# Thread: Global Maximum for area of a triangle.

1. ## Global Maximum for area of a triangle.

Hello,

I need to find the length of the base of a (isoceles) triangle (x) to get the maximum area in terms of the fixed sides (L)

so far I have found the critical points of the derivative

Area = (1/2) xh

Where h = ( L^2 - (x^2)/4)^1/2

L = Fixed lengths of triangle = constant

So when derivative = 0, we have
Area' = x = L* (2)^1/2

I just need to know how i can test if this is a global maximum of the fucntion.
do i just find the limits when x goes to 0 and infinity? What does it mean when i do?

Cheers

2. i think taking the second derivative allows you to ascertain whether it is a global min or max, like, if the second derivative is positive, i think its minimum and if the second deriv is negative, i think it will be a global max

3. No, that only tests for local max and min. To find Global max and min just calculate the area at each critical point and at the "endpoints", the base being 0 or $\displaystyle \infty$ (those are easy- the area is 0 so clearly not a maximum). The largest of those is the global maximum.

But I have no idea what you mean by "Area'= x= L*(2)^1/2" At the maximum area, Area'= 0, of course. Are you saying the maximum area occurs when x= L*(2)^1/2? If so, that is not what I got.

4. I'm getting the same answer as the original poster, i.e., $\displaystyle x=L \sqrt{2}$.

By the way, the the bounds of the possible values for x would be $\displaystyle 0<x<2L$. You may recall that the sum of the lengths of two sides of a triangle must be greater than the length of the third side.

In any case, you've determined that there is only one value of x where the derivative equals zero, and that the derivative is continuous, so we know that if the function has a global maximum, it will occur at $\displaystyle x=L\sqrt{2}$.

You have several ways to show that a local maximum occurs (which also proves that a global maximum occurs because there is only one critical point and the function is continuous). The method of finding the limit as $\displaystyle x \to 0^+$ and $\displaystyle x \to 2L^-$ will work. You just need to demonstrate that the areas at these endpoints are less than the area you get at $\displaystyle x=L \sqrt{2}$. Like HallsofIvy said, though, it's zero at the endpoints, so it's clear that the solution found is in fact a maximum.