 13√−−
13
= 208 + 104 13√−−≈ 44.844.
13
−13, (−2, 3)
(8 13√13 , −12 1313√ )
Figure : The function has a local minimum and a local maximum.
EXERCISE :
Use the problem-solving strategy for finding absolute extrema of a function to find the absolute extrema of the function
on the domain defined by and Hint
Answer
EXAMPLE : PROFITABLE GOLF BALLS
Pro- company has developed a profit model that depends on the number of golf balls sold per month (measured in thousands), and the number of hours per month of advertising , according to the function
where is measured in thousands of dollars. The maximum number of golf balls that can be produced and sold is , and the maximum number of hours of advertising that can be purchased is . Find the values of and that maximize profit, and find the maximum profit.
13.7.10 f(x, y)
13.7.3
f(x, y) = 4 − 2xy + 6 − 8x + 2y + 3x2 y2 0 ≤ x ≤ 2 −1 ≤ y ≤ 3.
13.7.4
T x
y
z = f(x, y) = 48x + 96y −x2− 2xy − 9 ,y2 (13.7.7)
z 50, 000
25 x y
Figure : (credit: modification of work by oatsy40, Flickr) Solution
Using the problem-solving strategy, step involves finding the critical points of on its domain. Therefore, we first calculate and then set them each equal to zero:
Setting them equal to zero yields the system of equations
The solution to this system is and . Therefore is a critical point of . Calculating gives
The domain of this function is and as shown in the following graph.
Figure : Graph of the domain of the function 13.7.11
1 f fx(x, y) fy(x, y),
(x, y) fx
(x, y)
fy = 48 − 2x − 2y
= 96 − 2x − 18y.
48 − 2x − 2y 96 − 2x − 18y= 0
= 0.
x = 21 y = 3 (21, 3) f f(21, 3)
f(21, 3) = 48(21) + 96(3) −212− 2(21)(3) − 9(3 = 648.)2 0 ≤ x ≤ 50 0 ≤ y ≤ 25
13.7.12 f(x, y) = 48x + 96y −x2− 2xy − 9 .y2
is the line segment connecting and and it can be parameterized by the equations for We then define
Setting yields the critical point which corresponds to the point in the domain of . Calculating gives
is the line segment connecting and , and it can be parameterized by the equations for . Once again, we define
This function has a critical point at , which corresponds to the point . This point is not in the domain of .
is the line segment connecting and , and it can be parameterized by the equations for . We define :
This function has a critical point at , which corresponds to the point which is not in the domain.
is the line segment connecting to , and it can be parameterized by the equations for . We define :
This function has a critical point at , which corresponds to the point , which is on the boundary of the domain. Calculating gives . We also need to find the values of at the corners of its domain. These corners are located at and :
L1 (0, 0) (50, 0), x(t) = t, y(t) = 0 0 ≤ t ≤ 50.
g(t) = f(x(t), y(t)) :
g(t) = f(x(t), y(t))
= f(t, 0)
= 48t + 96(0) −y2− 2(t)(0) − 9(0)2
= 48t − .t2
g'(t) = 0 t = 24, (24, 0) f f(24, 0) 576.
L2 (50, 25) x(t) = 50, y(t) = t 0 ≤ t ≤ 25
g(t) = f(x(t), y(t)) :
g(t) = f(x(t), y(t))
= f(50, t)
= 48(50) + 96t −502− 2(50)t − 9t2
= −9 − 4t − 100.t2
t = −29 (50, −29) f
L3 (0, 25) (50, 25) x(t) = t, y(t) = 25 0 ≤ t ≤ 50
g(t) = f(x(t), y(t))
g(t) = f(x(t), y(t))
= f(t, 25)
= 48t + 96(25) − − 2t(25) − 9(t2 252)
= − − 2t − 3225.t2
t = −1 (−1, 25),
L4 (0, 0) (0, 25) x(t) = 0, y(t) = t 0 ≤ t ≤ 25 g(t) = f(x(t), y(t))
g(t) = f(x(t), y(t))
= f(0, t)
= 48(0) + 96t − (0 − 2(0)t − 9)2 t2
= 96t − .t2
t = 163 (0, )163 f (0, )163 256
f(x, y) (0, 0), (50, 0), (50, 25) (0, 25)
f(0, 0) f(50, 0) f(50, 25) f(0, 25)
= 48(0) + 96(0) − (0 − 2(0)(0) − 9(0 = 0)2 )2
= 48(50) + 96(0) − (50 − 2(50)(0) − 9(0 = −100)2 )2
= 48(50) + 96(25) − (50 − 2(50)(25) − 9(25 = −5825)2 )2
= 48(0) + 96(25) − (0 − 2(0)(25) − 9(25 = −3225.)2 )2
The maximum critical value is , which occurs at . Therefore, a maximum profit of is realized when golf balls are sold and hours of advertising are purchased per month as shown in the following figure.
Figure : The profit function has a maximum at .
KEY CONCEPTS
A critical point of the function is any point where either , or at least one of and do not exist.
A saddle point is a point where , but 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
GLOSSARY
648 (21, 3) $648, 000 21, 000 3
13.7.13 f(x, y) (21, 3, 648)
f(x, y) ( , )x0 y0 fx( , ) = ( , ) = 0x0 y0 fy x0 y0 fx( , )x0 y0 fy( , )x0 y0 ( , )x0 y0 fx( , ) = ( , ) = 0x0 y0 fy x0 y0 ( , )x0 y0
D =fxx( , ) ( , ) − (x0 y0 fyy x0 y0 fxy( , )x0 y0 )2
critical point of a function of two variables discriminant
saddle point
CONTRIBUTORS
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.
