- The revenue as a function of x is equal to -x²/20 + 920x.
- The profit as a function of x is equal to -x²/20 + 840x - 6000.
- The value of x which maximizes profit is 8,400 and the maximum profit is $3,522,000.
- 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: https://brainly.com/question/13800671
#SPJ1
