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. a. Draw a table of cost relating to number of plates using 6 values of cost and its corresponding number of plates bought. b. What price in GH¢ should she sell the plates to maximize her revenue?

  a) see below

  b) GH¢85 per plate will maximize her revenue

Step-by-step explanation:

The relation "300 plates less 10 plates for each GH¢5 increase above GH¢20" can be modeled as ...

  p(c) = 300 -10(c -20)/5

  p(c) = 340 -2c = 2(170 -c)

a) A table and graph showing the relationship between cost (x) and plate sales (y) is attached.

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b) Revenue is the product of plate cost and number of plates sold.

 r(c) = c·p(c) = 2c(170 -c)

This is the equation of a a downward-opening parabola with zeros at c=0 and c=170. The vertex is on the line of symmetry, halfway between these zeros. That is, revenue is maximized for a plate cost of (170 +0)/2 = 85. (The attached table shows this.)

She should sell the plates at GH¢85 to maximize her revenue.

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