A piece of wire 15 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. (Give your answers correct to two decimal places.)

Answer 1

desmos.com/calculator/i7y6qygmi1

The graph shows the total area function and absolute max and min on closed interval from 0 to 15.

Answer 2

Let we cut x m for square, then (15 - x) m left for triangle. Then total area A = (x/4)² + (15 - x)²/9·√3/4 =

x²/16 + (15 - x)²·√3/36; dA/dx = x/8 - (15 - x)√3/18 = 0; 9x - 4(15 - x)√3 = 0; 9x + 4x√3 = 60√3;

x = 60√3/(9 + 4√3) = 6.524468 m

(b) So minimum total area min A will be when x = 60√3/(9 + 4√3) m = 6.524468 m

(a) Ma ximum then will be on one of the ends interval [0, 15]

x = 15; A = 225/16 = 14.0625

x = 0 A = 225√3/36 = 10.82532

So, maximum area if all wire we use for square