Question

C) A restaurant operator in Accra has found out that during the partial lockdown, if she sells a plate of her food for GH¢20 each, she can sell 300 plates, but for each GH¢5 she raises the price, 10 less plates are sold. What price in GH¢ should she sell the plates to maximize her revenue?

Answer 1

Maximum price = GH¢85

Explanation:

Let p be the number of times she raises the price as well as the number of times 10 less plates are sold

Revenue, R = number of plates sold * price sold

R = (300 - 10p)(20 + 5p)

R = 6000 + 1500p - 200p - 50p²

R = 6000 + 1300p - 50p²

50p² - 1300p - 6000 = 0

p² - 26p - 120 = 0

finding the axis of symmetry, p = -b/2a

p = -(-26)/2*1

p = 13

Maximum price = 13 * 5 + 20 = GH¢85

Number of plates sold = 300 - 10 * 13 = 170 plates

Maximum revenue = GH¢85 * 170 = GH¢14,450