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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

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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 ) = 2 x y 1 2 x 4 1 2 y 4 + 1

To determine

To calculate: The relative points of extremum and saddle point of the function f(x,y)=2xyx42y42+1.

Explanation

Given Information:

The function with two variables is f(x,y)=2xyx42y42+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:

Consider the given equation,

f(x,y)=2xyx42y42+1

Now, compute the first partial derivatives with respect to x keeping y as constant;

fx=2y2x3

And, compute the first partial derivatives with respect to y keeping x as constant

fy=2x2y3

Equating the above two equations to zero the critical point is obtained

fx=02y2x3=02y=2x3

Now, rearranging and solving the above equations,

yx=x3y3yx=(xy)(x2+y2+xy)(xy)(x2+y2+xy+1)=0

Hence, x=y.

Substitute, it in the equation fx=0 and the critical points are obtained (0,0),(1,1),(1,1)

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