<p>The wording of the question could be improved.  Assuming the question is asking for the rate of change of profit per year when demand is 50 units and has a rate of change of 3 units per year, you can apply the chain rule where you left off:</p>
<p><span class="math-container">$$\begin{align}
U&amp;=x\left({400-x \over 3}\right)-(200+60x)\\
\implies {\partial U \over \partial t}&amp;={\partial U \over \partial x}{\partial x \over \partial t}\\
&amp;=\left({400 \over 3}-{2\over 3}x-60\right)(3).
\end{align}$$</span></p>
<p>When <span class="math-container">$x=50$</span>, we thus get
<span class="math-container">$$\partial U/ \partial t ={\\\$} 120/\text{year}$$</span>.</p>

CC BY-SA 4.0
2023-04-05T23:48:51.233Z
2023-04-05T23:48:51.233Z
Golden_Ratio
<p>The demand equation for a company's product is <span class="math-container">$3p + x = 400$</span>, where <span class="math-container">$x(t)$</span> units can be sold at a price of <span class="math-container">$p$</span> each. If the demand increases at a rate of 3 units per year when the demand reaches 50 units, the fixed costs are 200, and the variable cost per unit is 60. What is the rate of change of the profit per year?</p>
<p>I can't understand this statement. I will put what I have understood:</p>
<p>Profit: <span class="math-container">$U$</span></p>
<p>Total Cost: <span class="math-container">$C$</span></p>
<p>Total revenue: <span class="math-container">$R$</span></p>
<p><span class="math-container">$U=R-C$</span></p>
<p><span class="math-container">$R=px$</span></p>
<p><span class="math-container">$C=200+60x$</span></p>
<p><span class="math-container">$U=R-C$</span></p>
<p><span class="math-container">$U=px-(200+60x)$</span></p>
<p>What I can't understand is the phrase &quot;If the demand increases at a rate of 3 units per year when the demand reaches 50 units&quot;</p>
<p>This means that <span class="math-container">$dx/dt=3$</span> then <span class="math-container">$\int_{50}^{x(t)} dx=2\int_{t=0\text{ or }1 \text{?}}^{t}dt$</span>?</p>
<p>Thanks.</p>

<supply-and-demand><demand><profits>
Finding the profit from a demand equation
eraldcoil
# Finding the profit from a demand equation

(Asked by: username_0 on 2023-04-02)

The demand equation for a company's product is $3p + x = 400$, where $x(t)$ units can be sold at a price of $p$ each. If the demand increases at a rate of 3 units per year when the demand reaches 50 units, the fixed costs are 200, and the variable cost per unit is 60. What is the rate of change of the profit per year?

I can't understand this statement. I will put what I have understood:

Profit: $U$

Total Cost: $C$

Total revenue: $R$

$U=R-C$

$R=px$

$C=200+60x$

$U=R-C$

$U=px-(200+60x)$

What I can't understand is the phrase "If the demand increases at a rate of 3 units per year when the demand reaches 50 units"

This means that $dx/dt=3$ then $\int\_{50}^{x(t)} dx=2\int\_{t=0\text{ or }1 \text{?}}^{t}dt$?

Thanks.

**Tags:** supply-and-demand, demand, profits

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## 1 answers

### Answer by username_1 on 2023-04-05. Score: 1 | Accepted answer

The wording of the question could be improved. Assuming the question is asking for the rate of change of profit per year when demand is 50 units and has a rate of change of 3 units per year, you can apply the chain rule where you left off:

$$\begin{align}
U&=x\left({400-x \over 3}\right)-(200+60x)\\
\implies {\partial U \over \partial t}&={\partial U \over \partial x}{\partial x \over \partial t}\\
&=\left({400 \over 3}-{2\over 3}x-60\right)(3).
\end{align}$$

When $x=50$, we thus get
$$\partial U/ \partial t ={\\\$} 120/\text{year}$$.


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