Optimization of the area of a triangle

[Jebus_Chris]

Homework Statement

What is the maximum area of an equilateral triangle and a square using only 20ft of wire?

Homework Equations

$$20=4x+3y$$
$$x=\frac{20-3y}{4}$$
$$A=x^2+\frac{1}{2}y^2\sqrt{3}$$

The Attempt at a Solution

So then
$$A=\frac{400-120y+9y^2}{16}+\frac{y^2\sqrt{3}}{4}$$
$$\frac{dA}{dy}=\frac{9y+4\sqrt{3}y-60}{8}$$
When you set the derivative to zero shouldn't y=0 since the max area should be when the square has the largest possible side length,x?

 

[LCKurtz]

When you write a solution you should indicate what your variables represent. If x is the length of the side of the square and y is the length of the side of the triangle, you should begin by saying so.

Your formula for the area of the triangle isn't quite right.

Once you fix that and get your area in terms of y you need to state the possible values of y: ? ≤ y ≤ ? that make sense in the problem.

Then you need to remember where possible extreme values of a function can occur:

1. Places where y' = 0 in the interval.
2. Places where y' doesn't exist in the interval, if any.
3. End points if the interval has them.

 

[Jebus_Chris]

I realize now when I set the derivative equal to zero I found the minimum.

Whats wrong with the triangle's area?
$$A=\frac{1}{2}bh=\frac{1}{2}y*\frac{y\sqrt{3}}{2}$$

 

[LCKurtz]

Nothing, now that you corrected it.