# Homework Help: Maximizing area of a triangle

1. Oct 25, 2006

### donjt81

So here is the problem. I did it but just wanted to make sure it is correct... does the solution look ok to you guys.

problem: an Isosceles triangle has its vertex at the origin and its base parallel to the x-axis with the vertices above the x-axis on the curve y = 27-x^2. find the largest area the triangle can have.

solution:So I assumed the base of the triangle will be the points (x, y) and (-x, y).

Therefore height of the triangle will be y
length of the base will be x+x = 2x

Area = 1/2 * base * height
= 1/2 * 2x * y
=xy

now we know that y = 27 - x^2
so this area eqn becomes
A = x(27 - x^2)
A = 27x - x^3
now
A' = 27 - 3x^2 = 0
so the critical points are x = 3 and x = -3

using the sign chart i figured out that i have to use x = 3 because that is the local maximum.

and when I plug x = 3 back into A = x(27 - x^2) we get A = 54 unit^2

Can some one verify that my solution is correct? I would really appreciate it.

2. Oct 25, 2006

### Andrew Mason

Correct analysis and correct result.

AM

3. Oct 25, 2006

### StatusX

One small point is that you should be more careful about what range you're trying to find a maximum over. For example, at x=-10, A=+730, so why isn't this a better solution? The reason is that you're looking only for solutions with 0<=x<=r, where r is the smallest positive root of 27-x^2. You should compare all local maxima in this range along with the values at the endpoints and pick the largest of these values, which will give what you found.