Question

At a price of $p the demand x per month (in multiples of 100) for a new piece of software is given by x 2 + 2xp + 4p 2 = 5200. Because of its popularity, the manufacturer is increasing the price at a rate of 70 /c per month. Find the corresponding rate of decrease in demand for the software when the software costs $10.

Answer 1

Answer:

The rate of decrease in demand for the software when the software costs $10 is -100

Step-by-step explanation:

Given the function of price $p the demand x per month as,

$x^{2}+2xp+4p^{2}=5200$

Also given that, the price is increasing at the rate of 70 dollar per month.

$\therefore \dfrac{dp}{dt}=70$.

To find rate of decrease in demand, differentiate the given function with respect to t as follows,

$\dfrac{d}{dt}\left(x^2+2xp+4p^2\right)=\dfrac{d}{dt}\left(5200\right)$

Applying sum rule and constant rule of derivative,

$\dfrac{d}{dt}\left(x^2\right)+\dfrac{d}{dt}\left(2xp\right)+\dfrac{d}{dt}\left(4p^2\right)=0$

Applying constant multiple rule of derivative,

$\dfrac{d}{dt}\left(x^2\right)+2\dfrac{d}{dt}\left(xp\right)+4\dfrac{d}{dt}\left(p^2\right)=0$

Applying power rule and product rule of derivative,

$2x^{2-1}\dfrac{dx}{dt}+2\left(x\dfrac{dp}{dt}+p\dfrac{dx}{dt}\right)+4\left(2p^{2-1}\right)\dfrac{dp}{dt}=0$

Simplifying,

$2x\dfrac{dx}{dt}+2\left(x\dfrac{dp}{dt}+p\dfrac{dx}{dt}\right)+8p\dfrac{dp}{dt}=0$

Now to find the value of x, substitute the value of p=$10 in given equation.

$x^{2}+2x\left(10\right)+4\left(10\right)^{2}=5200$

$x^{2}+20x+400=5200$

Subtracting 5200 from both sides,

$x^{2}+20x+400-5200=0$

$x^{2}+20x-4800=0$

To find the value of x, split the middle terms such that product of two number is 4800 and whose difference is 20.

Therefore the numbers are 80 and -60.

$x^{2}+80x-60x-4800=0$

Now factor out x from $x^{2}+80x$ and 60 from $60x-4800$

$x\left(x+80\right)-60\left(x+80\right)=0$

Factor out common term x+80,

$\left(x+80\right)\left(x-60\right)=0$

By using zero factor principle,

$\left(x+80\right)=0$ and $\left(x-60\right)=0$

$x=-80$ and $x=60$

Since demand x can never be negative, so x = 60.

Now,

$2x\dfrac{dx}{dt}+2\left(x\dfrac{dp}{dt}+p\dfrac{dx}{dt}\right)+8p\dfrac{dp}{dt}=0$

Substituting the value.

$2\left(60\right)\dfrac{dx}{dt}+2\left(60\left(70\right)+10\dfrac{dx}{dt}\right)+8\left(10\right)\left(70\right)=0$

Simplifying,

$120\dfrac{dx}{dt}+2\left(4200+10\dfrac{dx}{dt}\right)+5600=0$

$120\dfrac{dx}{dt}+8400+20\dfrac{dx}{dt}+5600=0$

Combining common term,

$140\dfrac{dx}{dt}+14000=0$

Subtracting 14000 from both sides,

$140\dfrac{dx}{dt}=-14000$

Dividing 140 from both sides,

$\dfrac{dx}{dt}=-\dfrac{14000}{140}$

$\dfrac{dx}{dt}=-100$

Negative sign indicates that rate is decreasing.

Therefore, the rate of decrease in demand of software is -100

Answer 2

The rate of decrease in demand for the software when the software costs $ 10 is $-\frac{10}{7}$.

How to predict the expected decrease in demand due to increase on product prices

In this question we must predict decreases in demand due to price increases, this can be done by applying implicit differentiation. First, we look for the value of $x$ at initial conditions: ($p = 10$)

$x^{2}+2\cdot x\cdot p + 4\cdot p^{2} = 5200$   (1)

$x^{2}+20\cdot x -4800 = 0$  

$x = 60$

Now we differentiate (1) with respect to $p$ and clear $\frac{dx}{dp}$ and determine the resulting expression:

$2\cdot x \cdot \frac{dx}{dp} + 2\cdot p \cdot \frac{dx}{dp} +2\cdot x + 8\cdot p = 0$

$\frac{dx}{dp} = -\frac{x+4\cdot p}{x+p}$

If we know that $x = 60$ and $p = 10$, then the rate of decrease in demand is:

$\frac{dx}{dp} = - \frac{60+4\cdot (10)}{60 + 10}$

$\frac{dx}{dp} = -\frac{10}{7}$

The rate of decrease in demand for the software when the software costs $ 10 is $-\frac{10}{7}$.  $\blacksquare$

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