Q. 69 - Precalculus Enhanced with Graphing Utilities

Question

The price $p$, in dollars, of a certain product and the quantity $x$ sold obey the demand equation $p=-\frac{1}{4}x+100, 0\le x\le 400.$

Suppose that the cost $C$, in dollars, of producing $x$ units is $C=\frac{\sqrt{x}}{25}+600.$ Assuming that all items produced are sold, find the cost $C$ as a function of the price $p$. 

Step-by-Step Solution

Answer

Hence, $C\left(p\right)=\frac{\sqrt{-4p+400}}{25}+600.$

Step 1. Given Information

Given that $p=-\frac{1}{4}x+100, 0\le x\le 400$ and $C=\frac{\sqrt{x}}{25}+600.$

Step 2. Solution

Here, $p=-\frac{1}{4}x+100, 0\le x\le 400$and $C=\frac{\sqrt{x}}{25}+600.$

We have to find the cost $C$ as a function of $p.$

Now,

$$\begin{gathered} p=-\frac{1}{4}x+100, 0\le x\le 400 \\ p-100=-\frac{1}{4}x \\ 4(p-100)=-x \\ x\left(p\right)=-4(p-400) \\ x\left(p\right)=400-4p, 0\le p\le 100. \end{gathered}$$

$C$ as a function of price $p$ is $C\circ x\left(p\right).$

So,

$$\begin{gathered} C\circ x\left(p\right)=C\left(x\right(p\left)\right) \\ C\circ x\left(p\right)=\frac{\sqrt{x\left(p\right)}}{25}+600 \\ C\circ x\left(p\right)=\frac{\sqrt{400-4p}}{25}+600 \\ C\left(p\right)=\frac{\sqrt{-4p+400}}{25}+600. \end{gathered}$$

Step 3. Final answer

Hence, $C\left(p\right)=\frac{\sqrt{-4p+400}}{25}+600.$