If a triangle has two sides with lengths 3 and 4, what should the length of the third side be in order to maximize the area of the triangle?
This is the type of problem that calculus was designed to solve:
- If a triangle has side lengths a and b, and the angle between them is θ, then the area of the triangle is ab sinθ
- To find the maximum, find the value of θ such that the derivative of the area is equal to zero: (ab sinθ)’ = ab cosθ = 0
- There are multiple solutions, but only 90° is a valid solution for an angle of a triangle – therefore it’s a right triangle and the third side c = sqrt (a2 + b2) = 5
But calculus is not really necessary for this problem.
We also know a formula for the area of a triangle to be 1/2 the base times the height.
Let’s say we pick the side of length 4 as the base. We can draw all sorts of triangles with a base of 4 and a second side of length 3, and we would quickly realize that the height of that triangle can never exceed the length of the remaining two sides (in this case, that means the height cannot be greater than 3). Since the base is fixed at 4, that means the maximum area of the triangle is when the height is 3, which again gives us a right triangle with side lengths 3, 4, and 5.