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Tom [10]
3 years ago
8

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. B

ecause 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.
Mathematics
1 answer:
WINSTONCH [101]3 years ago
6 0

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

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