Question

4.5 derivatives and graphs

1. 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?

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 \(x\):

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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.