Solve the problem.
-Assume that the profit generated by a product is given by $\mathrm { P } ( \mathrm { x } ) = 4 \sqrt { \mathrm { x } }$, where $x$ is the number of units sold. If the profit keeps changing at a rate of $\$ 400$ per month, then how fast are the sales changing when the number of units sold is 2000 ? (Round your answer to the nearest dollar per month.)
A) $\$ 4472 /$ month
B) $\$ 143,108 /$ month
C) $\$ 4 /$ month
D) $\$ 8944 /$ month

Question

Quizplus · Accepted Answer

We need to find how fast the sales are changing when the number of units sold is 2000. This means we need to find the derivative of the profit function with respect to the number of units sold.

$$\frac{\mathrm{dP}}{\mathrm{d}x} = 2\sqrt{x}$$

We know that the profit is changing at a rate of $\$400$ per month, so we can set up the following equation:

$$\frac{\mathrm{dP}}{\mathrm{d}t} = 400$$

We can relate this to the rate of change of the number of units sold by using the chain rule:

$$\frac{\mathrm{dP}}{\mathrm{d}t} = \frac{\mathrm{dP}}{\mathrm{d}x} \cdot \frac{\mathrm{d}x}{\mathrm{d}t}$$

Substituting the values we have:

\begin{align*}\frac{\mathrm{dP}}{\mathrm{d}t} &= \frac{\mathrm{dP}}{\mathrm{d}x} \cdot \frac{\mathrm{d}x}{\mathrm{d}t}\ 400 &= 2\sqrt{x} \cdot \frac{\mathrm{d}x}{\mathrm{d}t}\end{align*}

Now we can substitute $x= 2000$ to find the rate of change of the number of units sold:

$$400 = 2\sqrt{2000} \cdot \frac{\mathrm{d}x}{\mathrm{d}t}$$

Simplifying this:

$$200 = \frac{\mathrm{d}x}{\mathrm{d}t}$$

Therefore, the sales are changing at a rate of $\$200$ per month. Round this to the nearest dollar per month, we get $\$200$, which is closest to answer choice D ($\$8944/$month).

Solve the Problem P(x)=4x\mathrm { P } ( \mathrm { x } ) = 4 \sqrt { \mathrm { x } }P(x)=4x​

Solve the Problem $\mathrm { P } ( \mathrm { x } ) = 4 \sqrt { \mathrm { x } }$