Question

Suppose you have just been hired as the new production manager of Quest Steel. There are three plants currently producing ½ inch sheet steel of identical quality. Total cost of production in plant 1 is

C₁ = 1 + 10q₁,

Where q is measured in tones and C is in dollars. The total cost function in plant 2 is

C₂ = 10 + .5(q₂)² + 2q₂

Cost at plant 3 is given by

C₃ = 5 + (q₃)²

Total production Q = q₁ + q₂ + q₃

a. Traditionally, 10 tons of steel have been produced each period. Plant 1 has produced 2 tons, plant 2 has produced 4 tons and plant 3 has produced 4 tons. You have been assigned the task of developing ways to reschedule production among plants to reduce costs, given that the output level remains at 10 tons per day. What changes, if any, do you suggest?

b. You question whether output of 10 tons per day is profit maximizing. Suppose the current market price is $6 per ton, and the market is competitive. Would you change total output? If so, how?

c. If your suggestions are implemented, what will profits be?

Answer 1

a) The total output from all the three plants is 10 tons per day. The aim of the firms is to reduce the cost. $C_1=1+10q_1\\ C_2=10+0.5q_2^2+2q_2\\ C_3=5+q_3^2\\ q_1+q_2+q_3=10\\$

This implies:

$q_3=10-q_1-q_2\\\\ \text{Aim is to }\\ \underset{q_1,q_2}{min}C_1+C_2+C_3\\\\ \underset{q_1,q_2}{min}1+10q_1+10+0.5q_2^2+2q_2+5+(10-q_1-q_2)^2\\$$q_3=10-q_1-q_2\\\\ \text{Aim is to }\\ \underset{q_1,q_2}{min}\ C_1+C_2+C_3\\\\ \underset{q_1,q_2}{min}\ 1+10q_1+10+0.5q_2^2+2q_2+5+(10-q_1-q_2)^2\\$

Taking First order conditions with respect to q_1 and q_2:

$\frac{\partial (1+10q_1+10+0.5q_2^2+2q_2+5+(10-q_1-q_2)^2)}{\partial q_1}=0\\ 10-2(10-q_1-q_2) =0\\ q_1+q_2=5\cdots(1)\\\\ \frac{\partial (1+10q_1+10+0.5q_2^2+2q_2+5+(10-q_1-q_2)^2)}{\partial q_2}=0\\ q_2+2-2(10-q_1-q_2)=0\\ \implies 2q_1+3q_2=18\cdots(2)$

Solving equation 1 and 2 gives:

$q_2=8, q_1=-3, q_3=10-8+3=5$

Also by second order conditions, we can see that hessian is positive definite, hence the quantities found does minimize the cost functions.

But producing negative quantities is not feasible. So producing 0 from plant 1 is optimal. Now to keep the production at 10 tons a day, let us compare the cost of producing one unit at plant 2 and plant 3:

cost of producing 1 unit at plant 2=10+0.5+2=12.5 units

cost of producing 1 unit at plant 3=5+1=6 units

Thus production at plant 3 is cheaper.

So we eliminate the extra production from plant 2.

Thus plant 1 produces 0, plant 2 produces 5 tons and plant 3 produces 5 tons a day.

b) Profits from current production are -8. The firm needs to produce more in order to increase the revenue generted while keeping the cost minimal.

Now we need to eliminate the condition that q1+q2+q3=10. With the eliminated condition, the new firm problem is :

$\underset{q_1,q_2,q_3}{max}\pi(q_1,q_2,q_3)=\ 6*(q_1+q_2+q_3)-1-10q_1-10-0.5q_2^2-2q_2-5-q_3^2\\\\ \text{F.O.C }\\\\ \frac{\partial \pi(q_1,q_2,q_3)}{\partial q_1}=0\\\implies 6-10=0 \text{ (no solution) }\\\\ \frac{\partial \pi(q_1,q_2,q_3)}{\partial q_2}=0\\\implies 6-q_2-2=0 \\\implies q_2=4 \\\\\frac{\partial \pi(q_1,q_2,q_3)}{\partial q_3}=0\\\implies 6-2q_3=0 \\\implies q_3=3$

checking S.O.C will show that these are indeed the maximizing outcomes.

So the new suggestion would be q1=0, q2=4, q3=3, since cost function for firm 1 is a linear function in q1, it would be the least when q1=0

c) With the new quanitities suggested, profits would be:

Price*(q1+q2+q3)-c1(q1)-c2(q2)-c3(q3)=

6*7-1-10-0.5*4^2-2*4-5-3^2=$1