Question

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 \(\$ 10\) per foot. To separate the corrals, a fence that costs \(\$ 4\) per foot 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?

Short answer

The minimum cost is calculated by substituting the values of \(x\), \(y_1\), and \(y_2\) obtained from the Lagrange multiplier equations into the cost function. The dimensions that will minimize the cost of the fencing are obtained from the Lagrange multiplier equations.

Step-by-step solution

Formulate the Cost Function

First, denote the common length of the two corrals as \(x\) and the widths as \(y_1\) and \(y_2\) respectively. The cost of the outer fence is \$10 per foot, and hence, for the two corrals, it is \$10 times the perimeter, \[C = 10 * 2 * (x + y_1 + y_2)\] where \(x\) is common length and \(y_1\) and \(y_2\) are the widths. For the fence that separates the two corrals, the cost is \$4 per foot, therefore, the total cost for it is \[4 * x\] Hence, the total cost function to be minimized is \[C = 20x + 20y_1 + 20y_2 + 4x\]Simplify the expression and say \(C = 24x + 20(y_1 + y_2)\)

Formulate the Constraints

Given in the problem statement, the total area of the two corrals is 6000 square feet. Therefore, the constraining equation can be written as \[Area = x * y_1 + x * y_2 = 6000\]

Using Lagrange Multipliers

To find the minimum cost using Lagrange multipliers, equate the derivative of the function with the derivative of the constraint multiplied by \(\lambda\), a Lagrange multiplier. The three equations are: \[\frac{\partial C}{\partial x} = \lambda \frac{\partial Area}{\partial x}\]\[\frac{\partial C}{\partial y_1} = \lambda \frac{\partial Area}{\partial y_1}\] \[\frac{\partial C}{\partial y_2} = \lambda \frac{\partial Area}{\partial y_2}\] Solving these equations gives values for \(x\), \(y_1\), and \(y_2\) that minimize the cost.

Find the Minimum Cost

Substitute the obtained values of \(x\), \(y_1\), and \(y_2\) into the cost function to find the minimum cost.

Key concepts

Cost Minimization

Cost minimization is a critical concept in economics and business where one seeks to find the point at which costs to produce goods or services are the lowest. When dealing with problems such as the one faced by the rancher constructing corrals, cost minimization involves finding the dimensions of the corrals that will require the least amount of fencing, thereby keeping expenses as low as possible. Efficient resource allocation means achieving the desired outcome, such as the area of the corrals, while incurring the smallest possible costs.

By setting up a cost function that includes both the cost per foot of the outer fence and the dividing fence, along with the given constraints such as the fixed total area, we can apply calculus-based optimization techniques—namely, Lagrange multipliers—to determine the most cost-effective solution.

Rectangular Corral Area

The area of a rectangle, which in our scenario represents the corral, is given by the product of its length and width. For two adjacent rectangular corrals sharing a common divide, their total area will be the sum of the individual areas: hence, if the length is represented as \(x\) and the widths as \(y_1\) and \(y_2\), then the total area (\(A\)) is \(A = x \times y_1 + x \times y_2\). The rancher's requirement for a total area of 6000 square feet sets a constraint on the optimization problem—this area is not to be exceeded. Therefore, we must consider both the cost and area simultaneously when aiming for cost minimization, making it a constrained optimization problem.

Optimization with Constraints

Optimization with constraints deals with finding the best possible solution under given limitations. In mathematical terms, constraints are additional conditions that the solution to an optimization problem must satisfy. For the rancher's problem, the constraint is that the sum of the areas of the two corrals must equal a specific value, that is, 6000 square feet. Constraints make optimization problems more complex but also more reflective of real-world scenarios in which resources and conditions are often limited. The method of Lagrange multipliers is a powerful tool in such situations. It enables us to include those constraints directly in the optimization process by incorporating them into the cost function through an additional variable, the Lagrange multiplier.

Partial Derivatives

Partial derivatives are employed in mathematics to analyze the rates of change of multi-variable functions. They are fundamental to the process of optimization, especially when dealing with constraints. In the context of the rancher's problem, the cost function depends on three variables: \(x\), the common length, and \(y_1\), \(y_2\), the two widths. To minimize this function, we need to understand how changes in each of these variables affect the total cost.

By computing the partial derivatives of the cost function and setting them equal to the respective partial derivatives of the constraint function (multiplied by the Lagrange multiplier), we can solve for the optimal values of \(x\), \(y_1\), and \(y_2\). These partial derivatives give us the necessary conditions for a cost minimum under the constraint of the total corral area.