Rate of change - Implicit differentiation

Question

A price $p$ (in dollars) and demand $x$ for a product are related by $$\left(2x^2\right)-2xp+50p^2 = 20600$$ If the price is increasing at a rate of $2$ dollars per month when the price is $20$ dollars, find the rate of change of the demand.

I was a little confused on how to proceed with this question. Am I supposed to use implicit differentiation (with the $x$ serving the same purpose as a $y$) and then find the derivative of $x$?

This is the implicit differentiation I tried:

$$4x\frac{dx}{dp}-2\frac{dx}{dp}+100p = 0$$

$$4x\frac{dx}{dp}-2\frac{dx}{dp} = -100p$$

$$\frac{dx}{dp}(4x-2) = -100p$$

$$\frac{dx}{dp} = \frac{-100p}{4x-2}$$

I believe this is the derivative I am looking for (thought not entirely sure) but I am not sure what values of $p$ and $x$ to input, as I am supposed to get a numerical final answer.

Any help?

Answer 1

Guide:

$$2x^2-2xp+50p^2=20600$$

Differentiate with respect to $t$.

We have information about $x, p, \frac{dp}{dt}$, and you are interested in finding $\frac{dx}{dt}$.

Note that when $p=20$, you can compute the corresponding $x$.

Answer 2

The question is asking you to compute $\left.\dfrac{dx}{dt}\right|_{p=20}$ given that $\left.\dfrac{dp}{dt}\right|_{p=20}=2$ (where $t$ is time in months). The giveaway of that was the keyword “rate” and the unit “dollars per month.”

The first thing we need to do is implicitly differentiate so that our equation matches what we’ve been given.

$$\begin{align} 2x^2 - 2xp + 50p^2 &= 20600 \\ \frac{d}{dt}\left( 2x^2 - 2xp + 50p^2\right) &= \frac{d}{dt}(20600) \\ \frac{d}{dt}\left(2x^2\right) - \frac{d}{dt}(2xp) + \frac{d}{dt}\left(50p^2\right) &= 0 \\ 2\cdot2x\cdot\frac{dx}{dt} -2 \left( \frac{dx}{dt}p + x\frac{dp}{dt}\right) + 50\cdot2p\cdot\frac{dp}{dt} &= 0 \\ 4x\frac{dx}{dt} - 2p\frac{dx}{dt} - 2x\frac{dp}{dt} + 100p\frac{dp}{dt} &= 0 \\ \end{align}$$

Notice that before we can plug in what we’ve been given, we’re going to need to find $x$ when $p=20$.

$$\begin{align} 2x^2 - 2x(20) + 50(20)^2 &= 20600 \\ 2x^2 - 40x - 600 &= 0 \\ x^2 - 20x - 300 &= 0 \\ (x-30)(x+10) &= 0 \\ \end{align}$$

Let’s assume demand isn’t going to be negative, implying $x=30$ when $p=20$.

Next, we need to evaluate this expression by substituting the values that correspond to the situation. (I’m going to omit the evaluation bar for ease of notation.)

$$\begin{align} 4(30)\frac{dx}{dt} - 2(20)\frac{dx}{dt} - 4(30)(2) + 100(20)(2) &= 0 \\ 120\frac{dx}{dt} - 40\frac{dx}{dt} - 120 + 4000 &= 0 \\ 80\frac{dx}{dt} + 3880 &= 0 \\ 80\frac{dx}{dt} &= -3880 \\ \frac{dx}{dt} &= -48.5 \\ \end{align}$$

Don’t be alarmed by the negative! That just means that the demand is decreasing by $48.5$ per month.

Answer 3

Blindly follow the rule that $\boxed{{\rm d}f = \frac{\partial f}{\partial x} {\rm d}x + \frac{\partial f}{\partial p} {\rm d}p}$.

In this case:

$$ 2x^2-2xp+50p^2 = 20600 $$

$$ {\rm d} \left( 2x^2-2xp+50p^2 = 20600 \right) $$

$$ \frac{\partial}{\partial x} \left( 2x^2-2xp+50p^2 = 20600 \right) {\rm d}x + \frac{\partial}{\partial p} \left( 2x^2-2xp+50p^2 = 20600 \right) {\rm d}p$$

$$ \left( 4x -2p \right) {\rm d}x + \left(100p - 2x \right) {\rm d} p = 0 $$

$$ \frac{{\rm d}x}{{\rm d}p} = \frac{x-50 p}{2 x -p} $$