Answer:
Please check the equations provided. I'm adding this partial answer, but it is incorrect, and I don't understand why.
Explanation:
We could solve this either by graphing, or by differentiation to find the maximum profit. We'll do both.
The revenue equation, R=1800x-x^2 , tells us the the unit price is 1800 (from the "1800x") and that the revenue, R, falls as a function of -x^2. See the attached graph (Revenue). The term -x^2 forms an inverted parabola.
The Cost function, C = 400 + 1300x, tells us there is a fixed cost of 400, and a per unit cost of 1300.
Profit is the Revenue minus the Cost.
R - C = Profit
R = 1800x-x^2
C = 400 + 1300x
R - C = (1800x-x^2) - (400 + 1300x) = Profit
Profit = -x^2 - 1300x - 400 - 1300x
Profit = -x^2 -2600x - 400
We have the Revenue, Cost, and Profit equations, which I've graphed on the attached graph. They make no sense to me. The -x^2 term is suspect, as it forces the revenue curve to be an inverted parabola.
One could also take the first derivative of the Profit equation, and set it equal to zero, which will be the slope of the line at its maximum.
Profit' = -2x -2600
0 = -2x - 2600
2x = -2600
x = -1300
It doesn't make sense that the most profit would accrue at -1300 units. I hope someone can explain my error.
We can graph the profit and look for the maximum, or we can take the first derivative and set it = 0, the slope of the apex of the curve.
Graphing
See the attached Profit graph.
First Derivative
Profit = -x^2 + 1300x - 400
Profit' = -2x + 1300
0 = -2x + 1300
2x = 1300
x = 650
650 units will profit the maximum profit.
The maximum profit would therefore be:
Profit = -x^2 + 1300x - 400 for x = 650
Profit = -(650)^2 + 1300(650) - 400