Answer:
Therefore profit function is increasing on the interval (0,3200)
Step-by-step 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)
Answer:
The is profit function is increasing on (0, 3200).
Step-by-step 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)
