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AR15.COM
2/27/2008 12:57:24 PM EDT
How do I solve this problem, and also whats the answer?

When the price of a certain commodity is p dollars per unit, customers demand x hundred units of the commodity, where

x2 + 6p - p3=13

How fast is the demand x changing with respect to time when the price is $7 per unit and is changing at a rate of 0.23 dollars per month? Enter your answer to two decimal places.

2/27/2008 12:58:49 PM EDT
[#1]
Man....fuck math.
2/27/2008 1:05:23 PM EDT
[#2]
Easy one, .87=x , get both=p
2/27/2008 6:57:04 PM EDT
[#3]

Quoted:
How do I solve this problem, and also whats the answer?

When the price of a certain commodity is p dollars per unit, customers demand x hundred units of the commodity, where

x2 + 6p - p3=13

How fast is the demand x changing with respect to time when the price is $7 per unit and is changing at a rate of 0.23 dollars per month? Enter your answer to two decimal places.



The equation is a parabola.
Take its Derivative with respect to Time. Since the Price is Fluctuating it is not constant.

2xX' + 6p' - 3p^2p' = 0 (Chain Rule)

Solving for x:
2xX' = 3p^2p' - 6p'
--> X' = (3p^2p' - 6p')/2x

pluging in the values
p' = 0.23 dollars per month
p = 7 dollars
x = xhundred

we get:
X' = (3*7*7*0.23 - 6*0.23)/2xhundred
--> X' = (33.81 - 1.38)/2xhundred
-->X' = 0.16215/x

where x is some number (I already divided the numerator by 100).

Now we solve for X:
x2 + 6p - p3=13

since p = 7 dollars

So -->x^2 + 6*7 - 7^3 = 13
-->x^2 + 42 - 343 = 13
--> x^2 = 314
--> x = 17.72

We now plug this value for x into the equation:
X' = 0.16215/x
-->X' = 0.16215/17.72
-->X' = 0.00915 in Dollars per Month

Note: X' = dx/dt and p' = dp/dt