Using IQ map and Iso cost line, explain the concept of least cost combination of two inputs under given total cost outlay. What will be the effect on output when total cost outlay changes? [10+5]

Concept of Least-Cost Combination of Two Inputs Using Iso-Cost Line and Isoquant Map

The least-cost combination of inputs refers to the optimal mix of two or more inputs (e.g., labour and capital) that a firm can use to produce a given level of output at the minimum possible cost. This concept is derived from the isoquant-iso-cost framework, where a firm aims to minimize costs while maintaining a desired level of production.

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Key Concepts: Isoquant and Iso-Cost Line

1. Isoquant

An isoquant (equal quantity) is a curve that shows all possible combinations of two inputs (e.g., labour $L$ and capital $K$) that yield the same level of output. It represents the production possibilities of a firm.

• Shape: Typically convex to the origin (due to diminishing marginal rate of technical substitution (MRTS)).
• Higher isoquants represent greater output levels.
• Movement along an isoquant: If a firm substitutes one input for another while keeping output constant, it moves along the isoquant.

2. Iso-Cost Line

An iso-cost line represents all combinations of two inputs that cost the same total amount (i.e., the firm’s budget constraint).

• Equation: $C=wL+rK$, where:
  • $C$ = Total cost outlay
  • $w$ = Wage rate (price of labour)
  • $r$ = Rental rate (price of capital)
  • $L$ = Quantity of labour
  • $K$ = Quantity of capital
• Slope: $-\frac{w}{r}$ (negative because more of one input means less of the other for the same cost).
• Shift in iso-cost line: If total cost $C$ increases, the iso-cost line shifts outward (parallel to the original line).

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Least-Cost Combination: Graphical Explanation

The least-cost combination occurs at the point where the iso-cost line is tangent to the isoquant. At this point:

1. Slope of isoquant (MRTS) = Slope of iso-cost line $$MRT{S}_{L,K}=\frac{w}{r}$$

   • $MRT{S}_{L,K}$ = Marginal Rate of Technical Substitution (the rate at which labour can be substituted for capital while keeping output constant).
   • $\frac{w}{r}$ = Relative price of inputs.

2. Economic Interpretation:

   • The firm gets the maximum output for a given cost (cost minimization).
   • Any other combination on the same isoquant would cost more or produce less.

Graphical Illustration (Refer to the Figure Above)

• Isoquant $Q_1$: Represents a fixed output level.
• Iso-cost line $C_0$: Represents the firm’s total cost outlay.
• Point $E$ (Optimal Point): Where the iso-cost line is tangent to the isoquant.
  • At $E$, the firm uses $L^{\ast }$ labour and $K^{\ast }$ capital at the lowest possible cost.
• Point $A$: An inefficient combination (same output but higher cost).
• Point $B$: Unattainable with the given budget.

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Mathematical Derivation of Least-Cost Combination

Assume:

• Output $Q$ is fixed.
• The production function is $Q=f(L,K)$.
• Cost function: $C=wL+rK$.

Objective: Minimize $C$ subject to $Q=f(L,K)$.

First-Order Conditions (Using Lagrange Multipliers): $$\frac{\partial C}{\partial L}=w=\lambda \frac{\partial f}{\partial L}$$ $$\frac{\partial C}{\partial K}=r=\lambda \frac{\partial f}{\partial K}$$ Dividing the two equations: $$\frac{w}{r}=\frac{\frac{\partial f}{\partial L}}{\frac{\partial f}{\partial K}}=MRT{S}_{L,K}$$ This confirms that the slope of the isoquant equals the slope of the iso-cost line at the optimal point.

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Effect of Change in Total Cost Outlay on Output

When the total cost outlay $C$ changes, the iso-cost line shifts, affecting the firm’s production possibilities.

Case 1: Increase in Total Cost ($C\uparrow$)

• The iso-cost line shifts outward (parallel to the original line).
• The firm can now afford more inputs, moving to a higher isoquant.
• Result: Output increases (the firm produces more at a lower per-unit cost).

Case 2: Decrease in Total Cost ($C\downarrow$)

• The iso-cost line shifts inward.
• The firm can only afford fewer inputs, moving to a lower isoquant.
• Result: Output decreases (the firm produces less due to budget constraints).

Graphical Representation

Key Observations

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Numerical Example

Given:

• Wage rate ($w$) = Rs. 10 per unit of labour.
• Rental rate ($r$) = Rs. 20 per unit of capital.
• Total cost outlay ($C$) = Rs. 100.
• Production function: $Q=\min (5L,10K)$ (Leontief production function).

Find the least-cost combination for $Q=40$.

Step 1: Determine the Isoquant for $Q=40$

For a Leontief function, inputs are used in a fixed proportion: $$5L=10K⟹L=2K$$ So, any combination where $L=2K$ yields $Q=40$.

Step 2: Find the Optimal Combination

The cost function is: $$C=10L+20K=100$$ Substitute $L=2K$: $$10(2K)+20K=100⟹20K+20K=100⟹40K=100⟹K=2.5$$ Then, $L=2(2.5)=5$.

Optimal Input Combination:

• Labour ($L$) = 5 units
• Capital ($K$) = 2.5 units
• Total Cost = $10(5)+20(2.5)=50+50=100$ (as given).

Step 3: Effect of Increased Cost Outlay ($C=120$)

Now, $C=120$: $$10L+20K=120⟹L=2K$$ $$10(2K)+20K=120⟹40K=120⟹K=3$$ Then, $L=6$.

New Output: Since the production function is $Q=\min (5L,10K)$: $$Q=\min (5\times 6,10\times 3)=\min (30,30)=30$$ Wait! This seems incorrect because the isoquant should shift outward. Let’s correct this.

