Question

2 nur u A

Answer 1

P(x) = 10000 - 0.001x²

C(x) = 0.0002x² + 200x + 1000

a) Revenue function = R(x) = xP(x) = x(10000 - 0.001x²) = 10000x - 0.001x³

Solve for P(x) using below formula

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

= 10000x - 0.001x³ - (0.0002x² + 200x + 1000)

=  - 0.001x³ - 0.0002x² + 9800x - 1000

P(x) can be maximized by equating its first derivative to 0

Solving above equation we get

So at x = 1807.32 P(x) =?

$P(1807.32) = 10000-0.001\left(1807.32\right)^2 = 6733.59$

Therefore, the price that will maximize profit is 6733.59

b) Marginal profit at x = 1500

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

= 10000x - 0.001x³ - (0.0002x² + 200x + 1000)

=  - 0.001x³ - 0.0002x² + 9800x - 1000

Marginal profit is the derivative of the profit function, so take the derivative of P(x) and evaluate it at x = 1500.

So marginal profit = 3049.4

c) C(x) = 0.0002x² + 200x + 1000

the cost per unit U = C(x)/x

U = 0.0002x + 200 + 1000/x

So to find the minimum cost lets take deriavative of it

$U'=-\frac{1000}{x^2}+0.0002$

$U''=\frac{2000}{x^3}$

$U'=0\Rightarrow x=2236.06,x=-2236.06$

At x = 2236.06, U'' is positive So the function must be minimum