optomization

Question

A manufacture has been selling 1350 television sets a week at $510 each. A market survey indicates that for each $10 rebate offered to a buyer, the number of sets sold will increase by 100 per week. Find the function representing the demand p(x) , where is the number of the television sets sold per week and p(x) is the corresponding price. How large rebate should the company offer to a buyer, in order to maximize its revenue?  If the weekly cost function is 114750+170x, how should it set the size of the rebate to maximize its profit?

Answer 1

Let x = number of $10 rebates

 

p(x) = price of a TV set if there are x $10 rebates

 

       = 510 - 10x

 

R(x) = revenue = (price)(quantity)

 

                       = (510 - 10x)(1350 + 100x)

 

                       = -1000x² + 37500x + 688500

 

R'(x) = -2000x + 37500 = 0 when x = 18.75

 

So, maximum revenue occurs if the rebate is $187.50.

 

P(x) = profit = R(x) - C(x)

 

       = -1000x² + 37500x + 688500 - 170x - 114750

 

       = -1000x² + 37330x + 573750

 

P'(x) = -2000x + 37330 = 0 when x = 18.665

 

So, profit is maximized when the rebate is $186.65

 

Answer 2

Let x = price, s = sales (demand)

The relation between x and s is linear with a slope of -10

E.g. it passes through (510, 1350) and (500, 1450)

(s-1450)/(x-500)=-10

s-1450=5000-10x => s=6450-10x

 

The revenue is s*x = 6450x - 10x²

 

There derivative is -20x + 6450 and it is 0 where x = 322.50, i.e. the maximum revenue will occur when the price is 322.50 and the rebate is 510-322.50=187.50

 

The profit = 6450x -10x² - (170x + 114750) = -114750+6280x-10x²

The derivative is 6280-20x and it is 0 where x = 314, i.e.

when the price is set at 314 the profit will be maximized