Problem 47 - Calculus an Applied Approach

Question

Construction A rancher plans to use an existing stone wall and the side of a barn as a boundary for two adjacent rectangular corrals. Fencing for the perimeter costs dollar 10 per foot. To separate the corrals, a fence that costs dollar 10 per will divide the region. The total area of the two corrals is to be 6000 square feet. (a) Use Lagrange multipliers to find the dimensions that will minimize the cost of the fencing. (b) What is the minimum cost?

Step-by-Step Solution

Answer

To arrive at the short answer, you have to solve the system of equations you derived by setting the derivatives of the Lagrangian to zero. The solution to those equations would give you the dimensions of the corrals. The minimum cost is achieved when you substitute those dimensions into the original cost equation.

Step 1: Formulate the equations

Let the dimensions of the two corrals be \(x\) ft, \(y\) ft, and \(z\) ft, where \(x\) is common to both. So, the total cost, \(C\), is given by \(C= 10(3x + y + z)\) because the cost is $10 per foot. And, the total area \(A\), is \(A = 2xz + yz = 6000\) given.

Step 2: Use Lagrange Multipliers

We can solve these equations using Lagrange multiplier, where the method is used to find the local maximum and minimum of a function subject to equality constraints. The Lagrangian function will be \(L(x,y,z,λ) = 10(3x + y + z) + λ(2xz + yz - 6000)\). Now, differentiate \(L\) with respect to \(x\), \(y\), \(z\), and \(λ\) and set each equal to zero.

Step 3: Evaluate derivatives and solve

The derivatives are \(∂L/∂x = 30 + λ(2z) = 0, ∂L/∂y = 10 + λz = 0, ∂L/∂z = 10 + λ(2x + y) = 0\) and \(∂L/∂λ = 2xz + yz - 6000 = 0\). Solving these equations simultaneously, you will find the values of \(x\), \(y\), and \(z\).

Step 4: Find the minimum cost

Substitute the found values of \(x\), \(y\), and \(z\) into the cost equation. This will give the minimum cost.

Key Concepts

Optimization Problems

Optimization problems are all about finding the best possible solution under a given set of conditions. In this rancher exercise, our main goal is to minimize the cost of fencing while keeping a fixed area for the two corrals. Here are a few key things to remember about optimization problems:

• They usually involve finding either a maximum or a minimum value. In this case, we're looking for the minimum cost.
• Constraints such as the fixed area of 6000 square feet must always be considered when determining the optimal solution.
• Optimization problems can have various solutions, which make different assumptions or operate under different constraints.

When solving optimization problems, we often deal with a combination of equations—one that defines what we want to optimize (the cost here) and one (or more) that lay out the constraints. Being comfortable with setting up and solving such equations is a crucial part of coming up with an optimal solution.

Calculus

Calculus is the mathematical study of continuous change and is essential for solving optimization problems like the one in this exercise. This is because calculus provides us the tools to calculate rates of change and study functions in detail. Here's how calculus is used in this context:

• Differentiation helps us find the rates at which our functions change. By setting derivatives to zero, we can identify points of minimum cost (or maximum area, etc.).
• Understanding derivatives is key to using the Lagrange multiplier method, which finds the extremum of a function subject to constraints.
• Partial derivatives are particularly important when dealing with functions of multiple variables, as they allow us to isolate the effect of individual variables.

In this exercise, we differentiate the Lagrangian function concerning each variable to uncover points that minimize the overall function, respecting the given constraint. By setting each derivative equal to zero, it's possible to find values that give the lowest cost for fencing the corrals.

Constrained Optimization

Constrained optimization deals with finding the optimum value (either maximum or minimum) of a function subject to certain limitations or restrictions. In our example, the Lagrange multiplier method is used to tackle this problem. Here's what you should keep in mind about constrained optimization using this method:

• The main idea is to introduce a new variable, often called the "Lagrange multiplier," to incorporate the constraint directly into the optimization process.
• A Lagrangian function is formed by combining the objective function (like minimizing cost) and the constraint (fixed area in this case) using the Lagrange multiplier.
• By taking partial derivatives of this Lagrangian function, we account for both the objective and the constraint simultaneously, ensuring that our solutions satisfy both conditions.

The beauty of this approach lies in its power to simplify complex problems with constraints, turning them into manageable sets of equations, which can then be solved to find optimum solutions.