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Mathematics 13 Online
OpenStudy (anonymous):

please help.. how to solve this problem? When x units of a certain luxury commodity are produced, they can all be sold at a price of p thousand dollars per unit, where p=-6x+100. A. Express the revenue R(x) as a function of x. B. How much revenue is obtained when x=15 units are produced and sold?

OpenStudy (agent0smith):

Revenue is selling price multiplied by number of units sold.

OpenStudy (agent0smith):

Selling price is given. Units is x.

OpenStudy (anonymous):

how much is the Revenue or the selling price there??

OpenStudy (agent0smith):

They tell you the selling price in the question... it's p. Multiply it by x. That gives revenue.

OpenStudy (anonymous):

p=-6x+100 ?

OpenStudy (agent0smith):

Don't solve it. Multiply p, which is -6x+100, by x. x(-6x+100)

OpenStudy (agent0smith):

R(x) = x(-6x+100)

OpenStudy (agent0smith):

Revenue is price per unit (which is -6x+100) multiplied by number of units sold (which is x) So your revenue is R(x) = x(-6x+100)

OpenStudy (anonymous):

Let each unit = x\[1 unit = x \] Let number of units sold = p\[1000$/unit = p \] Note* = The money we make is in 1000s of dollars. Let the Revenue we make be \[x *(p)\] we know p \[p = -6x +100\] \[Hence, R = x(p) = x(-6x +100)\]

OpenStudy (agent0smith):

I'm sure you can at least try to simplify it on your own @phopXD

OpenStudy (anonymous):

Agent0 completed part a. just evaluate it at x = 15, multiply the answer by 1000$.

OpenStudy (anonymous):

ok..

OpenStudy (anonymous):

thanks mebs :)

OpenStudy (anonymous):

To evaluate at x = 15 \[R(15) = (15)(-6(15) +100) \]

OpenStudy (anonymous):

GET OUT 3:)

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