Question

Please Please Please help with this math problem

Answer 1

1. The revenue as a function of x is equal to -x²/20 + 920x.
2. The profit as a function of x is equal to -x²/20 + 840x - 6000.
3. The value of x which maximizes profit is 8,400 and the maximum profit is $3,522,000.
4. The price to be charged to maximize profit is $500.

How to express the revenue as a function of x?

Based on the information provided, the cost function, C(x) is given by 80x + 6000 while the demand function, P(x) is given by -1/20(x) + 920.

Mathematically, the revenue can be calculated by using the following expression:

R(x) = x × P(x)

Revenue, R(x) = x(-1/20(x) + 920)

Revenue, R(x) = x(-x/20 + 920)

Revenue, R(x) = -x²/20 + 920x.

Expressing the profit as a function of x, we have:

Profit = Revenue - Cost

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

P(x) = -x²/20 + 920x - (80x + 6000)

P(x) = -x²/20 + 840x - 6000.

For the value of x which maximizes profit, we would differentiate the profit function with respect to x:

P(x) = -x²/20 + 840x - 6000

P'(x) = -x/10 + 840

x/10 = 840

x = 840 × 10

x = 8,400.

For the maximum profit, we have:

P(x) = -x²/20 + 840x - 6000

P(8400) = -(8400)²/20 + 840(8400) - 6000

P(8400) = -3,528,000 + 7,056,000 - 6000

P(8400) = $3,522,000.

Lastly, we would calculate the price to be charged in order to maximize profit is given by:

P(x) = -1/20(x) + 920

P(x) = -1/20(8400) + 920

P(x) = -420 + 920

P(x) = $500.

Read more on maximized profit here: brainly.com/question/13800671

#SPJ1