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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

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BuyFindarrow_forward

Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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Applying the Second-Partials Test In Exercises 63–70, find the relative extrema and saddle points of the function.

f ( x , y ) = x 2 y 2 + 2 x + 4 y

To determine

To calculate: The relative extrema and saddle point of function f(x,y)=x2y2+2x+4y.

Explanation

Given Information:

The provided function is f(x,y)=x2y2+2x+4y.

Formula used:

Partial differentiation f(x,y) with respect to x by holding y constant as:

x[f(x,y)]=fx(x,y)

Partial differentiation f(x,y) with respect to y by holding x constant as:

y[f(x,y)]=fy(x,y)

The every partial differentiation of the function f(x,y),

2yxf(x,y)=2xyf(x,y)

The following procedure are used to calculate relative maximum and saddle point of given function f(x,y).

Step-1: Calculate first partial differentiation of function f(x,y) with respect to x and y.

Step-2 Calculate second partial differentiation of function f(x,y) with respect to x and y.

Step-3: Equate first partial differentiation of function f(x,y) with respect to x to zero.

Step-4: Equate first partial differentiation of function f(x,y) with respect to y to zero.

Step-5: Calculate critical point of function f(x,y) that is (a,b).

Step-6: Now test function f(x,y) at critical point (a,b) for extrema and saddle point that is describe in following table.

Critical point 22xf(x,y) 2x2f(x,y)2y2f(x,y)[2yxf(x,y)]2 Conclusion
(a,b) 22xf(a,b)>0 2x2f(a,b)2y2f(a,b)[2yxf(a,b)]2>0 The function f(x,y) has relative minimum at point (a,b)
(a,b) 22xf(a,b)<0 2x2f(a,b)2y2f(a,b)[2yxf(a,b)]2>0 The function f(x,y) has relative maximum at point (a,b)
(a,b) 2x2f(a,b)2y2f(a,b)[2yxf(a,b)]2<0 The function f(x,y) has saddle point (a,b,f(a,b))
(a,b) 2x2f(a,b)2y2f(a,b)[2yxf(a,b)]2=0 The test gives no information.

Calculation:

Consider the function,

f(x,y)=x2y2+2x+4y

Partial differentiation f(x,y) with respect to x by holding y constant,

xf(x,y)=x(x2y2+2x+4y)=x(x2)+(y2)x(1)+x(2x)+(4y)x(1)=2x+2

The first partial derivative of function f(x,y)=x2y2+2x+4y with respect to x,

xf(x,y)=2x+2

Now again partial differentiation above equation with respect to x by holding y constant,

22xf(x,y)=x(2x+2)=2

The first partial derivative of function f(x,y)=x2y2+2x+4y with respect to y by holding x constant

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