px-20p-6x-40 = 0, the rate of change of x is -8 crates/day.

[rafeeki92]

Given the equation
px-20p-6x-40 = 0, the rate of change of x is -8 crates/day.
Find the rate of p when x = 100.

 

[chrisr]

Re: AHH HELP PLEASE

The rate of change of p ?
differentiate with respect to time

d/dt(px-20p-6x-40) = 0.
(p)dx/dt + (x)dp/dt -(20)dp/dt -(6)dx/dt =0
-8p +(100)dp/dt-(20)dp/dt+48=0
(80)dp/dt=8p-48
dp/dt=(8p-48)/(80) = (p-6)/10 per day, whatever "p" is.

 

[BigGlenntheHeavy]

$\displaystyle Given: \ px-20p-6x-40 \ = \ 0 \ and \ \frac{dx}{dt} \ = \ -8: \ find \ \frac{dp}{dt} \ when \ x \ = \ 100.$

$\displaystyle p(x-20) \ = \ 6x+40, \ p \ = \ \frac{6x+40}{x-20}$

$\displaystyle \frac{dp}{dt} \ = \ \frac{(x-20)(6(dx/dt))-(6x+40)(dx/dt)}{(x-20)^{2}} \ = \ \frac{1280}{(x-20)^{2}}$

$\displaystyle Now, \ when \ x \ = \ 100, \ \frac{dp}{dt} \ \ = \ \frac{1280}{80^{2}} \ = \ \frac{1}{5}.$

 

[chrisr]

Also, as dp/dt = (p-6)/10,

you find p when x=100 by placing x=100 into px-20p-6x-40 = 0,
that allows you to solve in a second way for dp/dt,
by placing the resulting p into (p-6)/10.

 

[BigGlenntheHeavy]

$\displaystyle As \ always \ Chrisr, \ there \ is \ more \ than \ one \ way \ to \ skin \ a \ cat.$