Illustration 62
An industry’s total variable cost is given by the following:
TVC = 150V – 20V^2
+ 2V^3
Will the industry manufacture the commodity if price of the commodity is $80?
Solution
An industry manufactures a commodity if price of the commodity surpass its minimum average variable cost.
AVC = TVC = 150V – 20V^2
+ 2V^3
V V
= 150 – 20V + 2V^2 …..Equation (1)
AVC is minimised at the output level at which
d
(AVC) = 0
d
V
taking the derivative of AVC, we procure:
d
AVC = -20
+ 4V
d
V
Thus, AVC will be minim um when
-20 + 4V = 0
4V = 20
V = 20 / 4 = 5
Now substituting the value of V in the Equation for AVC we procure,
Minimum AVC = 150 - 20 * 5 + 2 * (5)^2
= 150 - 100 + 50 = 100
Therefore, price of $80 of the product is less than the minimum average variable cost the firm will not produce the product.
Illustration 63
An industry functioning in a purely rival environment is faced with a market price of $125 per unit of the commodity. The industry’s aggregate short run cost function is as follows: TC = 3000 + 200V – 10V^2 + 2V^3.
- Should the industry manufacture at this price in the short run?
- If the market price is $150 per unit, what will aggregate profits or losses be if the industry manufactures 5 units of productivity? Should the industry manufacture at this price?
- If the market price is greater than $150, should the industry manufacture at this price?
Solution
- An industry will continue manufacturing in the short run if price of the commodity
of $125 surpasses the minimum average variable cost. So we have first to ascertain
the minimum average variable cost.
It is to be noted that in the given cost function $3000 is fixed cost for the reason that it does not incorporate any productivity component V. Therefore,
TVC = 200V – 10V^2 + 2V^3
AVC = TVC = 200V – 10V^2 + 2V^3
V VAVC = 200 – 10V + 2V^2 …..Equation (1)
To ascertain the level of productivity at which average variable cost is minimum, we consider the first derivative of AVC function and set it equal to zero.
dAVC = - 10 + 4V
dVFixing dAVC = 0, we have
dV- 10 + 4V = 0
4V = 10
V = 10 / 4 = 2.5
Substituting the value of V in the AVC function Equation (1), we have
Minimum AVC = 200 – 10 (2.5) + 2 (2.5)^2
= 200 – 25 + 12.5 = 187.5
Since the price of $125 is less than the minimum average variable cost of $150, the industry will not manufacture in the short run for the reason that it will not even gain back variable costs.
- If the market price of the commodity is $150, it may persist manufacturing in the short run for the reason that it will be covering the variable costs entirely though it will not be gaining back any part of the fixed costs and thus incurring losses.
- It may be noted that at price $150, the industry shall manufacture 2.5 units. This
can be known by equating this price with marginal cost which is the profit optimising
situation under perfect rivalry. Therefore,
MC = dTVC = 200 – 20V + 6V^2
dV
MC = Price = 150200 – 20V + 6V^2 = 150
V = 2.5
ATC = 3000 + 200V – 10V^2 + 2V^3
V
= 3000 + 200 – 10V + 2V^2
VATC at productivity 2.5 = 6000 + 200 – 10 (2.5) + 2 (2.5)^2
2.5= 2400 + 200 – 25 + 12.5
= 2587.5
Losses at price $150:
π = TR – TC
TR = P.V – ATC.V
= (150 *2.5) – (2587.5 * 2.5)
= 375 – 6468.75
= - 6093.75
Therefore, the industry will be incurring losses equal to $3000 at price $150 per unit that is losses are almost double the total fixed cost.
Illustration 64
Presume that revenue and total cost of an industry are provided by the equations:
A = 120V and F = 20 + 10V^2, where V is the volume of output.
Using the TR, TC method, ascertain the profit optimisation level and total profits of the industry.
Solution
π = TR – TC
= 120V – 20 – 10V^2
Profits will be optimum at the level of productivity at which the first derivative of total profit function = 0. Therefore,
dπ = 120 – 20V
dV
setting dπ equal to zero, we have,
dV
120 – 20V = 0
20V = 120
V = 120 / 20 = 6
For the second order condition to be satisfied, the second order derivative of profit function must be negative. Taking the second derivative of the profit function, we have
d^2 π = -
20
dV^2
Therefore, the second order condition is fulfilled.
Substituting, V = 6 in the profit function, we have
π = 120*6 – 20 – 10 (6)^2
= 720 – 20 – 360
π = 340
Therefore, the profit optimising productivity will be 6 and that the total profit of the industry will be 340.
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