Let's break down the problem step-by-step.
### Step 1: Define the Given Information
1. Revenue Function: The revenue [tex]\( R \)[/tex] from selling [tex]\( x \)[/tex] units is given by:
[tex]\[
R = 1800x - 3x^2
\][/tex]
2. Sales Rate: Sales are increasing at a rate of 25 units per day. Mathematically, this can be written as:
[tex]\[
\frac{dx}{dt} = 25 \, \text{units/day}
\][/tex]
3. Number of Units sold: We are interested in the situation when [tex]\( x = 210 \)[/tex].
### Step 2: Derive the Rate of Change of Revenue with Respect to Units Sold
We need to find how the revenue [tex]\( R \)[/tex] changes with respect to time [tex]\( t \)[/tex], which is expressed as [tex]\(\frac{dR}{dt}\)[/tex]. To do this, we use the chain rule of differentiation:
[tex]\[
\frac{dR}{dt} = \frac{dR}{dx} \cdot \frac{dx}{dt}
\][/tex]
### Step 3: Find [tex]\(\frac{dR}{dx}\)[/tex]
First, we need to find the derivative of the revenue function [tex]\( R \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[
\frac{dR}{dx} = \frac{d}{dx}\left(1800x - 3x^2\right)
\][/tex]
Differentiating term by term, we get:
[tex]\[
\frac{dR}{dx} = 1800 - 6x
\][/tex]
### Step 4: Evaluate [tex]\(\frac{dR}{dx}\)[/tex] at [tex]\( x = 210 \)[/tex]
Substitute [tex]\( x = 210 \)[/tex] into [tex]\(\frac{dR}{dx}\)[/tex]:
[tex]\[
\frac{dR}{dx} \bigg|_{x=210} = 1800 - 6(210)
\][/tex]
Now, calculate the value:
[tex]\[
\frac{dR}{dx} = 1800 - 1260 = 540
\][/tex]
### Step 5: Find [tex]\(\frac{dR}{dt}\)[/tex]
Using the chain rule, multiply [tex]\(\frac{dR}{dx}\)[/tex] by [tex]\(\frac{dx}{dt}\)[/tex]:
[tex]\[
\frac{dR}{dt} = \left(1800 - 6x\right) \cdot \frac{dx}{dt}
\][/tex]
Substitute [tex]\( x = 210 \)[/tex] and [tex]\(\frac{dx}{dt} = 25 \)[/tex]:
[tex]\[
\frac{dR}{dt} = 540 \cdot 25
\][/tex]
Calculate the value:
[tex]\[
\frac{dR}{dt} = 13500
\][/tex]
### Final Answer
The revenue is increasing at a rate of [tex]$\boxed{13500}$[/tex] dollars per day when 210 units have been sold.
