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]
### 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]
