Question

21. A firm operates in a perfectly competitive market. The market price of its product is 4
birr and the total cost function is given by: TC= 1/3Q3-5Q2+50,
a. What level of output should the firm produce to maximize its profit?
b. Determine the level of profit at equilibrium.
c. What minimum price is required by the firm to stay in the market?

Answer 1

a) $\bold {Answer}$

Output, $Q = 11 \space units$

$\bold {Solution}$

Perfectly competitive firms are price takers, therefore the firm's price = market price = 4 birr.

$\therefore \space TR = 4Q \space birr$

Profit, $π = TR - TC$

$π = 4Q - \big(\dfrac {1}{3}Q^3 -5Q^2 +50 \big)$

$=4Q - \dfrac {1}{3} Q^3 +5Q^2 -50$

$=-\dfrac {1}{3}Q^3 +5Q^2 +4Q -50$

For maximum profit, we apply differential calculus:

$\dfrac {dπ}{dQ} = \dfrac {d}{dQ} \big(-\dfrac {1}{3} Q^3 +5Q^2+$

$4Q-50 \big)$

$= -Q^2 + 10Q + 4$

When π is maximum, $\dfrac {dπ}{dQ} = 0$

Therefore, $-Q^2 +10Q + 4 = 0$

$=> Q^2 -10 Q - 4 = 0$

$=> Q = \dfrac {-(-10) \pm \sqrt {(-10)^2 -4(1)(-4)}}{2(1)}$

$=> Q = \dfrac {10 \pm \sqrt {116}}{2}$

$=\dfrac { 20.770329614}{2}$

$= 10.385164807$

$\approx \space 11 \space$ $units$

b) $\bold {Answer}$

Profit, $π = 155.33 \space birr$

$\bold {Solution}$

At equilibrium, MR = MC and π is maximum. Therefore, at the equilibrium point, Q = 11 units.

$π = -\dfrac {1}{3} Q^3 + 5Q^2 + 4Q -50$

$= -\dfrac {1}{3}(11)^3 +5(11)^2 +4(11) -50$

$=- \dfrac {1}{3}(1331)+5(121)+44-50$

$= -443.66666667 + 605 -6$

$= 155.3333333333 \space birr$

$= 155.33 \space birr$

c) $\bold {Answer}$

Minimum price, $P = -18.75 \space birr$

$\bold {Solution}$

For a firm to continue operating, the price (AR) must at least cover all AVC. Therefore, the minimum acceptable price = minimum AVC.

Now, $TC = \dfrac {1}{3}Q^3 -5Q^2 + 50$

From the TC:

$TFC = 50 \space birr$

$TVC = \dfrac {1}{3}Q^3 -5Q^2$

$AVC = \dfrac {TVC}{Q}$

$= \dfrac {\dfrac {1}{3}Q^3-5Q^2}{Q}$

$= \dfrac {1}{3}Q^2 -5Q$

For minimum AVC, differential calculus is applied:

$\dfrac {d}{dQ}(AVC) = \dfrac {d}{dQ} \big(\dfrac {1}{3}Q^2 - 5 Q \big)$

$= \dfrac {2}{3} Q - 5$

At the minimum point:

$\dfrac {d}{dQ} (AVC) = 0$

$=> \dfrac {2}{3}Q - 5 = 0$

$=> 2Q - 15 =0$

$=> Q = \dfrac {15}{2}$

$\therefore \space Q = 7.5 \space units$

Thus, $AVC = \dfrac {1}{3}(7.5)^2 -5(7.5)$

$= 18.75 - 37.5$

$= -18.75 \space birr$

(We assume negative prices exist).

$\therefore \space P \geq -18.75 \space birr$