The mythical origins of the calculus of variations lie in "Queen Dido's Problem." Queen Dido was told she could rule the area covered by the hide of a cow. Dido had the hide cut into the thinnest string possible, then stretched the string to surround her new kingdom. Her problem was to determine how she should lay out the string to get the largest kingdom possible. That's a problem in the calculus of variations. The idea is to take a known relationship, a known constraint, and then find the value of the parameters that gives the largest or smallest value. For example, the path of the string is the variable, the length of Dido's string is the constraint, and the goal is to maximize the area.
The same kind of mathematics is used to minimize cost. A business might have costs that change as more raw materials are required, more electricity is used, more workers are needed, equipment is used more efficiently, and overhead operations are consolidated. Those costs will be modeled by an equation that shows how costs change with the number of units produced. The idea is to minimize the cost, and the calculus of variations shows how.
Although the proofs can be somewhat involved, the procedure is pretty simple. Take the first derivative of the cost function and set it equal to zero. Where it is equal to zero, the function has either a minimum or a maximum. Then take the second derivative of the cost function. Evaluate the second derivative at the point where the first derivative is equal to zero. If the second derivative at that point is positive, then the point is a minimum. If the second derivative is negative, then the function has a maximum at that point.
Farmer Bob needs to fence in 400 square meters of his field. He needs to build two opposite sides with barbed wire, which costs $1 a meter. The other two opposite sides will be a board fence costing $2 a meter. The length*width=400, and the cost is 2*length*$1 + 2*width*$2. Substituting, the cost can be rewritten as
cost = 2 * (400/width) + 4 * width.
The first derivative of the cost with respect to width is
-800/width^2 + 4, and setting this equal to zero, the width is 14.1 meters; that represents the width of either the most expensive or least expensive fence.
The second derivative is (-1/2)*(-800/width^3). When the width is 14.1 meters, the second derivative is 0.14, which is positive; therefore, 14.1 meters represents the minimum cost.
To complete the solution, you can calculate what that cost is: 800/14.1 + 4*14.1, or about $113.