Question

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.
Draw a table of cost relating to number of plates using 6 values of cost and its corresponding number of plates bought.

What price in GH¢ should she sell the plates to maximize her revenue?​

Answer 1

Answer:

Step-by-step explanation:

First, note this parameters from the question.

We let x = number of $5 increases and number of 10 decreases in plates sold.

Our Revenue equation is:

R(x) = (300-10x)(10+5x)

We expand the above equation into a quadratic equation by multiplying each bracket:

R(x) = 3000 + 1500x - 3000x - 1500x^2

R(x) = -1500x^2 - 1500x + 3000 (collect like terms)

Next we simplify, by dividing through by -1500

= 1500x^2/1500 - 1500x/1500 + 3000/1500

= X^2 - x + 2

X^2 - x + 2 = 0

Next, we find the axis of symmetry using the formula x = -b/(2*a) where b = 1, a = 1

X = - (-1)/2*1

X = 1/2

Number of $5 increases = $5x1/2 = $2.5

=$2.5 + $20 = $22.5 ticket price gives max revenue.