Answer:
 Maximum profit: 2000
Step-by-step explanation:
You want the maximum profit per week that can be obtained by production of x units at an average cost of c(x) = 500/x +1500 if the weekly demand equation is p = 1600 -x.
Cost
The total cost is the product of average cost and the number of units produced:
 C(x) = x·c(x) = 500 +1500x
Revenue
The revenue from sale of x units is the product of that number and their price:
 R(x) = px = (1600 -x)(x)
Profit
The profit from the sale of x units is the difference between revenue and cost:
 P(x) = R(x) -C(x)
 P(x) = (1600x -x²) -(1500x +500)
 P(x) = -x² +100x -500
The maximum profit will be had when the derivative of this is zero:
 P'(x) = 0 = -2x +100
 x = 100/2 = 50
That value of profit is ...
 P(x) = (-x +100)x -500
 P(50) = (-50 +100)(50) -500 = 2500 -500 = 2000
The maximum profit per week that can be achieved is 2000 from sale of 50 units.
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Additional comment
We can solve this without the need for derivatives by writing the profit equation in vertex form:
 P(x) = -(x -50)² +2000
This has a maximum of 2000 at x=50.
