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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 |
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Answer» 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` |
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