1. [SOLVED] Price demand exquation

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

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

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. $\displaystyle (2x*x'+2xp'+2px'+50pp')$ 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

2. 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.

3. so for 1) I would end up with

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

what do I do with the x'?

4. you can plug in the first equation for p=$30 then you find x. with this you can solve the last equation you have found above. 5. I'm confused, can you demonstrate what you mean? 6. Originally Posted by snappyslow22 I am having problems with the following problem that relates p(price) and demand (x) with the equation:$\displaystyle
x^2+2xp+25p^2=74,500
$1. If the price is increasing at a rate of$ 2 per mo. when the price is $30; how do I find the rate of change of the demand? I understand that I need to implicitly dif.$\displaystyle (2x*x'+2xp'+2px'+50pp') $but I'm not sure what to plug in after this step... 2. How do I find the rate of change of the price when demand is decreasing 6 units per mo when demand is 150? Any help would be greatly appreciated first find x when p=30$\displaystyle x^2+60x+25(30)^2 = 74500\displaystyle x^2+60x+25(900)-74500 =0 \displaystyle x^2+60x+100(225-745)=0\displaystyle x^2+60x+100(-520)=0\displaystyle x=\frac{-60\mp\sqrt{3600-(4(-52000))}}{2}\displaystyle x=\frac{-60\mp 10\sqrt{36+4(520)}}{2}\displaystyle x=\frac{-60\mp 10(2)\sqrt{529}}{2}\displaystyle x=\frac{-60\mp 20(23)}{2}\displaystyle x=\frac{20(-3+23)}{2}... or.....x=\frac{20(-3-23)}{2}$you find the value of x when p=30 now sub it in the formula that you mention before you have the value of x and p and p' you can find x' 7. x^2+2xp+25p^2=74,500 for p=30 you can find x from the equation when the price is increasing at a rate of$ 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.