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Unformatted text preview: price times quantity . So, R ( x ) = px . But we need to get the revenue as a function of x only. We solve for p in the equation we obtained in part (a): p =3 500 ( x3200) + 32 Substituting for p into the revenue function, we have R ( x ) = px = ±3 500 ( x3200) + 32 ² x (over for parts (c), (d) and (e)) 1 (c) The following is an acceptable graph of the revenue function: (d) Zooming and tracing on the maximum yvalue in the above graph, we see that x = 4270 maximizes revenue. This means 4270 tires sold per month will maximize monthly revenue. (e) We substitute the value we found for x into the expression we found for p in part (b): p =3 500 ( x3200) + 32 =3 500 (42703200) + 32 = 25 . 60 This represents the selling price the company should sell their tires at in order to maximize revenue. 2...
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 Spring '07
 carlson
 Math, Derivative, Revenue, Expression

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