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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 63–70, find the relative extrema and saddle points of the function.

f ( x , y ) = y 2 + x y + 3 y 2 x + 5

To determine

To calculate: The relative extrema and saddle point of function f(x,y)=y2+xy+3y2x+5.

Explanation

Given Information:

The provided function is f(x,y)=y2+xy+3y2x+5.

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 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)=y2+xy+3y2x+5

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

xf(x,y)=x[y2+xy+3y2x+5]=x(y2)+(y)x(x)+(3y)x(1)(2)x(x)+x(5)=y2

The first partial derivative of function f(x,y)=y2+xy+3y2x+5 with respect to x,

xf(x,y)=y2

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

22xf(x,y)=x[y2]=x[y]x[2]=1

The first partial derivative of function f(x,y)=y2+xy+3y2x+5 with respect to y by holding x constant.

yf(x,y)=y[y2+xy+3y2x+5]=y(y2)+(x)y(y)+(3)y(y)(2)y(x)+x(5)=2y+x+3

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

22yf(x,y)=y[2y+x+3]=(2)y(y)+y(x)+y(3)=2

The first partial derivative of function f(x,y)=y2+xy+3y2x+5 with respect to x,

xf(x,y)=y2

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

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