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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

Use a graphing calculator or computer (or Newton’s method or a computer algebra system) to find the critical points of f(x, y) = 12 + 10y – 2x2 – 8xyy4 correct to three decimal places. Then classify the critical points and find the highest point on the graph.

To determine

To find: The critical points of the function f(x,y)=12+10y2x28xyy4 using the graph and then the highest point of the function.

Explanation

Result used:

Second Derivative Test:

“Suppose the second partial derivatives of f are continuous on a disk with center (a,b), and suppose that fx(a,b)=0 and fy(a,b)=0 (that is (a,b) is a critical point of f).

Let D=D(a,b)=fxx(a,b)fyy(a,b)[fxy(a,b)]2

(a) If D>0 and fxx(a,b)>0, then f(a,b) is a local minimum.

(b) If D>0 and fxx(a,b)<0, then f(a,b) is a local maximum.

(c) If D<0, then f(a,b) is not a local maximum or minimum and it is called a saddle point”.

Given:

The function is, f(x,y)=12+10y2x28xyy4.

Calculation:

Take the partial derivative in the given function with respect to x and obtain fx.

fx=x(12+10y2x28xyy4)=x(12)+x(10y)x(2x2)x(8xy)x(y4)=0+02(2x)8y(1)0=4x8y

Thus, fx=4x8y. (1)

Take the partial derivative in the given function with respect to y and obtain fy.

fy=y(12+10y2x28xyy4)=y(12)+y(10y)y(2x2)y(8xy)y(y4)=0+10(1)08x(1)4y3=108x4y3

Thus, fy=108x4y3. (2)

Solve the equations (1) and (2) and find the values of x and y.

From the equation (1),

4x8y=04x=8yx=2y

Substitute x=2y in the equation (2),

108(2y)4y3=010+16y4y3=02(5+8y2y3)=05+8y2y3=0

Find the value of y by using the graph of 5+8y2y3=0.

Use online graphing calculator and draw the graph of 5+8y2y3=0 as shown below in Figure 1.

From Figure 1 it can be observed that the values of y are 1.542,0.717 and 2.260.

Substitute y=1.542 in x=2y and obtain the critical point as follows.

x=2(1.542)=3.084

Substitute y=0.717 in x=2y and obtain the critical point as follows.

x=2(0.717)=1.434

Substitute y=2.260 in x=2y and obtain the critical point as follows.

x=2(2.260)=4.520

Therefore, the critical points of the function f(x,y)=12+10y2x28xyy4 are (3.084,1.542),(1.434,0.717) and (4.520,2.260).

Find the second derivatives as follows.

Take the partial derivative of the equation (1) with respect to x and obtain fxx.

2fx2=x(4x8y)=4(1)0=4

Hence, 2fx2=4

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Chapter 14 Solutions

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