Problem 68 - Applied Calculus

Question

An isocost line (iso means "same") shows the different combinations of labor and capital (the value of factory buildings, machinery, and so on) a company may buy for the same total cost. An isocost line has equation $$ w L+r K=C \quad \text { for } \quad L \geq 0, \quad K \geq 0 $$ where \(L\) is the units of labor costing \(w\) dollars per unit, \(K\) is the units of capital purchased at \(r\) dollars per unit, and \(C\) is the total cost. Since both \(L\) and \(K\) must be non-negative, \(a n\) isocost line is a line segment in just the first quadrant. a. Write the equation of the isocost line with \(w=8, r=6, \quad C=15,000,\) and graph it in the first quadrant. b. Verify that the following \((L, K)\) pairs all have the same total cost. $$ (1875,0),(1200,900),(600,1700),(0,2500) $$

Step-by-Step Solution

Answer

The isocost line is \( 8L + 6K = 15,000 \), and all given pairs are on this line.

Step 1: Identify Values from Equation

The isocost equation is given by \( wL + rK = C \). We need to substitute the given values into this equation: \( w = 8 \), \( r = 6 \), and \( C = 15,000 \). The equation we will use becomes \( 8L + 6K = 15,000 \).

Step 2: Find Intercepts for Graphing

To find the intercepts, first set \( L = 0 \) and solve for \( K \): \[ 8(0) + 6K = 15,000 \] \[ 6K = 15,000 \] \[ K = 2500 \] Next, set \( K = 0 \) and solve for \( L \): \[ 8L + 6(0) = 15,000 \] \[ 8L = 15,000 \] \[ L = 1875 \]These intercepts are \( (1875, 0) \) and \( (0, 2500) \).

Step 3: Graph the Isocost Line

Plot the intercepts \((1875, 0)\) and \((0, 2500)\) on a coordinate plane with \( L \) on the horizontal axis and \( K \) on the vertical axis. Draw the line connecting these two points. This line represents all combinations of \( L \) and \( K \) that have the same total cost of \( C = 15,000 \).

Step 4: Verify Given Pairs

Check each given pair by substituting \( L \) and \( K \) into the isocost equation:- For \((1875, 0)\): \[ 8(1875) + 6(0) = 15,000 \] Simplifies to \( 15,000 = 15,000 \). True. - For \((1200, 900)\): \[ 8(1200) + 6(900) = 15,000 \] Simplifies to \( 15,000 = 15,000 \). True.- For \((600, 1700)\): \[ 8(600) + 6(1700) = 15,000 \] Simplifies to \( 15,000 = 15,000 \). True.- For \((0, 2500)\): \[ 8(0) + 6(2500) = 15,000 \] Simplifies to \( 15,000 = 15,000 \). True.All pairs are verified as having the same total cost.

Key Concepts

Labor and Capital Allocation

The isocost line is a fundamental concept in economics, especially when analyzing the allocation of resources such as labor and capital. This line represents all the different combinations of labor (L) and capital (K) that a company can afford given a certain budget or total cost, marked as (C). Let's break this down.When considering labor and capital allocation, the company must decide how much of its resources to spend on labor (people or hours worked) versus capital (such as machinery or buildings). These decisions are guided by two key ideas:

• Cost per Unit: Labor costs (L) are determined by the wage rate (w), while capital costs (K) are influenced by the rental rate (r).
• Total Cost: The overall spending limit is denoted by (C), the total available funds for these resources.

In our example, this decision-making turns into a mathematical equation: \[ 8L + 6K = 15,000 \] This equation guides the combinations of (labor, capital) that maintain the same total cost. Businesses use these combinations to maximize productivity within their financial constraints.

Intercepts in Graphing

Finding the intercepts is an essential skill when plotting an isocost line on a graph. These intercepts help you understand exactly where your line crosses the labor and capital axes, giving a visual representation of different resource allocations.Solving for intercepts means determining where one of the variables (either L or K) is zero, giving us pure amounts of just capital or labor:

• Labor Intercept: Set K = 0, and solve for L: \[ 8L = 15,000 \] gives \[ L = 1875 \]. This means if you use only labor, you can employ 1875 units before your budget is exhausted.
• Capital Intercept: Set L = 0, and solve for K: \[ 6K = 15,000 \] gives \[ K = 2500 \]. Here, using only capital, you can obtain 2500 units.

These intercepts, plotted as points (1875, 0) and (0, 2500) on a graph, provide the endpoints for the isocost line. This line signifies that, regardless of how labor and capital are mixed, they can be exchanged along the line without affecting the total cost.

Cost Equation Verification

Verifying the equation for labor and capital pairs ensures that all combinations indeed lie on the same isocost line and stay within the given budget. When each pair is considered correctly, it confirms the line's validity, i.e., each point corresponds to a combination where the total cost remains constant.The verification process involves substituting each (L, K) pair into the isocost equation and checking if it satisfies:

• For (1875, 0): \[ 8(1875) + 6(0) = 15,000 \] Correct, total cost = 15,000.
• For (1200, 900): \[ 8(1200) + 6(900) = 15,000 \] Correct, total cost = 15,000.
• For (600, 1700): \[ 8(600) + 6(1700) = 15,000 \] Correct, total cost = 15,000.
• For (0, 2500): \[ 8(0) + 6(2500) = 15,000 \] Correct, total cost = 15,000.

Each computation shows that these pairs rely on the same total cost line, confirming consistency and accuracy in resource allocation.