Twenty feet of wire is use to form an equilateral triangle a

Twenty feet of wire is use to form an equilateral triangle and a square. How much of the wire should be use for the equilateral triangle and how much of the wire should be use for the square such that the sum of the areas of the equilateral triangle and the square is maximum.

Solution

Let S be the number of feet of wire used to form the triangle.

Then (20 - S) is number of feet of wire used to form the square.

Each side of the square is (20 - S)/4 feet long.

The area of the square is ((20 -S)/4)^2 = (400 - 40S + S^2)/16 square feet
Each side of the triangle is S/3 feet.

By Pythagoras (the altitude of the triangle)^2 + (S/6)^2 = (S/3)^2
so the altitude = Ssqrt (1/12)

the area of the triangle = (1/2)(S3)(Ssqrt(1/12) = S^2/6)sqrS(1/12)
the total area of both figures is
(400 - 40S + S^2)/16 +S^2/6)sqrt(1/12)

Graphing this function shows that the maximums are at the extreme values forS, so if S= 20 (all the wire in the triangle) the area is 14.1 square feet
if S= 0 (all the wire in the square) the area is 25 square feet.
So the maximum area is enclosed when all 20 feet of the wire are made into a square.
there foreThe total area is at a MINIMUM at t 11.3 feet .