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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 ) = x 2 + y 2 + 8 x 12 y 3

To determine

To calculate: The relative extrema and saddle points of the function f(x,y)=x2+y2+8x12y3.

Explanation

Given Information:

The provided function is f(x,y)=x2+y2+8x12y3.

Formula used:

Second partial test to determine relative extrema.

If the function f have continuous second partial derivative in the open region (a,b) for which fx(a,b)=0 and fy(a,b)=0.

Steps to test relative extrema of the function f,

Consider the quantity, d=fxx(a,b)fyy(a,b)[fxy(a,b)]2

1. The function f contains a relative minimum at (a,b) if the quantity d>0 and fxx(a,b)>0.

2. The function f contains a relative maximum at (a,b) if the quantity d>0 and fxx(a,b)<0.

3. The point (a,b,f(a,b)) is a saddle point if the quantity d<0.

4. If the quantity d=0 then the test gives no information.

In any case, if d>0 then fxx(a,b) and fyy(a,b) must contain same sign. The derivative fxx(a,b) can be replaced by the derivative fyy(a,b) in the first two steps.

Calculation:

Consider the function,

f(x,y)=x2+y2+8x12y3

The first partial derivatives of the function f with respect to x is,

fx(x,y)=x[x2+y2+8x12y3]=2x+8

Equate fx(x,y)=0,

2x+8=02x=8x=4

The first partial derivatives of the function f with respect to y is,

fy(x,y)=y[x2+y2+8x12y3]=2y12

Equate fy(x,y)=0,

2y12=02y=12y=6

So, the critical point is (4,6)

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