   Chapter 7.5, Problem 5E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Applying the Second-Partials Test In Exercises 1-18, find the relative extrema and saddle points of the function. See Examples f ( x , y ) = x 2 + y 2 + 1

To determine

To calculate: The relative extrema and saddle point of the function f(x,y)=x2+y2+1.

Explanation

Given Information:

The given function is f(x,y)=x2+y2+1.

Formula used:

Partial derivatives of the function with respect to x as well as y,

fx=f(x,y)xfy=f(x,y)y

Double partial derivatives of the function;

fxx=2f(x,y)x2fyy=2f(x,y)y2fxy=2f(x,y)xy

At the critical point if the value of D=fxx(a,b)fyy(a,b)[fxy(a,b)]2,

D>0 and r<0, then declare the point as a maximum.

D>0 and r>0, then declare the point as a minima.

D<0, then this critical point will not be a extremum point.

D=0, then say double partial derivative fails and need further investigation on this point

Calculation:

Considering the given equation,

f(x,y)=x2+y2+1

The critical point of the function f(x,y) is given by

fx=0   and   fy=0

Now, find the partial derivative with respect to x,

fx=0x(x2+y2+1)=0(2x)2x2+y2+1    =0                       x =0

Now, find the partial derivative with respect to y,

fy=0y(x2+y2+1)=02yx2+y2+1      =0 y=0

Hence, the critical point is (0,0)

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