5. Extrema of Multivariable Functions


Example 6 Identify Critical Points


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Extrema of Multivariable Functions

Example 6 Identify Critical Points


Use the Test for Relative Extrema to classify the critical points for



f x, y y4 32 y x3 x2
saddle points.
as relative maximum, relative minimum, or

Solution To find the critical points, we need to compute the first partial derivatives of the function. The first partial derivatives are



x y
f x, y 3x2 2x f x, y 4 y3 32

Set each partial derivative equal to zero to find the critical points. Let’s start with the partial derivative with respect to x:





0  3x2  2x
0 x 3x 2
x  0 or 3x  2  0

3
x 2
Factor the right side. Set each factor equal to zero to solve for x.

Now the partial derivative with respect to y:



0  4 y3  32
0  4 y3  8
0  y3  8
8  y3
2  y
Factor the right side. Set each factor equal to zero. Only the factor in parentheses can be zero.

Cube root both side to solve for y.






3
Combining these together we get two critical points, 0, 2 and 2 ,2 .
We’ll need the second partial derivatives to apply the Test for Relative Extrema to each critical point.



fxx x, y 6x 2
f x, y 12 y2
fxy x, y 0




yy
Evaluate each second partial derivative at the critical points to find the value of D.






fxx x, y 6x 2

f yy x, y 12 y
2

fxy x, y 0

D

Type

0, 2

-2

48

0

-96

Saddle point

2 ,2
3

2

48

0

96

Relative minimum

With each critical point classified, the corresponding z values are


f 0, 2 24 32 2 03 02 48

3 3 3 27
f 2 , 2  24  32  2  2 3  2 2   1300  48.15

When a business produces several products, a multivariable profit function may be calculated to find the production levels that maximize profit.


Example 7 Maximize Profit


A small startup company produces speakers and subwoofers for computers that they sell through a website. After extensive research, the company has developed a revenue function,





R x, y x 110 4.5x y 155 2 y
thousand dollars

where x is the number of subwoofers produced and sold in thousands and y is the number of speakers produced and sold in thousands. The corresponding cost function is


C x, y 3x2 3y2 5xy 5 y 50 thousand dollars

Find the production levels that maximize revenue.


Solution By subtracting the cost from the revenue, we get the profit function,


P x, y R x, y C x, y
x 110  4.5x y 155  2 y 3x2  3y2  5xy  5 y  50
 110x  4.5x2 155 y  2 y2  3x2  3y2  5xy  5 y  50
 7.5x2  5 y2  5xy 110x 150 y  50

The first partial derivatives are




Px x, y 15x 5 y 110
Py x, y 10 y 5x 150

The critical point are found by setting the partial derivatives equal to zero. This results in a system of equations that is solved using the Elimination Method.



15x  5 y 110  0 15x  5 y  110
10 y  5x 150  0 5x 10 y  150

Multiply the second equation by -3 and add it to the first equation to eliminate x:


15x  5 y  110



15x  30 y  450

 25 y  340 




y  13.6

If we substitute this value in the second equation, we get a value for x,




5x 10 13.6 150

5x
136
 150

5x  14
x  2.8


The critical point is at 2.8,13.6. This point could be a relative maximum, a relative minimum, or a saddle point. The Test for Relative Extrema helps us to distinguish whether the point is a relative maximum. The second partial derivatives are



Pxx x, y 15
Pyy x, y 10
Pxy x, y 5

Each of these derivatives is a constant so any critical points have




D 151052 125
  

Pxx
Pyy
Pxy


Since
D  0
and
Pxx  0 , the critical point is a relative maximum. The

profit at these production levels is



P 2.8,13.6 7.52.82 513.62 52.813.6110 2.8150 13.6 50
 6326


At a production level of 2.8 thousand subwoofers and 13.6 thousand speakers, the company will lose 6326 thousand dollars. The company will need to reassess their business model in order to make positive profit.





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