Question

(1)A firm in a perfectly competitive market sells all its product (Q) at a constant price (P) of $60. Suppose the total cost function (TC) for this firm is described by the following equation: 2 3 TC(Q) 128+ 690-14Q + Q (a)Form the profit function and determine the output that maximizes the firm's profit. Evaluate the second order condition to assure that profit is maximized at this level of output. (b)Derive the marginal revenue (MR) and the marginal cost(MC). Graph both curves for Q 1, 2...12 Use the graph to find the level of output that maximizes the firm's profit. (c )Solve algebraically for the profit maximizing output using the MR and MC derived in part (b). (d)How much profit/loss does the firm earn by producing the profit maximizing output?

Can you label each one please .

Answer 1

Answer:

(a) Given: TC=128+69Q-14Q²+Q³

TR=P*Q=60Q

Thus, Profit ($\pi$)=TR-TC=60Q-128-69Q+14Q²-Q³=-9Q+14Q²-Q³-128

Output is maximized when $\frac{\partial \pi }{\partial Q}=0=>\frac{\partial }{\partial Q}(-9Q+14Q^2-Q^3-128)=0=>-9+28Q-3Q^2=0$

i.e. $3Q^2-28Q+9=0$

This equation is a quadratic function where a=3, b=-28 and c=9

We know that the solution of quadratic equation is given by $\frac{-b\pm \sqrt{b^2-4ac} }{2a}$

Thus, $Q=\frac{-b\pm \sqrt{b^2-4ac} }{2a}=\frac{-(-28)\pm \sqrt{(-28)^2-4*(3)*(9)} }{2*3}$

i.e. $Q=\frac{28\pm 26}{6}=9 \, or \, \frac{1}{3}$

Hence the profit maximizing output is Q=9 because output can't be in fraction.

Second order condition: $\frac{\partial^{2} \pi }{\partial Q^2}=>\frac{\partial }{\partial Q}(-9&plus;28Q-3Q^2)=28-6Q=28-6*9=-26<0$

Hence, Q=9 is the profit maximizing output.

(b) Given: TC=128+69Q-14Q²+Q³ and TR=P*Q=60Q

Thus, MC=69-28Q+3Q² and MR=60

Price=MR=S60 | MR MC=69-2sQ+3Q^2 MC নিলনালিঃ

MR +MC 170 160 150 140 MR and MC 0 1 2 3 4 5 8 9 10 11 12 13 6 7 Output

The graph shows that at Q=9, MC=MR. Thus 9 is the profit maximizing output.

(c) The profit maximizing condition is MC=MR=>69-28Q+3Q²=60=>3Q²-28Q+9=0 which is the same quadratic equation solved above. Thus profit maximizing output, Q=9

(d) Thus plugging Q=9 in the profit function we get the maximum profit=-9Q+14Q²-Q³-128=-9*9+14*9²-9³-128=-81+1134-729-128=196