110quiz2solutions

# 110quiz2solutions - price times quantity So R x = px But we...

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Math 110 Quiz 2 Sept. 7, 2006 Name: 1. A tire company currently sells 3200 tires a month at a selling price of \$32. Past experience indicates that for each \$3 increase in selling price, the monthly sales will decrease by 500. (a) Determine an equation relating the selling price p and the number x of tires sold. (b) Determine a model (function) for the monthly revenue R ( x ). (c) Use a graphing utility to graph R ( x ). (d) Use a graphing utility to determine (to the nearest 10) the number of tires sold that will maximize revenue. (e) To the nearest \$0 . 10, what selling price p will maximize revenue? Solutions: (a) We recognize that the relationship between the price p and the number of tires sold x is linear . We consider the points (3200 , 32) and (2700 , 35), where the ﬁrst number represents x and the second represents price. The slope is then m = 32 - 35 3200 - 2700 = - 3 500 . Using p - p 1 = m ( x - x 1 ), we have p - 32 = - 3 500 ( x - 3200) (b) Remember that revenue is always

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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 ( x-3200) + 32 Substituting for p into the revenue function, we have R ( x ) = px = ±-3 500 ( x-3200) + 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 y-value 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 ( x-3200) + 32 =-3 500 (4270-3200) + 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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110quiz2solutions - price times quantity So R x = px But we...

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