[SOLVED] Price demand exquation

**snappyslow22** · Jun 26th 2009, 09:16 AM

I am having problems with the following problem that relates p(price) and demand (x) with the equation:

$ x^2+2xp+25p^2=74,500 $

1. If the price is increasing at a rate of $ 2 per mo. when the price is $ 30; how do I find the rate of change of the demand?

I understand that I need to implicitly dif. $(2x*x'+2xp'+2px'+50pp')$ but I'm not sure what to plug in after this step...

2. How do I find the rate of change of the price when demand is decreasing 6 units per mo when demand is 150?

Any help would be greatly appreciated

**calc101** · Jun 26th 2009, 09:31 AM

You plug in the following:

p' = $2
p = $30

2. You implicitly differentiate, and then plug in:
x' = 6
and equate the equation to: 150

I am not certain of my answers.

**snappyslow22** · Jun 26th 2009, 09:36 AM

so for 1) I would end up with

$ 2x*x'+2x(2)+2(30)+50(3)(2)=0 $
$ 2x*x'+4x+360=0 $

what do I do with the x'?

**borulce** · Jun 26th 2009, 09:49 AM

you can plug in the first equation for p=$30 then you find x.
with this you can solve the last equation you have found above.

**snappyslow22** · Jun 26th 2009, 10:04 AM

I'm confused, can you demonstrate what you mean?

**Amer** · Jun 26th 2009, 10:07 AM

Originally Posted by snappyslow22

I am having problems with the following problem that relates p(price) and demand (x) with the equation:

$ x^2+2xp+25p^2=74,500 $

1. If the price is increasing at a rate of $ 2 per mo. when the price is $ 30; how do I find the rate of change of the demand?

I understand that I need to implicitly dif. $(2x*x'+2xp'+2px'+50pp')$ but I'm not sure what to plug in after this step...

2. How do I find the rate of change of the price when demand is decreasing 6 units per mo when demand is 150?

Any help would be greatly appreciated

first find x when p=30

$x^2+60x+25(30)^2 = 74500$

$x^2+60x+25(900)-74500 =0$

$x^2+60x+100(225-745)=0$

$x^2+60x+100(-520)=0$

$x=\frac{-60\mp\sqrt{3600-(4(-52000))}}{2}$

$x=\frac{-60\mp 10\sqrt{36+4(520)}}{2}$

$x=\frac{-60\mp 10(2)\sqrt{529}}{2}$

$x=\frac{-60\mp 20(23)}{2}$

$x=\frac{20(-3+23)}{2}... or.....x=\frac{20(-3-23)}{2}$

you find the value of x when p=30 now sub it in the formula that you mention before you have the value of x and p and p' you can find x'

**borulce** · Jun 26th 2009, 10:17 AM

x^2+2xp+25p^2=74,500
for p=30 you can find x from the equation when the price is increasing at a rate of $ 2 per mo.

last equation you found 2x*x'+4x+360=0 now you have x and you can find the rate of change of the demand namely x'
I hope I could make it clear.

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