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Mathematical Applications for the ...

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

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BuyFindarrow_forward

Mathematical Applications for the ...

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

A firm sells 100 TVs per month at $300 each, but market research indicates that it can sell 1 more TV per month for each $2 reduction of the price. At what price will the revenue be maximized?

To determine

To calculate: The price at which TV should be sold to maximize the revenue where the firm sells 100 TV per month each at $300 and the firm can sell 1 more TV per month for $2 each reduction in price.

Explanation

Given Information:

The firm sells per month 100 TV and it can sell one more at $2 reduction in price for each per month.

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 a firm sells 100 TV per month at $300 each and it can sell 1 more TV per month for each $2 reduction in price.

Let x represent the additional units of TV sold by the firm.

Since, the number of units sold is 100+x and the price is 3002x.

Then, revenue is the product of number of units sold and the price.

Thus, the revenue function is given by

R(x)=(100+x)(3002x)=30,000+100x2x2

Calculate the first derivative of the function R(x)=30,000+100x2x2.

R(x)=1004x

Set the first derivative equal to 0.

1004x=0

Solve for x.

100=4xx=1004=25

Now, calculate the values of the first derivative on either side of the critical point x=25

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