# Thread: calculus application to economics

1. ## calculus application to economics

A manufacture has been selling 1650 television sets a week at 480 each. A market survey indicates that for each 19 rebate offered to a buyer, the number of sets sold will increase by 190 per week.
a) Find the demand function , where is the number of the television sets sold per week.

b) How large rebate should the company offer to a buyer, in order to maximize its revenue?
c) If the weekly cost function is , how should it set the size of the rebate to maximize its profit?

2. Originally Posted by xcelxp
A manufacture has been selling 1650 television sets a week at 480 each. A market survey indicates that for each 19 rebate offered to a buyer, the number of sets sold will increase by 190 per week
....
b) How large rebate should the company offer to a buyer, in order to maximize its revenue?
c) If the weekly cost function is , how should it set the size of the rebate to maximize its profit?

Hello, xcelxp,

to b) and c) only:

let h be the number how often the rebate was offered, then the revenue can be calculated by:
Code:
r(h) = (1650 + 190·h)·(480 - 19·h) =
2
- 3610·h  + 59850·h + 792000
The graph of this function is a parabola which opens downward that means that the maximum is the r-value of the vertex.
Calculate the first derivative of r:
Code:
 r'(h) = 59850 - 7220·h

r'(h) = 0 ==> h =  315
⎯⎯⎯
38
If the rebate is 315/38 * 19 = $157.50 that means if the TV set is sold for$322.50 then the revenue has a maximum.

to c)

the profit is the revenue minus the cost:
Code:
p(h) = (1650+190·h)·(480-19·h)-(132000+160·(1650+190·h))
Expand all brackets, calculate the first derivative, let the derivative be zero, solve for h and you'll get:
Code:
     155
h = ⎯⎯⎯
38
Plug in this value into the function p(h) to get the maximum profit: p_max = \$4560625
Number of sold TV sets: 2425 pieces.

EB