Find the price that will maximize the profit

Question

Given the functions for price (p) and cost (C):

$p=75-.1\sqrt{x}$

$C=30x+400$

Find the price that will maximize the profit.

---

Solution:

I will denote profit by $P$. Then $P=R-C$ where $R$ is the revenue, and $R=xp$, the price function multiplied by $x$. We need to figure out what the function for profit is, find the value of $x$ that maximizes it, and then plug that value of $x$ into our price function. This will give us the price that maximizes the profit.

So lets find the value of $x$ that maximizes $P$ by finding the critical points to $P$:

$P=xp-C=x(75-.1\sqrt{x})-(30x+400)$

$=75x-\frac{x^{\frac{3}{2}}}{10}-30x-400$

$=45x-\frac{x^{\frac{3}{2}}}{10}-400$

$\rightarrow P'=45-\frac{3}{2}\frac{x^{\frac{1}{2}}}{10}-0$

$=45-\frac{3x^{\frac{1}{2}}}{20}$

We are finding the critical points, so we have to set this to zero and solve for $x$:

$45-\frac{3x^{\frac{1}{2}}}{20}=0$

$\rightarrow 45=\frac{3x^{\frac{1}{2}}}{20}$

$\rightarrow \frac{900}{3}=x^{\frac{1}{2}}$

$\rightarrow 300^2=x$

Lets plug this value of $x$ into our price function $p$:

$p(300^2)=75-.1\sqrt{300^2}$

$=75-.1(300)$

$=75-30=45$

So the price that will maximizes the profit is $p=45$

Answer 1

As Andrei's question comment indicates, what you've done is OK, but I have just a couple of small issues. First, you have a minor mistake in your question text where you state

So lets find the value of $x$ that minimizes $P$ by finding the critical points to $P$:

The "minimizes" should be "maximizes" instead. Also, you should confirm you've found a maximum, as opposed to a minimum or a saddle point. Note that

$$P''(x) = -\frac{3}{40x^{1/2}} \implies P''(300^2) = -0.00025 \lt 0 \tag{1}\label{eq1A}$$

As stated in the second derivative test section of Wikipedia's Derivative test section, this confirms $x = 300^2$ gives a maximum profit.