The often-quoted prime example of a problem in variational calculus is stated in Queen Dido's Problem. The mythical queen was given the ability to rule over whatever land she could cover with the hide of a cow. She took the cowhide and cut it into thin strips, tying them together to make a long loop. Her problem was to take that loop and use it to outline the largest possible kingdom. That's a problem in the calculus of variations.
In the late 1700s, Isaac Newton set about the problem of designing a body that would have minimum resistance when flying through the air. This is the earmark of variational problems. There is a quantity --- air resistance --- that depends on another set of parameters --- the shape of the surface. The idea is to find the set of parameters that minimizes or maximizes the quantity. The first step is to express the quantity to be minimized as a function of the parameters of the problem. That first step requires you to apply your knowledge of the problem; then you can apply the techniques of variational calculus.
A rancher is bringing in some sheep he wants to keep separate from the rest of his animals. He has a straight irrigation canal he will use as one side of an enclosure, and he has one kilometer of fence for the remainder. He wants to fence off the largest possible pasture. How should he lay out his fence?
First, you apply your background knowledge to express the quantity to be maximized. The area is equal to the length multiplied by width of the enclosed area. In addition, the length + 2 multiplied by width equals one kilometer. That leads to the equation:
area = width - 2 multiplied by width^2.
Now the expression is ready for you to apply variational calculus techniques.
The simplest form of optimization is the one in this problem. The quantity to be maximized is an expression of a single variable, and all you need to do is find one maximum. The simple process is to take the derivative of the quantity to be maximized with respect to the variable. In this case, the derivative of the area with respect to width.
d(area)/d(width) = 1 - 4 multiplied by width.
Set that equal to zero; this will identify an inflection point --- a place where the original function changes its direction. Setting the derivative of this problem equal to zero, you find that the width equals 1/4 of a kilometer.
The process outlined in the previous paragraph identifies an inflection point in the curve of the quantity to be maximized, but it hasn't yet distinguished a maximum from a minimum from a "knee" in the curve. Another test is necessary. Take the second derivative of the curve. If it is negative at the point where the first derivative is zero, the curve is at a maximum. If the second derivative is positive there, then the curve has a minimum. If the second derivative is zero, then there's neither a maximum or minimum of the original curve at that point.
For the example problem, the second derivative is -4 --- less than zero --- which means that the area is maximized when the width is 1/4 of a kilometer. Just to finish off the problem, if the width is 1/4 km, the length is 1/2 km, and the area is 1/8 km^2.
There are scores, if not hundreds, of textbooks on variational calculus. The example here represents the simplest type of problem. Other problems that can be solved with variational calculus include such things as: finding the 1-gallon package that uses the least amount of material; finding the path light takes through a lens; determining the right amount of inventory to keep on hand to minimize total costs; designing shape to insert in front of a jet engine to make the air intake as smooth as possible. Those examples, of course, are representative of a much larger field of problems accessible to the techniques of the calculus of variations.