30cm long wire is bent to form 2 shorter sides of a right-angled triangle. What is the maximum area?

Question

30cm long wire is bent to form 2 shorter sides of a right-angled triangle. What is the maximum area?

Answer 1

Let the sides be a and b.  Then, a + b = 30, and we are trying to maximize (a*b)/2.  Intuition tells us that the product of two numbers with a fixed sum is maximized when those two numbers are equal, but let's try to prove this.

Method 1:

The arithmetic mean-geometric mean inequality (AM-GM) states that:

[(a+b)^2]/8 >= (a*b)/2 for any positive integers a and b (try to prove this yourself).

In this case, it says that:

900/8 >= (a*b)/2, so

112.5 >= Area.

This gives us an upper bound for the area of the triangle, which can clearly be obtained by taking a = b = 15.  In addition, equality holds in AM-GM if and only if a = b, so the only time that this maximal area of 112.5 can be achieved is when the two legs have equal length.

Method 2:

a + b = 30, so b = 30 - a.  Then, a * b = a * (30 - a) = 30a - a^2.  If we consider the area as a function of a, then 2*Area = f(a) = 30a - a^2, so to maximize f(a) we let f '(a) = 30 - 2a = 0.  This gives that a = 15, so the maximal area of (30 * 15 - 15^2)/2 = 112.5 is achieved with an isosceles right triangle.

Answer 2

The maximum area is (30/2)^2/2 = 225/2 = 112.5 cm^2

112.5 cm^2