A piece of wire 40 meters long is to be split into two length. One of the lengths will form an equilateral triangle and the other length will form a square. Cal

Question

A piece of wire 40 meters long is to be split into two length. One of the lengths will form an equilateral triangle and the other length will form a square. Calculate the minimum possible total area of the two figures, correct to one decimal place. Please explain step by step 

Answer 1

 X is the portion of the wire used for the equilateral.

40-x is the portion of the wire used for the square. 

 

The combined perimeter is 40

 

sqrt(3)/4*(X/3)^2 +  ((40-X)/4)^2 = A(x) is the area

 

sqrt(3)/36 * x^2 + (160 - 80x + x^2)/16 = A(x)

 

 

sqrt(3)/36 * x^2 + (1/16)x^2 - 5x + 10 = A(x)

 

8*( 1 + sqrt(3))/144 * x^2  - 5x + 10 = A(x)

 

0 = dA/dx = A'(x) = (1+ sqrt(3))/9 * x - 5

 

x = 45/(1+ sqrt(3))  = 45(1 - sqrt(3)) / ( 1 + sqrt(3))(1 - sqrt(3))

 

                               = 45( 1 - sqrt(3))/ ( 1 - 3)

 

                                = 45 ( 1 - sqrt(3))/(-2)

 

                                = 45( sqrt(3) - 1)/2 

 

                                which is approximately 16.471143....