Chapter 14.4, Problem 21E

### 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

# Nutrition A new food is designed to add weight to mature beef cattle. The weight in pounds is given by W =   13 x y ( 20   − x   −   2 y ) , where x is the number of units of the first ingredient and y is the number of units of the second ingredient. How many units of each ingredient will maximize the weight? What is the maximum weight?

To determine

To calculate: The number of units of each ingredient that will maximize the weight and the maximum weight if the new food is designed to add weight to mature beef cattle and the weight is given by W=13xy(20x2y) which is in pounds. In that x is the number of units of the first ingredient or y is the number of units of the second ingredient.

Explanation

Given Information:

The weight in pounds is given by W=13xy(20âˆ’xâˆ’2y).

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.

Product rule for function f(x)=u(x)â‹…v(x), where u and v are differentiable functions of x, then fâ€²(x)=v(x)â‹…uâ€²(x)+u(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 provided function is W(x,y)=13xy(20âˆ’xâˆ’2y).

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

Thus,

âˆ‚Wâˆ‚x=013y(20âˆ’xâˆ’2y)+13xy(âˆ’1)=0260yâˆ’13yxâˆ’26y2âˆ’13xy=0260yâˆ’26y2âˆ’26xy=0

Simplify it further,

âˆ‚Wâˆ‚x=026y(10âˆ’yâˆ’x)=0

Thus, y=0 or y=10âˆ’x

And,

âˆ‚Wâˆ‚y=013x(20âˆ’xâˆ’2y)+13xy(âˆ’2)=0260xâˆ’13x2âˆ’26xyâˆ’26xy=0260xâˆ’13x2âˆ’52xy=0

Simplify it further,

âˆ‚Wâˆ‚y=013x(20âˆ’xâˆ’4y)=0

Thus, x=0 or x=20âˆ’4y

Consider the equation, y=0

Since, x=20âˆ’4y, thus, x=20âˆ’4(0)=20.

Now, consider the equation, x=0

Since, y=10âˆ’x, thus, y=10âˆ’0=10

Thus, the critical points are (0,0), (0,10) and (20,0).

Also, consider the equations y=10âˆ’x and x=20âˆ’4y.

Substitute 10âˆ’x for y in x=20âˆ’4y.

x=20âˆ’4(10âˆ’x)x=20âˆ’40+4xâˆ’3x=âˆ’20x=203

Since, y=10âˆ’x, thus, y=10âˆ’203=103.

Thus, (203,103) is also a critical point.

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).

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

Wxx=âˆ‚2Wâˆ‚x2=âˆ‚âˆ‚x(âˆ‚Wâˆ‚x)=âˆ‚âˆ‚x(26y(10âˆ’yâˆ’x))=âˆ’26y

And,

Wyy=âˆ‚2Wâˆ‚y2=âˆ‚âˆ‚y(âˆ‚Wâˆ‚y)=âˆ‚âˆ‚y(13x(20âˆ’xâˆ’4y))=âˆ’52x

And,

Wxy=âˆ‚2Wâˆ‚yâˆ‚x=âˆ‚âˆ‚y(âˆ‚Wâˆ‚x)=âˆ‚âˆ‚y(26y(10âˆ’yâˆ’x))=260âˆ’52yâˆ’26x

Thus, Wxx=âˆ’26y, Wyy=âˆ’52x and Wxy=260âˆ’52yâˆ’26x

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