Problem 86 - Applied Calculus

Question

The intersection of an isocost line \(w L+r K=C\) and an isoquant curve \(K=a L^{b}\) (see pages 18 and 31 ) gives the amounts of labor \(L\) and capital \(K\) for fixed production and cost. Find the intersection point \((L, K)\) of each isocost and isoquant. [Hint: After substituting the second expression into the first, multiply through by \(L\) and factor.] $$ 5 L+4 K=120 \text { and } K=180 \cdot L^{-1} $$

Step-by-Step Solution

Answer

The intersection point is \((12, 15)\).

Step 1: Substitute Isoquant into Isocost

We are given the isoquant equation as \( K = 180 \cdot L^{-1} \). Substitute this expression for \( K \) into the isocost equation \( 5L + 4K = 120 \). This results in the equation: \[ 5L + 4(180 \cdot L^{-1}) = 120 \].

Step 2: Simplify the Equation

Multiply through the entire equation by \( L \) to get rid of the fraction. This gives: \[ 5L^2 + 4 \times 180 = 120L \]. Simplify this to: \[ 5L^2 = 120L - 720 \].

Step 3: Rearrange to Form a Quadratic

Rearrange the equation to form a standard quadratic equation: \[ 5L^2 - 120L + 720 = 0 \].

Step 4: Solve the Quadratic Equation

Use the quadratic formula \( L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve for \( L \), where \( a = 5 \), \( b = -120 \), and \( c = 720 \). Calculate: \( L = \frac{120 \pm \sqrt{(-120)^2 - 4 \cdot 5 \cdot 720}}{2 \cdot 5} \).

Step 5: Calculate the Discriminant

Calculate the discriminant \( \Delta = (-120)^2 - 4 \cdot 5 \cdot 720 \). \( \Delta = 14400 - 14400 = 0 \), so \( 120 \pm 0 \) is our solution for \( L \).

Step 6: Find the Value of L

Since the discriminant is zero, there is one repeated root for \( L \). Solve for \( L \): \( L = \frac{120}{10} = 12 \).

Step 7: Calculate K Using Found L

Substitute \( L = 12 \) back into the isoquant equation to find \( K \): \( K = 180 \cdot 12^{-1} = 15 \).

Step 8: Verify the Intersection

Check the values in the isocost equation: \( 5(12) + 4(15) = 120 \). Simplifying this gives \( 60 + 60 = 120 \), confirming the solution is correct.

Key Concepts

Understanding Quadratic Equations

Quadratic equations are vital in solving many economic problems, like finding the intersection of isocost and isoquant lines. These equations are typically in the form \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are constants where \( a eq 0 \). The quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), is used to solve these equations.
Understanding how to apply this formula to our problem involves substituting our specific values (like \( a = 5 \), \( b = -120 \), \( c = 720 \)), and calculating the discriminant \( b^2 - 4ac \). We find the discriminant tells us if we have real roots, repeated roots, or complex roots. In our case, it simplifies to zero, indicating one real and repeated root for \( L \). That means there's only one value of \( L \) where the isoquant and isocost intersect.
This process helps us to isolate to where the two functions meet, giving a unique solution in real-world applications like economics and production planning.

The Role of Labor and Capital

In economics, labor and capital are essential factors of production. By definition, labor refers to the human work needed to produce goods or services, while capital includes all kinds of tools and machinery that aid production.
The intersection of an isocost line, representing constant cost, and an isoquant curve, symbolizing constant output, helps determine the most efficient mix of labor and capital. In our example, we work with equations telling us how much labor \( L \) and capital \( K \) are needed at a specific cost and production level.
By solving these equations, we find that precisely 12 units of labor and 15 units of capital meet both the production goal and budget constraints. This process highlights the importance of economic planning to balance resources efficiently.

Applying Calculus in Economic Models

Calculus is an essential tool for optimal decision-making in economics, as it allows us to analyze and model changing conditions. Applied calculus helps in maximizing or minimizing functions by understanding rates of change and slopes of curves.
In solving the intersection of isocost and isoquant functions, calculus assists in modeling the economic environment by finding efficient resource allocations. The substitution and simplification steps in our problem mirror calculus techniques. Here, calculus offers the foundational support to rearrange and solve equations for real variables like labor and capital.
This allows economists and businesses to predict outcomes, determine where costs will be minimized or production maximized, and make informed decisions based on these models. The insights gained through these calculations are critical for strategic planning and optimization in the economic sphere.