Problem 85 - Applied Calculus

Question

85-86.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.] $$ 3 L+8 K=48 \text { and } K=24 \cdot L^{-1} $$

Step-by-Step Solution

Answer

The intersection point is \((L, K) = (8, 3)\).

Step 1: Substitute Isoquant into Isocost

Start with the two given equations: the isocost line equation \(3L + 8K = 48\) and the isoquant curve equation \(K = 24 \cdot L^{-1}\). Substitute the second expression, \(K = 24 \cdot L^{-1}\), into the first equation to eliminate \(K\). This results in \(3L + 8(24 \cdot L^{-1}) = 48\).

Step 2: Simplify the Equation

Simplify the equation \(3L + 8(24 \cdot L^{-1}) = 48\). This becomes \(3L + \frac{192}{L} = 48\).

Step 3: Eliminate the Fraction

Multiply every term by \(L\) to eliminate the fraction: \(3L^2 + 192 = 48L\).

Step 4: Rearrange into Quadratic Form

Rearrange the equation obtained in the previous step: \(3L^2 - 48L + 192 = 0\).

Step 5: Solve the Quadratic Equation

Use the quadratic formula \(L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to solve the quadratic equation \(3L^2 - 48L + 192 = 0\). Here, \(a = 3\), \(b = -48\), and \(c = 192\). Calculate the discriminant: \((-48)^2 - 4 \cdot 3 \cdot 192\) to find the value of \(L\).

Step 6: Calculate L

The discriminant is \(2304 - 2304 = 0\). Since the discriminant is zero, there is one real solution: \(L = \frac{48}{6} = 8\).

Step 7: Calculate K using Isoquant Equation

Substitute \(L = 8\) back into the isoquant equation \(K = 24 \cdot L^{-1}\): \(K = 24 \cdot \frac{1}{8} = 3\).

Step 8: Confirm the Solution

Check if \(L = 8\) and \(K = 3\) satisfy the original isocost equation: \(3 \times 8 + 8 \times 3 = 24 + 24 = 48\). This confirms the values found are correct.

Key Concepts

Understanding Calculus Problem Solving in Economics

Calculus is a fundamental tool that helps us solve various problems in economics and other scientific fields. The primary task in calculus problem solving is to manage complex relationships and analyze how changes in one quantity affect another. In the given exercise, calculus aids in finding the intersection of an isocost line and an isoquant curve. Understanding how to solve such problems involves:

• Identifying what needs to be solved, in this case, the intersection points (L, K).
• Substituting one equation into the other to simplify the process, as done in step 1 by inserting the isoquant into the isocost equation.
• Utilizing mathematical techniques to tackle different forms of equations, as seen in step 2 where simplification and step 3 where fractions are eliminated.

These steps showcase the practical application of calculus in breaking down economic models into solvable parts, leading to a clear solution that fits the given constraints.

Tackling Quadratic Equations

Quadratic equations are often encountered in a variety of scenarios, including the world of economic theory. A quadratic equation has the form ax² + bx + c = 0, and solving these equations can reveal important information about economic models. In the exercise, after substituting the isoquant into the isocost line, the resulting equation is a quadratic: \[3L^2 - 48L + 192 = 0\].To solve it:

• Rearrange terms into standard quadratic form, highlighting the need for a clear arrangement to apply the quadratic formula.
• Use the quadratic formula: \(L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find the roots of the equation, which involves calculating the discriminant \(b^2 - 4ac\).
• A zero discriminant, as in the exercise, indicates one unique real solution, simplifying the task as opposed to handling multiple roots.

These steps emphasize the systematic, logical approach needed to address quadratic equations, ensuring precise solutions in economic contexts.

Exploring Economic Theory: Isocosts and Isoquants

Economic theory utilizes models like isocost and isoquant curves to understand relationships between resources in production. An isocost line represents all combinations of labor and capital that cost the same, given by the equation \(w L + r K = C\). On the other hand, an isoquant curves represent different combinations of inputs that produce the same level of output, like \(K = aL^b\).In the exercise provided:

• The isocost line \(3L + 8K = 48\) corresponds to resource costs, while the isoquant \(K = 24 \cdot L^{-1}\) signifies the production level.
• The task is to find their intersection, which reveals the optimal combination of inputs (L and K) for given cost and production constraints.
• This intersection signifies where cost minimization meets output maximization, a vital concept in economic production theory.

Understanding the intersection of these curves through calculus and quadratic solutions provides insights into efficient production practices, applying valuable economic principles into practical scenarios.