Theory of Price and Output Determination | Numerical Problems and Solutions

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THEORY OF PRICE AND OUTPUT DETERMINATION (NUMERICAL QUESTIONS AND THEIR SOLUTIONS)

This article contains all the important numerical problems with their solutions in chapter 4, the theory of price, and the output determination of economics of class 12.

PROBLEM 1

A firm’s total revenue is Rs. 1,200 and total cost is Rs. 1,000. Find out total profit.

SOLUTION
Total revenue (TR) = Rs. 1,200
Total cost (TC) = Rs. 1,000
Total fixed cost (TFC) = Rs. 800
We know that
Total profit (π) = TR – TC
= 1,200 – 1,000
= Rs. 200

PROBLEM 2

Calculate the profit of the firm when P = Rs. 30, Q = 10 units and AC = Rs. 25.

SOLUTION
Given
Price (P) = Rs. 30
Output (Q) = 10 units
Average cost (AC) = Rs. 25
We know that
TR = P × Q
= 30 × 10
= Rs. 300

Again,
$A C = \frac{T C}{Q}$

Or, $T C = A C \times Q$
= 25 × 10
= Rs. 250

Profit (π) = TR – TC
= 300 – 250
= Rs. 50

PROBLEM 3

Derive MR and MC functions from the TR and TC functions given below.
$T R = 10 Q - 4 Q^{2}$
$T C = 50 + 6 Q^{2}$

SOLUTION
Given
$T R = 10 Q - 4 Q^{2}$
$T C = 50 + 6 Q^{2}$
We know

$M R = \frac{d (T R)}{d Q}$
$M R = \frac{d (10 Q - 4 Q^{2})}{d Q}$
$M R = 10 - 8 Q$

And

$M C = \frac{d (T C)}{d Q}$
$M C = \frac{d (50 + 6 Q^{2})}{d Q}$
$M C = 12 Q$

∴ MR = 10 - 8Q and MC = 12Q

PROBLEM 4

Calculate equilibrium level of output of the firm when marginal revenue (MR) = 300 - 0.5Q and marginal cost (MC) = 50 + 2Q.

SOLUTION
Given
$M R = 300 - 0.5 Q$
$M C = 50 + 2 Q$
For equilibrium level of output,
$M R = M C$

Or, $300 - 0.5 Q = 50 + 2 Q$
Or, $300 - 50 = 0.5 Q + 2 Q$
Or, $250 = 2.5 Q$
Or, $Q = \frac{250}{2.5}$
$Q = 100 units$

$PROBLEM 5$

The total revenue and total cost functions of a perfectly competitive firm are given as

$T R = 10 Q$
$T C = 100 + 2 Q + 0.01 Q^{2}$
Determine level of output that maximizes profit.

SOLUTION
We have,
$T R = 10 Q$
$T C = 100 + 2 Q + 0.01 Q^{2}$

$MR = \frac{d}{dQ} (TR)$
$MR = \frac{d}{dQ} (10Q)$
$MR = 10$

$MC = \frac{d}{dQ} (TC)$
$MC = \frac{d}{dQ} (100 + 2Q + 0.01Q^2)$
$MC = \frac{d(100)}{dQ} + \frac{d(2Q)}{dQ} + \frac{d(0.01Q^2)}{dQ}$
$MC = 0 + 2 + 0.02Q$

Condition for profit maximization
$MC = MR$
Or, $2 + 0.02Q = 10$
Or, $0.02Q = 8$
Or, $Q = 400 \text{ units}$

Hence, total profit is maximized at $Q = 400 \text{ units}$ .

PROBLEM 6

Let the revenue function TR = (20Q – Q^2) and cost function C = Q^2 + 8Q + 4. Find profit-maximizing level of output and price.

SOLUTION
Given
Revenue function: $T R = (20 Q - Q^{2})$
Cost function: $C = Q^{2} + 8 Q + 4$
We know that

$M R = \frac{d (T R)}{d Q}$
$M R = \frac{d (20 Q - Q^{2})}{d Q}$
$M R = 20 - 2 Q$

$M C = \frac{d (C)}{d Q}$
$M C = \frac{d (Q^{2} + 8 Q + 4)}{d Q}$
$M C = \frac{d (Q^{2})}{d Q} + \frac{d (8 Q)}{d Q} + \frac{d (4)}{d Q}$
$M C = 2 Q + 8 + 0$
$M C = 2 Q + 8$

For profit maximizing output or equilibrium
$M R = M C$
Or, $20 - 2 Q = 2 Q + 8$
Or, $20 - 8 = 2 Q + 2 Q$
Or, $12 = 4 Q$
Or, $Q = \frac{12}{4} = 3 units$

Price (P) = 20 – Q
$P = 20 - 3$ $P = R s .17$

Hence, profit-maximizing level of output is 3 units and price is Rs. 17.

PROBLEM 7

A firm’s marginal cost and marginal revenue functional are given as follows:

$T C = 50 + 6 Q^{2}$
$P = 100 - 4 Q$
$M C = 12 Q$
$M R = 100 - 8 Q$
Calculate profit maximizing output and price, and total profit.
SOLUTION

Given
$M C = 12 Q$
$M R = 100 - 8 Q$

For profit maximization (Equilibrium of the firm)
$M C = M R$

Or, $12 Q = 100 - 8 Q$
Or, $12 Q + 8 Q = 100$
Or, $20 Q = 100$
$Q = 5 units$

Price (P) = 100 – 4 × 5
$P = 100 - 20$
$P = R s .80$

Profit (π) = TR - TC
$T R = P \times Q = 80 \times 5 = R s .400$
$T C = 50 + (6 \times 5^{2}) = 50 + (6 \times 25) = 50 + 150 = R s .200$
$π = 400 - 200 = R s .200$

Hence, profit maximizing output, $Q$ is 5 units, profit maximizing price, $P$ is Rs. 80 and maximum profit, $π$ is Rs. 200.

PROBLEM 8

Let, cost function: C = 100 + 12Q^2 and demand function: P = 200 – 8Q. Find profit maximizing output and price, and maximum profit.

SOLUTION

Given
Cost function: $T C = 100 + 12 Q^{2}$
Demand function: $P = 200 – 8 Q$

We know that
$MC = \frac{d(TC)}{dQ}$
$MC = \frac{d(100 + 12Q^2)}{dQ}$
$MC = 24Q$

$TR = P \times Q$
$TR = (200 - 8Q) Q$
$TR = 200Q - 8Q^2$

$MR = \frac{d(TR)}{dQ}$
$MR = \frac{d(200Q - 8Q^2)}{dQ}$
$MR = 200 - 16Q$

For equilibrium condition,
$MR = MC$
$200 - 16Q = 24Q$
Or, $200 = 40Q$
Or, $Q = 5$

Price (P)
$P = 200 - 8Q$
$P = 200 - 8 \times 5$
$P = 200 - 40$
$P = Rs. 160$

Profit (π)
$\pi = TR - TC$
$\pi = P \times Q - (100 + 12Q^2)$
$\pi = 160 \times 5 - (100 + 12 \times 5^2)$
$\pi = 800 - (100 + 12 \times 25)$
$\pi = 800 - (100 + 300)$
$\pi = 800 - 400$
$\pi = Rs. 400$

Neupane Nirmal

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