A company selling widgets has found that the number of items sold [tex]\(x\)[/tex] depends upon the price [tex]\(p\)[/tex] at which they're sold, according to the equation [tex]\(x=\frac{90000}{\sqrt{4p+1}}\)[/tex].

Due to inflation and increasing health benefit costs, the company has been increasing the price by [tex]\(\$4\)[/tex] per month. Find the rate at which revenue is changing when the company is selling widgets at [tex]\(\$130\)[/tex] each.

[tex]\[
\begin{array}{l}
\text{First, take the derivative of } R = x \cdot p \text{ using the quotient rule. This gives us:} \\
d\frac{R}{dt} = \frac{(4p+1)^{\frac{1}{2}}(90000) - (90000p)\left(\frac{1}{2}\right)(4p+1)^{-\frac{1}{2}}(4)}{4p+1} \cdot \frac{dp}{dt}. \\
\text{Substitute } p = 130 \text{ and } \frac{dp}{dt} = 4 \text{ to find } d\frac{R}{dt}.
\end{array}
\][/tex]



Answer :

To approach this problem, we need to understand how the revenue changes with respect to time when the price is increasing. We'll go through it step-by-step:

### Step 1: Define the variables and given functions
- Number of items sold [tex]\( x \)[/tex]: Given by [tex]\( x = \frac{90000}{\sqrt{4p + 1}} \)[/tex]
- Price [tex]\( p \)[/tex]: Current price at which the widgets are sold.

### Step 2: Define the revenue function [tex]\( R \)[/tex]
Revenue [tex]\( R \)[/tex] is defined as:
[tex]\[ R = p \cdot x \][/tex]

Substitute [tex]\( x \)[/tex] into the revenue equation:
[tex]\[ R = p \cdot \frac{90000}{\sqrt{4p + 1}} \][/tex]

### Step 3: Simplify the revenue function
The revenue function becomes:
[tex]\[ R = \frac{90000p}{\sqrt{4p + 1}} \][/tex]

### Step 4: Differentiate the revenue function with respect to [tex]\( p \)[/tex]
To find the rate of change of revenue with respect to price, we need to differentiate [tex]\( R \)[/tex] with respect to [tex]\( p \)[/tex]:
[tex]\[ \frac{dR}{dp} = \frac{d}{dp} \left(\frac{90000p}{\sqrt{4p + 1}}\right) \][/tex]

Using the quotient rule for differentiation:
[tex]\[ \frac{d}{dp} \left(\frac{90000p}{\sqrt{4p + 1}}\right) = \frac{\left(\sqrt{4p + 1}\right)(90000) - (90000p)\left(\frac{1}{2}\right)(4p + 1)^{-\frac{1}{2}}(4)}{(4p + 1)} \][/tex]

### Step 5: Simplify the derivative
Combine the terms to simplify:
[tex]\[ \frac{dR}{dp} = \frac{90000\sqrt{4p + 1} - \frac{360000p}{2\sqrt{4p + 1}}}{4p + 1} \][/tex]
[tex]\[ \frac{dR}{dp} = \frac{90000\sqrt{4p + 1} - 180000p/\sqrt{4p + 1}}{4p + 1} \][/tex]

Combine under a common denominator:
[tex]\[ \frac{dR}{dp} = \frac{90000(4p + 1) - 180000p}{(4p + 1)\sqrt{4p + 1}} \][/tex]
[tex]\[ \frac{dR}{dp} = \frac{360000p + 90000 - 180000p}{(4p + 1)\sqrt{4p + 1}} \][/tex]
[tex]\[ \frac{dR}{dp} = \frac{180000p + 90000}{(4p + 1)\sqrt{4p + 1}} \][/tex]

### Step 6: Evaluate the derivative at [tex]\( p = 130 \)[/tex]
Now, we substitute [tex]\( p = 130 \)[/tex]:
[tex]\[ \frac{dR}{dp} \Bigg|_{p=130} = \frac{180000(130) + 90000}{(4(130) + 1)\sqrt{4(130) + 1}} \][/tex]
[tex]\[ = \frac{23400000 + 90000}{521 \cdot \sqrt{521}} \][/tex]

Combine the constants and simplify:
[tex]\[ \frac{dR}{dp} \Bigg|_{p=130} = \frac{23490000}{521 \cdot \sqrt{521}} \][/tex]

### Step 7: Use the chain rule to find [tex]\( \frac{dR}{dt} \)[/tex]
Given that the price increases by [tex]$4 per month (\frac{dp}{dt} = 4): \[ \frac{dR}{dt} = \frac{dR}{dp} \Bigg|_{p=130} \cdot \frac{dp}{dt} \] \[ \frac{dR}{dt} = \frac{23490000}{521 \cdot \sqrt{521}} \cdot 4 \] Simplify the result: \[ \frac{dR}{dt} = \frac{93960000 \cdot \sqrt{521}}{271441} \] Thus, the rate at which the revenue is changing when the company is selling the widgets at $[/tex]130 each is:

[tex]\[ \frac{93960000 \cdot \sqrt{521}}{271441} \, \text{dollars per month}\][/tex]