Question

A manufacturer sells video games with the following cost and revenue functions​ (in dollars), where x is the number of games sold. Determine the​ interval(s) on which the profit function is increasing. Upper C (x )equals 0.17 x squared minus 0.00016 x cubed Upper R (x )equals 0.362 x squared minus 0.0002 x cubed

Answer 1

The is profit function is increasing on (0, 3200).

Explanation:

Given

C(x) = 0.17x² - 0.00016x³

R(x) = 0.362x² - 0.0002x³

The profit function is given by

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

P(x) = (0.362x² - 0.0002x³) - (0.17x² - 0.00016x³)

P(x) = 0.362x² - 0.0002x³ - 0.17x² + 0.00016x³

P(x) = 0.192x² - 0.00004x³

The derivative of the profit function is given by

P'(x) = 0.384x - 0.00012x²

Determine the critical number of P(x) to get interval where the profit function is increasing,

Set P'(x) = 0

0.384x - 0.00012x² = 0

x(0.384 - 0.00012x) = 0

x = 0 or 0.384 - 0.00012x = 0

0.384 - 0.00012x = 0

x = 0.384/0.00012 = 3200

Therefore the is profit function is increasing on (0, 3200)

Answer 2

Therefore profit function is increasing on the interval (0,3200)

Explanation:

The cost function of the manufacturer C(x) is given as:

[TeX]C(x)= 0.17x^{2}-0.00016x^{3}[/TeX]

The Revenue function is also given as:

[TeX]R(x)= 0.362x^{2}-0.0002x^{3}[/TeX]

Profit=Revenue-Cost

Therefore:

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

[TeX]= 0.362x^{2}-0.0002x^{3}-[0.17x^{2}-0.00016x^{3}][/TeX]

[TeX]= 0.362x^{2}-0.0002x^{3}-0.17x^{2}+0.00016x^{3}[/TeX]

The Profit Function, [TeX]P(x)= 0.192x^{2}-0.00004x^{3}[/TeX]

To determine the point where the function is increasing, we take the derivative and examine it's critical points.

The derivative of the profit function is:

[TeX]P^{'}(x)= 0.384x-0.00012x^{2}[/TeX]

Set the derivative equal to zero.

[TeX]0.384x-0.00012x^{2}=0[/TeX]

[TeX]x(0.384x-0.00012x)=0[/TeX]

x=0 or [TeX]0.384-0.00012x=0[/TeX]

x=0 or [TeX]0.384=0.00012x[/TeX]

x=0 or x=3200

Now let's choose 2000 and 4000 as test points.

[TeX]P^{'}(2000)= 0.384(2000)-0.00012(2000)^{2}=288[/TeX]

[TeX]P^{'}(4000)= 0.384(4000)-0.00012(4000)^{2}=-384[/TeX]

Therefore profit function is increasing on the interval (0,3200)