Chapter 7, Problem 65RE

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

10th Edition
Ron Larson
ISBN: 9781305860919

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 63–70, find the relative extrema and saddle points of the function. f ( x , y ) = ( x 2 + y 2 + 1 ) 1 / 4

To determine

To calculate: The relative extrema and saddle point of function f(x,y)=(x2+y2+1)1/4.

Explanation

Given Information:

The provided function is f(x,y)=(x2+y2+1)1/4.

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 quotient rule of differentiation with respect to x by holding y constant,

x[f(x,y)g(x,y)]=[g(x,y)xf(x,y)f(x,y)xg(x,y)][g(x,y)]2

The quotient rule of differentiation with respect to y by holding x constant,

y[f(x,y)g(x,y)]=[g(x,y)yf(x,y)f(x,y)yg(x,y)][g(x,y)]2

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

Calculation:

Consider the function,

f(x,y)=(x2+y2+1)1/4

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

xf(x,y)=x(x2+y2+1)1/4=14(x2+y2+1)3/4x(x2+y2+1)=14(x2+y2+1)3/4(2x)=x2(x2+y2+1)3/4

The first partial derivative of function f(x,y)=(x2+y2+1)1/4 with respect to x,

xf(x,y)=x2(x2+y2+1)3/4

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

22xf(x,y)=x(x2(x2+y2+1)3/4)=2(x2+y2+1)3/4x(x)(x)x2(x2+y2+1)3/44(x2+y2+1)3/2=2(x2+y2+1)3/4(1)(2x)34(x2+y2+1)1/4(2x)4(x2+y2+1)3/2=(x22y22)4(x2+y2+1)7/4

The first partial derivative of function f(x,y)=(x2+y2+1)1/4 with respect to y by holding x constant.

yf(x,y)=y(x2+y2+1)1/4=14(x2+y2+1)3/4y(x2+y2+1)=14(x2+y2+1)3/4(2y)=y2(x2+y2+1)3/4

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

22yf(x,y)=y(y2(x2+y2+1)3/4)=2(x2+y2+1)3/4y(y)(y)y2(x2+y2+1)3/44(x2+y2+1)3/2=2(x2+y2+1)3/4(1)(2y)34(x2+y2+1)1/4(2y)4(x2+y2+1)3/2=(y22x21)4(x2+y2+1)7/4

The first partial derivative of function f(x,y)=(x2+y2+1)1/4 with respect to x,

xf(x,y)=x2(x2+y2+1)3/4

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

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