Problem 37 - Calculus an Applied Approach

Question

A manufacturer has an order for 2000 units of all-terrain vehicle tires that can be produced at two locations. Let \(x_{1}\) and \(x_{2}\) be the numbers of units produced at the two plants. The cost function is modeled by \(C=0.25 x_{1}^{2}+10 x_{1}+0.15 x_{2}^{2}+12 x_{2}\) Find the number of units that should be produced at each plant to minimize the cost.

Step-by-Step Solution

Answer

The optimal production quantities of the all-terrain vehicle tyres should be about 352 for \(x_{1}\) and about 1648 for \(x_{2}\). The associated minimum cost is about $26704.

Step 1: Expression of the Total Number of Units

Express the total number of units as \(x_{1} + x_{2} = 2000\)

Step 2: Formulate the Lagrange's function

The function is \(L(x_{1}, x_{2}, λ) = 0.25x_{1}^{2} + 10x_{1} + 0.15x_{2}^{2} + 12x_{2} - λ(x_{1} + x_{2} - 2000)\)

Step 3: Solve for partial derivatives and equal them to zero

For \(x_1\): \(\frac{\partial L}{\partial x_1}=0.5x_{1}+10-λ=0\), and for \(x_2\): \(\frac{\partial L}{\partial x_2}=0.3x_{2}+12-λ=0\), and for \(λ\): \(\frac{\partial L}{\partial λ}= x_1+ x_{2}-2000=0\)

Step 4: Solve the system of equations obtained in step 3

First, equate the first two equations from step 3 to get: \(0.5x_{1} + 10 = 0.3x_{2} + 12\). Simplifying gives the equation \(0.2x_{1} - 0.3x_{2} = 2\). Then, combine this equation with the third equation from step 3, \(x_{1} + x_{2} = 2000\), to find the values of \(x_{1}\) and \(x_{2}\).

Step 5: Find the minimum

Substitute the obtained values of \(x_{1}\) and \(x_{2}\) in the original cost function to verify if it equals the minimum cost.

Key Concepts

Lagrange Multipliers

In optimization problems that involve constraints, Lagrange multipliers provide a systematic way to find the maximum or minimum values of a function. This technique introduces an auxiliary variable, λ, which helps incorporate constraints directly into the objective function. Consider a situation where a function needs to be maximized or minimized subject to one or more constraints. Instead of directly working with the constraints and the function separately, we create a new function called the Lagrangian.
The Lagrangian, generally denoted by L(x, λ), combines the original function and the constraints using the Lagrange multiplier λ. For our tire production problem, this is given as:

• Objective Function: Represents the cost to be minimized, which is the cost function in our exercise.
• Constraint: Represents the total number of units that must sum to 2000.

By solving the system of equations derived from the Lagrangian, we find the stationary points, giving us values for the production numbers that minimize the total cost while satisfying the constraint.

Cost Function Minimization

Cost function minimization is a fundamental aspect of operations management and is crucial when trying to minimize production costs while meeting certain requirements, such as production quotas. The cost function represents the total cost associated with producing a certain number of units, and its minimization involves finding the least expensive way to produce those units.
Here, the cost function is given by: \[ C = 0.25 x_{1}^{2} + 10 x_{1} + 0.15 x_{2}^{2} + 12 x_{2} \] where \(x_{1}\) and \(x_{2}\) are the units produced at two different plants.

• The goal is to find the values of \(x_{1}\) and \(x_{2}\) that result in the minimum value of C, while still producing exactly 2000 units in total.

This involves balancing the cost contributions from both plants to achieve the lowest possible combined expense. Cost function minimization uses calculus and optimizational techniques, such as taking derivatives and applying constraints, to arrive at the optimal solution.

System of Equations

Solving a system of equations is a method used to find values that satisfy multiple conditions simultaneously. In our tire production problem, the system of equations arises from applying the method of Lagrange multipliers.
We derive several key equations by taking the partial derivatives of the Lagrangian function. The primary equations from our exercise are:

• \( 0.5x_{1} + 10 = λ \)
• \( 0.3x_{2} + 12 = λ \)
• \( x_{1} + x_{2} = 2000 \)

These equations represent equilibrium conditions where both the constraint and the cost functions are optimally balanced.
By solving these simultaneously, we can determine the values of \(x_{1}\), \(x_{2}\), and \(λ\) that minimize the cost while fulfilling the production requirement. Efficiently solving this system may involve algebraic manipulation, substitution, or graphical methods, depending on the complexity.

Partial Derivatives

Partial derivatives play a critical role in optimizing multivariable functions, particularly in constrained optimization problems. When we deal with a function of several variables, the partial derivative with respect to one variable gives us the rate of change of the function with respect to that variable, while keeping other variables constant.
In our problem of tire production, we take the partial derivatives of the Lagrangian function with respect to \(x_{1}\), \(x_{2}\), and \(λ\). This process allows us to:

• Identify how changes in production units at each plant affect the overall cost.
• Formulate conditions (equations) that each must satisfy for cost minimization.

The goal is to find where each of these derivatives equals zero, which indicates stationary points that are potential minima (or maxima).
This strategic use of partial derivatives helps in assessing the interaction of costs at both production facilities, ensuring that the distribution of production leads to the lowest cost possible. Through careful calculation and comparison of these derivatives, we reach the solution to the optimization problem.