[SPLIT] A piece of wire 10m long is cut into two pieces....

[xtrmk]

A piece of wire 10m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. How should the wire be cut so that the enclosed area is a minimum? Find the minimum area.

Area of a square= s^2
Area of eq. triangle = [root(3)/4]s^2

How do I proceed from here?

 

[galactus]

I'll get you started with this one. OK?.

As can be seen from the diagram, the area of the square and triangle is \(\displaystyle \L\\s^{2}+\frac{\sqrt{3}}{4}w^{2}\)

Because x is the perimeter of the square and y is the perimeter of the triangle:

\(\displaystyle \L\\s=\frac{x}{4}\) and \(\displaystyle w=\frac{y}{3}\)

\(\displaystyle \L\\A=(\frac{x}{4})^{2}+\frac{\sqrt{3}}{4}(\frac{y}{3})^{2}\)

Now, the length of the wire is \(\displaystyle x+y=10\Rightarrow{y=10-x}\)

Sub into A:

\(\displaystyle \L\\A=(\frac{x}{4})^{2}+\frac{\sqrt{3}}{4}(\frac{10-x}{3})^{2}\)

I mislabeled the side of the triangle as x. That should be a w.

Now, simplify, differentiate, set to 0 and solve for x. You should be able to find your dimensions from there. Okey-doke.