Chapter 14, Problem 3T

### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

Chapter
Section

### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# Let z = 6 x 2 + x 2 y + y 2 − 4 y +   9 . Find the pairs ( x , y ) that are critical points for z and then classify each as a relative maximum, a relative minimum, or a saddle point.

To determine

To calculate: The pairs (x,y) that are critical points for z and then classify each as a relative maxima, relative minima, and saddle points of z=6x2+x2y+y24y+9.

Explanation

Given Information:

The provided function is, z=6x2+x2y+y2âˆ’4y+9.

Formula used:

To calculate relative maxima and minima of the z=f(x,y),

1. Find the partial derivatives âˆ‚zâˆ‚x and âˆ‚zâˆ‚y.

2. Find the critical points, that is, the point(s) that satisfy âˆ‚zâˆ‚x=0 and âˆ‚zâˆ‚y=0.

3. Then find all the second partial derivatives and evaluate the value of D at each critical point, where D=(zxx)(zyy)âˆ’(zxy)2=âˆ‚2zâˆ‚x2â‹…âˆ‚2zâˆ‚y2âˆ’(âˆ‚2zâˆ‚xâˆ‚y)2.

(a) If D>0, then relative minimum occurs if zxx>0 and relative maximum occurs if zxx<0.

(b) If D<0, then neither a relative maximum nor a relative minimum occurs.

For a function f(x,y), the partial derivative of f with respect to x is calculated by taking the derivative of f(x,y) with respect to x and keeping the other variable y constant and the partial derivative of f with respect to y is calculated by taking the derivative of f(x,y) with respect to y and keeping the other variable x constant. The partial derivative of f with respect to x is denoted by fx and with respect to y is denoted by fy.

For a function z(x,y), the second partial derivative,

1. When both derivatives are taken with respect to x is zxx=âˆ‚2zâˆ‚x2=âˆ‚âˆ‚x(âˆ‚zâˆ‚x).

2. When both derivatives are taken with respect to y is zyy=âˆ‚2zâˆ‚y2=âˆ‚âˆ‚y(âˆ‚zâˆ‚y).

3. When first derivative is taken with respect to x and second derivative is taken with respect to y is zxy=âˆ‚2zâˆ‚yâˆ‚x=âˆ‚âˆ‚y(âˆ‚zâˆ‚x).

4. When first derivative is taken with respect to y and second derivative is taken with respect to x is zyx=âˆ‚2zâˆ‚xâˆ‚y=âˆ‚âˆ‚x(âˆ‚zâˆ‚y).

Power of x rule for a real number n is such that, if f(x)=xn then fâ€²(x)=nxnâˆ’1.

Chain rule for function f(x)=u(v(x)) is fâ€²(x)=uâ€²(v(x))â‹…vâ€²(x).

Constant function rule for a constant c is such that, if f(x)=c then fâ€²(x)=0.

Coefficient rule for a constant c is such that, if f(x)=câ‹…u(x), where u(x) is a differentiable function of x, then fâ€²(x)=câ‹…uâ€²(x).

Calculation:

Consider the function, z=6x2+x2y+y2âˆ’4y+9.

Recall that, for a function f(x,y), the partial derivative of f with respect to x is calculated by taking the derivative of f(x,y) with respect to x and keeping the other variable y constant and the partial derivative of f with respect to y is calculated by taking the derivative of f(x,y) with respect to y and keeping the other variable x constant.

Use the power of x rule for derivatives, the constant function rule, the chain rule, and the coefficient rule,

Thus,

âˆ‚zâˆ‚x=012x+2xy=02x(6+y)=0

Thus, x=0 or y=âˆ’6.

And,

âˆ‚zâˆ‚y=0x2+2yâˆ’4=0

Consider the equation, x=0

Substitute 0 for x in x2+2yâˆ’4=0.

0+2yâˆ’4=0y=2

Consider the equation, y=âˆ’6

Substitute âˆ’6 for y in x2+2yâˆ’4=0.

x2+2(âˆ’6)âˆ’4=0x2âˆ’12âˆ’4=0x2=16x=Â±4

Thus, the critical points are (0,2), (4,âˆ’6) and (âˆ’4,âˆ’6).

Recall that, for a function z(x,y), the second partial derivative, when both derivatives are taken with respect to x is zxx=âˆ‚2zâˆ‚x2=âˆ‚âˆ‚x(âˆ‚zâˆ‚x), when both derivatives are taken with respect to y is zyy=âˆ‚2zâˆ‚y2=âˆ‚âˆ‚y(âˆ‚zâˆ‚y), when first derivative is taken with respect to x and second derivative is taken with respect to y is zxy=âˆ‚2zâˆ‚yâˆ‚x=âˆ‚âˆ‚y(âˆ‚zâˆ‚x)

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