   Chapter 10, Problem 34RE ### Mathematical Applications for the ...

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

#### Solutions

Chapter
Section ### Mathematical Applications for the ...

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

# In Problems 31-36, cost, revenue, and profit are in dollars and x is the number of units.Profit How many units (x) will maximize profit if R ( x ) = 46 x − 0.01 x 2 and C ( x ) = 0.05 x 2 + 10 x + 1100 ?

To determine

To calculate: The number of units of the product to obtain maximum profit where the revenue function is R(x)=46x0.01x2 and the cost function is C(x)=0.05x2+10x+1100 where x is the number of units produced and sold.

Explanation

Given Information:

The provided revenue function is R(x)=46x0.01x2 and the cost function is C(x)=0.05x2+10x+1100.

Formula used:

To find relative maxima and minima of a function,

1. Set the first derivative of the function to zero, f(x)=0, to find the critical values of the function.

2. Substitute the critical values into f(x) and calculate the critical points.

3. Calculate the sign of the function to the left and right of the critical values.

(a) If f(x)<0 and f(x)>0 on the left and right side respectively of a critical value, then the point is relative minimum.

(b) If f(x)>0 and f(x)<0 on the left and right side respectively of a critical value, then the point is relative maximum.

Calculation:

Consider the provided revenue function is R(x)=46x0.01x2 and the cost function is C(x)=0.05x2+10x+1100 where x is the number of units produced and sold.

Since, profit is the difference of revenue and cost.

Thus, the profit function is the difference of revenue and cost function.

Let P represent profit.

Thus,

P(x)=46x0.01x20.05x210x1100=0.06x2+36x1100

Calculate the first derivative of the function P(x)=0.06x2+36x1100.

P(x)=0

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