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ocduit itpscbtlayerHomework.aspx?homew. X ob Apps D Apple D Bing G Google D Yahoo D ABI MasterMind.. MAT 210 Online Summer 18 Homework: HW 2.7 Score: 0 of 1 pt Bus Econ 2.7.11 Sav 5 of 9 (2 complete) HW Score: 22.22%, 2 of 9 Question Help Until recently, hamburgers at the city sports arena cost $6.20 each. The food concessionaire sold an average of 4,500 hamburgers on game night. When the price was raised to $6.60, hamburger sales dropped off to an average of 3,500 per night (a) Assuming a linear demand curve, find the price of a hamburger that will maximize the nightly hamburger revenue. (b) If the concessionaire had fixed costs of $1,500 per night and the variable cost is $0.80 per hamburger, find the price of a hamburger that will maximize the nightly hamburger profit. (a) Assuming a linear demand curve, find the price of a hamburger that will maximize the nightly hamburger revenue. The hamburger price that will maximize-the nightly hamburger revenue is Round to the nearest cent as needed.) Enter your answer in the answer box and then click Check Answer Clear All Check Answer remaining 544

Explanation / Answer

Hence the points are (6.20,4500) and (6.60,3500)

Slope of line = (y2-y1)/(x2-x1) = (3500-4500)/(6.60-6.20) = -2500

Hence the equation will be

y = -2500x + C

Now since the point (6.20,4500) lie on the line we get

4500 = -2500(6.20) + C, C = 20000

y = -2500x + 20000

Revenue = y * x = (-2500x + 20000) * x = -2500x^2 + 20000x

Taking the derivative of the revenue function wrt x we get

d/dx(Revenue) = d/dx(-2500x^2 + 20000x) = -5000x + 20000

x = 4

Hence for maximizing the revenue, we need to sold the burger at $4 per piece

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