Suppose a bakery determines that the demand for its muffins has changed

a) Assuming that the demand function is linear, determine an equation for p = D(x), where p is the price in dollars and x is the number of muffins.

Demand can be represented by a linear demand function X (D) = a + bP, where

x(D) = demand of good x

P = price of good x

Solving to get b,

B = change in Q / change in P

B = (X2 - X1) / (P2 - P1) = (350 -300) / (0.75 -0.85) = -500

Solving to get a,

350 = a - 500 (0.75)

725= a

X (D) = a + bP

D (x) = 725 - 500P

P =1.45 - 1/500x

b) Graph both cost and revenue on the same set of axes.

Total Cost: C(x) = 0.24x + 122

Total Revenue = P (X) = 1.45x - 1/500x^2

c) Algebraically determine the break-even point(s).

Profit Function is given by,

Profit = TR - TC

= (1.45x - 1/500x^2) - (0.24x + 122)

Profit = 1.21x - 1/500x^2 - 122

Solving for the Break-even output

1.21x - 1/500x^2 - 122 = 0

x^2 - 605x + 61000 = 0

Using the quadratic formula, we get,

X1= 477.16 and X2 = 127.84

There are two levels of quantity at each of which profit is equal to zero.

Computing for Break-even price using the formula P = TR/Q,

X= 477.16

TR = 1.45x - 1/500x^2 = 1.45 (477.16) - 1/500 (477.16) ^2 = 236.52

TC = 0.24x + 122 = 0.24 (477.16) + 122 = 236.52

P = 236.52 / 477.16 = 0.50

Profit = TR - TC = 236.52 - 236.52 = 0

d) How many muffins must be sold to maximize revenue? Determine the maximum revenue. Locate that spot on your graph. At what price must the muffins be sold to maximize revenue?

MR = 1.45 - 1/250x

MC = 0.24

Equating MR = MC

1.45 - 1/250 = 0.24

X = 302.5

Substituting x for p,

P = 1.45 - 1/500x = 1.45 - 1/500 (302.5) = 0.845