4.5 derivatives and graphs\n1. suppose that when your company sells shoes for $x per pair, your company will…

4.5 derivatives and graphs\n1. suppose that when your company sells shoes for $x per pair, your company will make a total profit of ( p(x)=-100x^{2}+10200x - 20000 ). find where ( p(x) ) is increasing and where it is decreasing. in order to maximize total profit, how much should your company charge per pair of shoes?

4.5 derivatives and graphs\n1. suppose that when your company sells shoes for $x per pair, your company will make a total profit of ( p(x)=-100x^{2}+10200x - 20000 ). find where ( p(x) ) is increasing and where it is decreasing. in order to maximize total profit, how much should your company charge per pair of shoes?

Answer

Explanation:

Step1: Find the derivative of (P(x))

The derivative of (P(x)=- 100x^{2}+10200x - 20000) using the power rule ((x^{n})^\prime=nx^{n - 1}) is (P^\prime(x)=-200x + 10200).

Step2: Find the critical points

Set (P^\prime(x)=0), so (-200x + 10200=0). Solving for (x): [ \begin{align*} -200x&=-10200\ x& = 51 \end{align*} ]

Step3: Determine the intervals of increase and decrease

We consider the intervals ((-\infty,51)) and ((51,\infty)). Take a test - point in ((-\infty,51)), say (x = 0). Then (P^\prime(0)=-200\times0+10200 = 10200>0). So (P(x)) is increasing on the interval ((-\infty,51)). Take a test - point in ((51,\infty)), say (x = 100). Then (P^\prime(100)=-200\times100 + 10200=-20000 + 10200=-9800<0). So (P(x)) is decreasing on the interval ((51,\infty)).

Since the function (P(x)) changes from increasing to decreasing at (x = 51), by the first - derivative test, the function (P(x)) has a maximum at (x = 51).

Answer:

(P(x)) is increasing on the interval ((-\infty,51)) and decreasing on the interval ((51,\infty)). To maximize the total profit, the company should charge ($51) per pair of shoes.