A manufacture has been selling 1300 television sets a week at $420 each. A market survey indicates that for each $16 rebate offered to a buyer, the number of sets sold will increase by 160 per week.

If the weekly cost function is 91000+140x, what value of the rebate maximizes the profit? dollars

If y is the number of rebates, then the revenue is

r(y) = (420-16y)(1300+160y)
assuming only enough sets are made to sell out, then x = 1300+160y

so, the profit is revenue less cost

p(y) = r(y) - c(1300+160y)
= (420-16y)(1300+160y) - (91000+140(1300+160y))
= -2560y^2 + 24000y + 273000

This has a maximum at y=75/16

so, a rebate of $75 will maximize the profit

as always, check my math and reasoning.

To find the value of the rebate that maximizes the profit, we need to identify the profit function and then differentiate it with respect to the rebate.

Let's start by determining the profit function. The profit can be calculated as the revenue minus the cost.

Revenue = Selling price per unit * Number of units sold
Cost = Fixed cost + Variable cost per unit * Number of units sold

Given:
Selling price per unit = $420
Number of units sold = 1300 + 160 * (rebate amount / $16)

It is given that the weekly cost function is 91000 + 140x, where x represents the number of units sold.

Substituting the values into the revenue and cost functions, the profit function becomes:

Profit = (420 * (1300 + 160 * (rebate / 16))) - (91000 + 140 * (1300 + 160 * (rebate / 16)))

Now, let's simplify the profit function:

Profit = (420 * 1300) + (420 * 160 * (rebate / 16)) - 91000 - (140 * 1300) - (140 * 160 * (rebate / 16))

Profit = 546000 + 134400 * (rebate / 16) - 91000 - 182000 - 140 * rebate

Profit = 546000 + 8400 * rebate / 16 - 273000 - 140 * rebate

Profit = 8400 * rebate / 16 - 140 * rebate + 273000

Now, we can differentiate the profit function with respect to the rebate:

dProfit/dRebate = (8400 / 16) - 140

Simplifying further:

dProfit/dRebate = 525 - 140

dProfit/dRebate = 385

To find the maximum profit, we need to set the derivative equal to zero:

385 = 0

This implies that there is no maximum profit point, and the profit will continue to increase with increasing rebate amounts.

Therefore, there is no specific value of the rebate that maximizes the profit.

To find the value of the rebate that maximizes profit, we need to determine the profit function first.

We know that the manufacturer sells 1300 television sets a week at $420 each, so the revenue function is given by:
Revenue = Number of sets sold * Selling price = (1300 + 160x) * $420

The cost function is given as:
Cost = 91000 + 140x

The profit function is then:
Profit = Revenue - Cost
Profit = (1300 + 160x) * $420 - (91000 + 140x)

To find the value of x that maximizes profit, we need to differentiate the profit function with respect to x and set it equal to zero.

Profit' (derivative of Profit with respect to x) = 0

Differentiating the profit function:

Profit' = (1300 + 160x) * $420 - (91000 + 140x)
Profit' = 546000 + 67200x - 91000 - 140x
Profit' = -23800 + 53200x

Setting Profit' equal to zero:

-23800 + 53200x = 0
53200x = 23800
x = 23800 / 53200
x ≈ 0.448

So, the value of x that maximizes profit is approximately 0.448.

Now, we need to find the corresponding value of the rebate.

The market survey indicates that for each $16 rebate offered to a buyer, the number of sets sold will increase by 160 per week. This means that for each unit increase in x, the rebate increases by $16.

Since x ≈ 0.448, the corresponding value of the rebate is approximately 0.448 * $16 ≈ $7.168.

Therefore, the value of the rebate that maximizes profit is approximately $7.168.