Maximum and Minimum / Optimization...?

Question

I am confused as to how to go about this problem. We are studying optimization in calculus. Thank you for any help!

A piece of wire 10 m 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 total area enclosed is (a) maximum and (b) a minimum.

Answer 1

Let L be the length of wire used for the triangle and ( 10 - L ) the length used in the square. The areas are then

( ( 1 / 4 ) ( 10 - L ) )² + ( 1 / 2 ) ( L ) ( L sin 60 )

where ( 1 / 4 ) ( 10 - L ) is the length of one side of the square, L the base of the triangle, and L sin 60 the altitude of the triangle.

Multiply it out and simplify to get a quadratic equation in L. Then to find the max and min, set the first derivative of that function equal to zero and solve for L.