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A wire of length 28 m is to becut into two pieces. One of the pieces is to be made into a square and theother into a circle. What should be the length of the two pieces so that thecombined area of the square and the circle is minimum? |
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Answer» Let r be the radius of the circle and x be the side of square. `:. ` Circumference of circle + perimeter of square = 28 cm ` rArr 2 pi r + 4x =28` `rArr x = (14 - pi r)/2 ` ….(1) and let area of circle + area of square = A `:. A = pi r^(2) + x^(2)` ` = pi r^(2) + ((14 - pi r)/2)^(2)` [From equation (1)] ` = pi r^(2)+ 1/4 (pi^(2)r^(2)- 28 pi r+ 196)` ` rArr (dA)/(dr) = 2 pir+1/4(2pi^(2)r-28pi)` and ` (d^(2)A)/(dr^(2) = 2 pi + 1/4 (2pi^(2))` For maxima/minima ` (dA)/(dr) = 0` ` rArr pir+1/4 (2pi^(2)r-28pi)=0` ` rArr pir+pi^(2)r-14pi =0` ` rArr r(4+pi)=14` ` rArr r=14/(pi+4)` at ` r=14/(pi+4)` ` (d^(2)A)/(pi r^(2)) gt 0` ` rArr ` A is minimum. Now ` 2pi r = 2pi((14)/(pi+4)) = (28pi)/(pi+4)` and ` 4x = 28 - 2pi r ` ` = 28 - (28pi)/(pi+4) = (112)/(pi+4)` Therefore, the length of two pieces of wire ` = (28pi)/(pi+4) cm and (112)/(pi+4)` cm. |
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