Chapter 14.4, Problem 19E

### 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

# Profit Suppose that the quarterly profit from the sale of Kisses and Kreams is given by P ( x , y )   =   10 x + 6.4 y   — 0.001 x 2 —   0.025 y 2 dollars where x is the number of pounds of Kisses and y is the number of pounds of Kreams. Selling how many pounds of Kisses and Kreams will maximize profit? What is the maximum profit?

To determine

To calculate: The number of pounds of Kisses and Kreams that will maximize the profit and the maximum profit. Suppose that the quarterly profit from the sale of Kisses and Kreams is given by P(x,y)=10x+6.4y0.001x20.025y2 dollars, where x is the number of pounds Kisses and y is the number of pounds of Kreams.

Explanation

Given Information:

The provided function is P(x,y)=10x+6.4y0.001x20.025y2.

Formula used:

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

(1) Find the partial derivatives zx and zy.

(2) Find the critical points, that is, the point(s) that satisfy zx=0 and zy=0.

(3) Then find all the second partial derivatives and evaluate the value of D at each critical point, where D=(zxx)(zyy)(zxy)2=2zx22zy2(2zxy)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=2zx2=x(zx).

(2) When both derivatives are taken with respect to y is zyy=2zy2=y(zy).

(3) When first derivative is taken with respect to x and second derivative is taken with respect to y is zxy=2zyx=y(zx).

(4) When first derivative is taken with respect to y and second derivative is taken with respect to x is zyx=2zxy=x(zy).

Power of x rule for a real number n is such that, if f(x)=xn then f(x)=nxn1.

Chain rule for function f(x)=u(v(x)) is f(x)=u(v(x))v(x).

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

Coefficient rule for a constant c is such that, if f(x)=cu(x), where u(x) is a differentiable function of x, then f(x)=cu(x).

Calculation:

Consider the function P(x,y)=10x+6.4y0.001x20.025y2.

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

Thus,

Px=0100.001(2x)=00.002x=10x=5,000

And,

Py=06.40.025(2y)=00.05y=6.4y=128

Thus, the critical point is (5,000,128)

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