Question

A bakery sells specialty handmade loaves of bread. Daily fixed costs of product on are $125, while the marginal cost per loaf is $1.60. A bit of experimenting with their pricing structure has determined that 40 loaves will be sold if the selling price per loaf is $7.00; while 60 loaves will be sold if the selling price per loaf is $5.50. Assuming a linear price-demand relationship :

(1) Show that the maximum daily profit is $110.20 when the price per loaf is set at $5.80.

(2) If the marginal cost per loaf is reduced to $1.30, the potential profit should increase. Should the selling price be maintained at $5.80? What is the maximum possible profit with the reduced marginal cost?

(3) The bakery would like to find a way to increase the maximum possible profit to $140. How low would the marginal cost need to be in order to allow a profit of $140? Prepare complete solutions to the above.

Answer 1

(a) For 40 loaves of bread, the cost is equal to $7.00

For 60 loaves of bread, the cost is equal to $5.50

Slope = (5.50-7.00)/(60-40) = -1.50/20 = -0.075

Hence the price function can be modeled as

pー7ー-0.075(z-40)

p = 7 + 40(0.075) _ 0.075·r = 10-01 752.

Total Cost = Fixed Cost + Variable Cost = 125 + 1.60x

Revenue = (Number of loaves) * (Price of loaves) = (x)*(10-0.075x) = 10x - 0.075x^2

Profit = Revenue-Cost = 10x-0.07522-125-1.60x =ー0.075x2 + 8.4.r 125

Profit-( (-0.075r+8.4r- 125)- -0.150r 8.4

8.4 56 0.150

10-56(0.075) = 5.89 Price per loaf

Profit =-0.075(56). 8.4(56)-125 $110.2

b)

Total Cost = Fixed Cost + Variable Cost = 125 + 1.30x

Revenue = (Number of loaves) * (Price of loaves) = (x)*(10-0.075x) = 10x - 0.075x^2

Profit Revenue - Cost 10r 0.075225 1.300.0752+ 8.7.r- 125

(-0.075r+8.7r - 125)- -0.150r 8.7

8.7 T 58 0.150

Selling price should be changed to 10 - 0.075 * 58 = 5.65 $

Profitー-0.075(58)-+ 8.70(58)-125 = 127.3

c)

The marginal cost should be lower down to $1 in order to increase the profit to $140

Let the marginal cost be p

So, the total cost function will be 125 + px

Revenue = (10-0.075x)*x = 10x - 0.075x^2

The maxima will come at the point when

10-p 0.150

Now substitute this value in the profit function and equate the profit function to 140$

This will yield the value of p, the marginal cost equal to $1

Note - Post any doubts/queries in comments section.