A manufacture has been selling 1750 television sets a week at dollar 480 each. A

Question

A manufacture has been selling 1750 television sets a week at dollar 480 each. A market survey indicates that for each dollar 16 rebate offered to a buyer, the number of sets sold will increase by 160 per week. Find the demand function p(x), where x is the number of the television sets sold per week. p(x) = Preview How large rebate should the company offer to a buyer, in order to maximize its revenue? Dollar Preview If the weekly cost function is 140000 + 160x, how should it set the size of the rebate to maximize its profit? dollar Preview

Explanation / Answer

Ans- a
slope of demand function = change in sales/change in rebate
slope of demand function = 160 / 16
slope of demand function = 10

Since when rebate is 0, then demand is 1750, then
demand(r) = 10r + 1750

Ans-b
Formula for revenue-
Revenue(r) = demand(r) * (price - rebate)
Revenue(r) = (10r + 1750)(480 - r)
Revenue(r) = 4800r + 840000 - 10r^2 - 1750r
Revenue(r) = 3050r + 840000 - 10r^2
Revenue(r) = -10r^2 + 3050r + 840000

To find the rebate that maximizes revenue, get the first derivative and solve for r
Revenue'(r) = -20r + 3050 = 0
20r = 3050
r = 3050/20
r = 152.5
So revenue is maximized when rebate is 152.5 dollars.

Ans-c

Formula for profit -
profit(r) = demand(r) * (price - cost - rebate)
profit(r) = (10r + 1750)(480 - 160 - r)
profit(r) = (10r + 1750)(320 - r)
profit(r) = 3200r + 560000 - 10r^2 - 1750r
profit(r) = 1450r + 560000 - 10r^2
profit(r) = -10r^2 + 1450r + 560000

To find the rebate that maximizes profit, get the first derivative and solve for r
profit'(r) = - 20r + 1450 = 0
20r = 1450
r = 72.50
So profit is maximized when rebate is 72.50 dollars.

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