Problem 16 - Applied Calculus

Question

A firm manufactures a commodity at two different factories. The total cost of manufacturing depends on the quantities, \(q_{1}\) and \(q_{2},\) supplied by each factory, and is expressed by the joint cost function, $$C=f\left(q_{1}, q_{2}\right)=2 q_{1}^{2}+q_{1} q_{2}+q_{2}^{2}+500$$ The company's objective is to produce 200 units, while minimizing production costs. How many units should be supplied by each factory?

Step-by-Step Solution

Answer

Factory 1 should produce 50 units, and Factory 2 should produce 150 units.

Step 1: Understand the Joint Cost Function

The joint cost function is given by \(C = f(q_{1}, q_{2}) = 2q_{1}^{2} + q_{1}q_{2} + q_{2}^{2} + 500\). This function represents the total cost of producing units \(q_1\) at Factory 1 and units \(q_2\) at Factory 2.

Step 2: Define the Production Constraint

The company needs to produce a total of 200 units. Thus, we have the constraint \(q_1 + q_2 = 200\).

Step 3: Solve Using the Method of Lagrange Multipliers

To minimize the cost function subject to the production constraint, use the method of Lagrange Multipliers. Introduce a Lagrange multiplier \(\lambda\) and form the Lagrangian: \[L(q_1, q_2, \lambda) = 2q_{1}^{2} + q_{1}q_{2} + q_{2}^{2} + 500 + \lambda(200 - q_1 - q_2)\]

Step 4: Calculate the Partial Derivatives

Find the partial derivatives of the Lagrangian with respect to \(q_1\), \(q_2\), and \(\lambda\):\[\frac{\partial L}{\partial q_1} = 4q_1 + q_2 - \lambda = 0\]\[\frac{\partial L}{\partial q_2} = q_1 + 2q_2 - \lambda = 0\]\[\frac{\partial L}{\partial \lambda} = 200 - q_1 - q_2 = 0\]

Step 5: Solve the System of Equations

Substitute and solve the system of equations:1. \(4q_1 + q_2 = \lambda\)2. \(q_1 + 2q_2 = \lambda\)3. \(q_1 + q_2 = 200\)Subtract equation 2 from equation 1: \(4q_1 + q_2 - (q_1 + 2q_2) = 0\) simplifying to \(3q_1 - q_2 = 0\), i.e., \(q_2 = 3q_1\).

Step 6: Apply the Production Constraint

Use the production constraint \(q_1 + q_2 = 200\) with \(q_2 = 3q_1\) to solve for \(q_1\):\[q_1 + 3q_1 = 200\]\[4q_1 = 200\]\[q_1 = 50\]Substitute \(q_1\) back to find \(q_2\):\[q_2 = 3(50) = 150\]

Step 7: Verify the Cost Minimization

Verify that substituting \(q_1 = 50\) and \(q_2 = 150\) satisfies both the constraint and minimizes the cost function. The system of equations reflects the minimum cost configuration given the constraint.

Key Concepts

Joint Cost Function

The joint cost function is a fundamental concept in optimizing production costs. In this exercise, the joint cost function is represented by the formula \(C = f(q_{1}, q_{2}) = 2q_{1}^{2} + q_{1}q_{2} + q_{2}^{2} + 500\). This equation combines the costs incurred from producing units at two different factories. Here, \(q_1\) and \(q_2\) refer to the quantities produced at Factory 1 and Factory 2, respectively. The expression \(2q_{1}^{2}\) accounts for the cost that scales with the square of the quantity produced at Factory 1, indicating an increasing cost at a higher production rate. Likewise, \(q_{2}^{2}\) represents the cost structure for Factory 2 in a similar fashion. The term \(q_{1}q_{2}\) represents an interaction effect or cost that arises when both factories are producing at the same time. The constant 500 represents a fixed cost that does not change with the level of output. This setup allows the company to understand their cost structure and optimize production such that costs can be minimized.

Production Constraint

A production constraint refers to a condition or limitation that must be satisfied in the production process. In this context, the company wants to manufacture exactly 200 units in total. This constraint is mathematically represented as \(q_1 + q_2 = 200\). It means that the sum of units produced by both factories must equal 200. Production constraints are crucial for ensuring that the solution to our optimization problem remains feasible and applicable in real-world scenarios. Constraints can be equality constraints, as in this case, or they can be inequalities, which set upper or lower bounds on production levels. By applying a constraint to our problem, we ensure not only that the firm's production target is met but also help in exploring the cost implications of allocating production differently between the two factories.

Partial Derivatives

Partial derivatives are a key tool in calculus, used to analyze how a function changes as one of its input variables changes, while keeping others constant. In the context of using Lagrange Multipliers, we find the partial derivatives of the Lagrangian function to determine conditions for optimality. The Lagrangian \(L(q_1, q_2, \lambda)\) is constructed by combining the joint cost function with the production constraint, using a Lagrange multiplier \(\lambda\). We then compute:

• \(\frac{\partial L}{\partial q_1} = 4q_1 + q_2 - \lambda = 0\)
• \(\frac{\partial L}{\partial q_2} = q_1 + 2q_2 - \lambda = 0\)
• \(\frac{\partial L}{\partial \lambda} = 200 - q_1 - q_2 = 0\)

These equations give us conditions that must hold at the minimum cost. Each partial derivative equation represents a balance of forces: the marginal change in cost due to producing an additional unit must equally influence constraints represented by \(\lambda\). This mathematical balancing act ensures that we find a cost-efficient allocation of production that respects the production constraint.

System of Equations

A system of equations is a collection of two or more equations involving the same set of variables. Solving this system is essential to finding the values of \(q_1\) and \(q_2\) that minimize cost while satisfying the production constraint. From the partial derivatives calculated, we derive these equations:

• The first equation: \(4q_1 + q_2 = \lambda\)
• The second equation: \(q_1 + 2q_2 = \lambda\)
• The third equation: \(q_1 + q_2 = 200\)

To solve this system:1. Equalizing the first two equations helps us remove \(\lambda\) and focus on \(q_1\) and \(q_2\). Subtracting leads to \(3q_1 - q_2 = 0\), hence \(q_2 = 3q_1\).2. Substitute \(q_2 = 3q_1\) in the production constraint \(q_1 + q_2 = 200\) yielding \(4q_1 = 200\) and thus \(q_1 = 50\).3. Using \(q_1 = 50\), determine \(q_2\): \(q_2 = 150\).Solving the system provides the optimal production quantities that minimize cost while satisfying the total output required, assisting strategic decision-making for efficient resource allocation.