Calculus Optimization help?

Question

An isosceles triangle has a perimeter of 6 cm. What is the maximum possible area of such a triangle?

Answer 1

I have a hunch the maxumim area will be that of the eqilateral triange with each side being 3, but for the sake of math...

Drawing a diagram is a good way to understand, but since I can't do that here...

if it's isosclees two sides are the same, you can label each one as "x". If the perimeter is 6, you can label the third side as 6-2x. Now, the equation for the area of a triangle is 1/2(base(height)). the base can easily be the 6-2x side, the height can be found by splitting the triange down the middle and using the pythagorean theorm:

height^2 = (3-x)^2 + x^2 so height = the square root of that. ( can see this getting messy to try and type...)

So, you have A = 1/2(6-2x)[(3-x)^2 + (x^2)]^1/2.

Now the messy part is taking the derivative of all that, which seems to require a double chain rule and a product rule....

remember, product rule is dxy+xdy.

A' = -1[(3-x)^2 + x^2]^1/2 + [3-x][1/2[(3-x)^2 + x^2]^-1/2[2(3-x)(-1)+ 2x]

Simplifying....

= (6-4x)/2 [(3-x)^2 + x^2]^1/2 [(3-x)^2 + x^2]^-1/2

and further...

= (6-4x)/2 (like terms and the -1/2 and 1/2 cancel eachother out)

set that equal to zero and x=3/2.

So....your sides are 3/2, 3/2, and 3. (Guess i was wrong in my prediction). plugged back into the original A= equation, the area is...

(3-3/2)[(3-3/2)^2 + (3/2)^2]^1/2

= 3/2(9/4 + 9/4)^1/2

= 3(square root of 6)


Man...That's a load of work for such a simple ansswer :P

Answer 2

Hi. Let y be the lenght of the base, and x the lenght of any of the sides (the equal ones). Also, let h be the altitude of the triangle to the base. Lengths in cm. So, you know that:
1.- 2x+y=6
2.- h^2=x^2-(y/2)^2
3.- y*h/2=Area
Now, you can express the Area function only in terms of "y", using equations (1) and (2). So, to find the maximum possible area, you have to optimize the Area function. That can be done taking the derivative of the Area function, then set it equal to 0 and find "y", find the Area, and check the second derivative to see if that was a maximum. I think you should use some kind of numeric approach to solve equations (for example, try Matlab, or Maple, or make yourself a code).
This is a rough answer, because I think you should look deeper into the problem, and learn. For example, ask yourself why is valid (and if it`s possible) to use the way I described to find the maximum with that functions

Regards.

Answer 3

dukebdevil93 is almost certainly right, but I'd like to work it out. Let x be the equal sides, so the base must be 6-2x, and half the base is 3-x.

An altitude splits the base in half and gives us a right triangle, where

h = √( x² - [3-x]²) = √( x² - 9 + 6x - x²) = √(6x - 9)

so the area, A, is (3-x) • √( 6x - 9), and we want to maximize it, so we find the derivative and set it equal to 0. We could use the product rule, or we could square the 3-x and move it under the √. Doing that gives us

A = √( [9 - 6x + x²][6x - 9])
A = √( 6x^3 - 45x² + 108x - 81)
A' = (18x² - 90x + 108) / [2√ all that stuff]
0 = 9x² - 45x + 54 / √stuff
0 = x² - 5x + 6
0 = (x - 2)(x - 3)

x = 3 for the equal sides leaves nothing for the base, so
x = 2 is our solution, and dukebdevil was right.

Answer 4

General plan:

1. Express the area of the triangle as a function of a single variable. This will turn out to be differentiable.

2. Solve for where the derivative equals zero. This will turn out not to happen, or else only at the endpoints of the interval on which the function is defined.

3. So the maximum is at one of the endpoints of the interval. It will be easy to check which one that is.

Answer 5

the max would be found when the triangle is at it's "widest" possible setting so, it would be a n equalateral triangle of sides 2,2,2. In this case the height is sqrt(3) and the area is 1/2*2*sqrt(3) = sqrt(3)

this is the max, the min goes all the way down to zero (imagine a really skinny triangle of sides almost 3, almost3, and barely more than zero)