Answer to Question #124160 in Calculus for joe chegg

Question

A wire of length 100 centimetres is cut into two pieces. One piece is bent to form a square.

The other piece is bent to form an equilateral triangle. Find the dimensions of the two pieces of wire so that the sum of the areas of the square and the triangle is minimized.

Answer 1

Let the amount cut for the square be x, amount for the triangle - 100-x

Side length of the square is x/4, area=(x/4)²=x²/16

Area of the triangle with side length a is $a^2\sqrt 3$ /4

Side length of the triangle is (100-x)/3

Area of the triangle=$(\sqrt3/4)$ ((100-x)/3)²=$\sqrt3(100-x)^2/36$

Total area=area of square+area of triangle

A(x)=x²/16+$\sqrt3(100-x)^2/36$

Differentiate:

A'(x)=x/8-$\sqrt3(100-x)/18$

solve the equation A'(x)=0

x/8-$\sqrt3(100-x)/18=0$

$x/8=\sqrt3(100-x)/18$

$18x=8\sqrt3(100-x)$

$18x=800\sqrt3 -8\sqrt3x$

x=$(800\sqrt3)/(18+8\sqrt3)$ =43.5

For minimum area, 43.5m should be used for square and 100-43.5=56.5 should be used for the equilateral triangle