change in price of a comodity (derivative)

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change in price of a comodity (derivative)

Postby Integr88 » Fri 08 April, 2011 18:24

The price of a commodity is increasing at $3 per month and the price function is `p(x) = 100 - 0.002x^2` . Sales are 9,407.
How quickly is revenue changing?
Give your answer to the nearest whole number.

If` p(x) = (971 - 3x^2)` find `(dR)/dt` if `dx/dt = 7` when x = 2.4. Give the exact answer.
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Re: change in price of a comodity (derivative)

Postby barnamah » Fri 08 April, 2011 18:36

fort the first question of the sales `p(x) = 100 - 0.002x^2` find the rate of change when sales are at 9407
First find the derivatives of the `p(x) = 100 - 0.002x^2`
`(dp)/dx=2(-0.002)x^(2-1)=-0.004x`
so we have our equation rate as `p'=0.004x` to find the rate when x= 9407 we sub it into the equation
`p'=-0.004(9407)=-37.638` so it is decreasing

Question 2
for the 2nd question ` p(x) = 971 - 3x^2` find `(dR)/dt` if `dx/dt = 7` when x = 2.4.
the question asks for `(dR)/dt` and it is `(dR)/dt=(dR)/dx*(dx)/dt`
We have to find `(dR)/dx` which means take the derivative of the ` p(x) = 971 - 3x^2`
`(dR)/dx=-6x`
and we have `dx/dt = 7`. in our formula `(dR)/dt=(dR)/dx*(dx)/dt` sub in the values and we have:
`(dR)/dt=-6x*7=-42x`

so `(dR)/dt=-42x`
Explore and know. That is asked.
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