Correction: For a Leontief function, output is proportional to the limiting input. If we increase $C$, the firm can produce more by scaling up inputs proportionally.

Let’s assume the firm wants to produce $Q=50$ (higher output).

For $Q=50$: $$5L=50⟹L=10$$ $$10K=50⟹K=5$$ Now, calculate the required cost: $$C=10(10)+20(5)=100+100=200$$ If the firm has $C=120$, it cannot produce $Q=50$. Instead, it must find the maximum output possible with $C=120$.

Finding Maximum Output with $C=120$: The firm will choose inputs such that the cost is exhausted: $$10L+20K=120⟹L=12-2K$$ Substitute into $Q=\min (5L,10K)$: $$Q=\min (5(12-2K),10K)=\min (60-10K,10K)$$ To maximize $Q$, set $60-10K=10K⟹60=20K⟹K=3$. Then, $L=12-2(3)=6$.

Maximum Output: $$Q=\min (5\times 6,10\times 3)=\min (30,30)=30$$ Conclusion:

• With $C=100$, the firm produces $Q=40$ (as initially).
• With $C=120$, the firm can produce $Q=30$ (which is less than before). This seems counterintuitive!

Realization: The mistake arises because the Leontief function is not scalable in the way assumed. Instead, let’s consider a Cobb-Douglas production function for a more realistic example.

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Revised Numerical Example (Cobb-Douglas Function)

Given:

• Production function: $Q=L^{0.5}{K}^{0.5}$
• Wage rate ($w$) = Rs. 4 per unit of labour.
• Rental rate ($r$) = Rs. 4 per unit of capital (for simplicity, $w=r$).
• Total cost outlay ($C$) = Rs. 100.

Find the least-cost combination for $Q=20$.

Step 1: Express Cost in Terms of One Input

From the production function: $$Q=L^{0.5}{K}^{0.5}⟹20=L^{0.5}{K}^{0.5}⟹LK=400$$ Cost function: $$C=4L+4K=100⟹L+K=25$$

Step 2: Substitute and Solve

From $L+K=25$, $K=25-L$. Substitute into $LK=400$: $$L(25-L)=400⟹25L-L^2=400⟹L^2-25L+400=0$$ This quadratic has no real solution, indicating an error in assumption. Let’s adjust $Q$.

Let $Q=10$: $$LK=100$$ $$L+K=25$$ Substitute: $$L(25-L)=100⟹25L-L^2=100⟹L^2-25L+100=0$$ Solving: $$L=\frac{25\pm \sqrt{625-400}}{2}=\frac{25\pm \sqrt{225}}{2}=\frac{25\pm 15}{2}$$ Thus:

• $L=20$, $K=5$
• $L=5$, $K=20$

Check Cost: For $L=20$, $K=5$: $$C=4(20)+4(5)=80+20=100$$ For $L=5$, $K=20$: $$C=4(5)+4(20)=20+80=100$$ Both combinations yield the same cost, but the MRTS condition must hold.

Step 3: Apply MRTS Condition

For Cobb-Douglas, $MRT{S}_{L,K}=\frac{MPl}{MPk}=\frac{K}{L}$. Set $MRTS=\frac{w}{r}=\frac{4}{4}=1$: $$\frac{K}{L}=1⟹K=L$$ Thus, the optimal combination is $L=K=12.5$.

Verify Output: $$Q=(12.5{)}^{0.5}(12.5{)}^{0.5}=12.5\approx 11.18$$ This is less than 10, indicating another miscalculation.

Correct Approach: The least-cost combination for a given $Q$ is found by setting $MRTS=\frac{w}{r}$.

For $Q=10$: $$LK=100$$ $$\frac{K}{L}=1⟹K=L$$ Thus: $$L^2=100⟹L=10,K=10$$ Cost: $$C=4(10)+4(10)=80$$ But our budget is $C=100$. This means we can produce more with the remaining budget.

Find Maximum $Q$ with $C=100$: From $C=4L+4K=100⟹L+K=25$. Substitute into $Q=L^{0.5}{K}^{0.5}$: $$Q=(L(25-L){)}^{0.5}$$ Maximize $Q$ by maximizing $L(25-L)$: $$\frac{d}{dL}[L(25-L)]=25-2L=0⟹L=12.5,K=12.5$$ Maximum Output: $$Q=(12.5\times 12.5{)}^{0.5}=12.5\approx 11.18$$ Cost: $$C=4(12.5)+4(12.5)=100$$

Effect of Increased Cost ($C=120$): Now, $L+K=30$. Optimal inputs: $$L=K=15$$ New Output: $$Q=(15\times 15{)}^{0.5}=15$$ Conclusion:

• With $C=100$, maximum $Q=11.18$.
• With $C=120$, maximum $Q=15$.
• Output increases as total cost outlay increases.

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Final Answer Summary

1. Least-Cost Combination:

   • Occurs where the iso-cost line is tangent to the isoquant.
   • Mathematically, $MRT{S}_{L,K}=\frac{w}{r}$.
   • Graphically, the optimal point is where the slopes of the isoquant and iso-cost line are equal.

2. Effect of Change in Total Cost Outlay:

   • Increase in $C$: Iso-cost line shifts outward → Firm moves to a higher isoquant → Output increases.
   • Decrease in $C$: Iso-cost line shifts inward → Firm moves to a lower isoquant → Output decreases.

3. Numerical Insight:

   • For a given production function, increasing the total cost outlay allows the firm to purchase more inputs, leading to higher output levels (assuming the firm operates efficiently).

This framework is fundamental in cost minimization and production efficiency, guiding firms in optimal resource allocation.