Question

A wire of length 36cm is cut into the two pieces, one of the pieces is turned in the form of a square and other in form of an equilateral triangle. Find the length of each piece so that the sum of the areas of the two be minimum

Answer 1

Correct Answer - `(144 sqrt3)/(4 sqrt3 + 9) cm, (324)/(4 sqrt3 + 9) cm`
Let the perimeter of the square be x cm
Then the perimeter of the triangle is `(36 -x)cm`
`:.` side of the square `= (x)/(4)` cm
And, side of the triangle `= (1)/(3) (36 -x) cm`
`:. A = (x^(2))/(16) + (sqrt3)/(4) (12 - (x)/(3))^(2) = (x^(2))/(16) + (sqrt3)/(4) (144 + (x^(2))/(9) - 8x)`
`rArr A = ((sqrt3)/(36) + (1)/(16)) x^(2) - 2 sqrt3x + 36 sqrt3`
`rArr (dA)/(dx) = ((4 sqrt3 + 9))/(144) xx 2x - 2sqrt3 and (d^(2)A)/(dx^(2)) = (4 sqrt3 + 9)/(72) gt 0`
`:. (dA)/(dx) = 0 hArr x = (144 sqrt3)/((4 sqrt3 + 9)) cm`