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