Question

A manufacturer sells video games with the following cost and revenue functions (in dollars) where x is the number of games sold, for 0 less than or equal to x less than or equal to 3300.
C(x)= 0.32x^2-0.00004x^3
R(x)= 0.848x^2-0.0002x^3

Determine the intervals on which the profit function is increasing.

Answer 1

Answer:

Therefore the profit interval is (0,2200).

Step-by-step explanation:

Given cost and revenue function are respectively,

$C(x)= 0.32x^2-0.00004x^3$

and

$R(x)= 0.848x^2-0.0002x^3$

where x is the number of game sold, $0\leq x\leq 3300$.

Therefore the profit function is

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

     $=0.848x^2-0.0002x^3-(0.32x^2-0.00004x^3)$

     $=0.848x^2-0.0002x^3-0.32x^2+0.00004x^3$

     $=0.528x^2 -0.00016x^3$

To determine the profit interval where profit function increasing, we have to find the critical point of P(x). We set the first order derivative equal to 0.

Therefore,

$P'(x) = 1.056 x - 0.00048 x^2 =0$

⇒x(1.056-0.00048 x) =0

$\Rightarrow x = 0$    or,     $x= \frac{1.056}{0.00048} =2200$

Now choose two point one is less than 2200 and other is greater than of 2200.

We choose x= 1000 and x= 2000

Now

$P'(1000) = 1.056 \times 1000- 0.00048 (1000)^2 = 576$  >0

$P'(2000) = 1.056 \times 2000- 0.00048 (2000)^2 = -1152$ <0

Therefore the profit interval is (0,2200).