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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

Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement.

If f(x, y) has two local maxima, then f must have a local minimum.

To determine

Whether the statement “If f(x,y) has two local maxima, then f must have a local minimum.” is true or false.

Explanation

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”.

Reason:

The given statement is false since the below example disproves the given statement.

If the function f(x,y)=(x21)2(x2yx1)2 .

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

fx=x[(x21)2(x2yx1)2]=x[(x21)2]x[(x2yx1)2]=2(x21)(2x)2(x2yx1)(2xy1)

Thus, fx=2(x21)(2x)2(x2yx1)(2xy1) (1)

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

fy=y[(x21)2(x2yx1)2]=y[(x21)2]y[(x2yx1)2]=02(x2yx1)(x2)=2x2(x2yx1)

Thus, fy=2x2(x2yx1) (2)

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

From the equation (2),

2x2(x2yx1)=02x2=0or(x2yx1)=0x=0ory=x+1x2

From the result there are two possibilities such as either x=0ory=x+1x2 .

Suppose x=0 then the equation (1),

fx(x,y)=2(x21)(2x)2(x2yx1)(2xy1)fx(0,y)=2(01)(2(0))2(001)(2(0)y1)=02(1)(1)=2

Thus, there are no critical points for x=0 .

Therefore, y=x+1x2 substitute in the equation (1),

fx(x,y)=2(x21)(2x)2(x2yx1)(2xy1)fx(x,x+1x2)=4(x)(x21)2(x2(x+1x2)x1)(2x(x+1x2)1)=4x(x21)(2x2+2x+2)(2x+2x1)=4x(x21)

Thus, the value of x=±1 .

Therefore, the critical points of the function f(x,y)=(x21)2(x2yx1)2 are (1,2)and(1,0)

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