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A wire of length 25m is to be cut into two pieces. One of the wires is to be made into a square and the other into a circle. What should be the lengths of the two pieces so that the combined area of the square and the circle is minimum ? |
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Answer» Let x meters and `(25 -x)` metres be the required lengths. Let a be the side of the square formed and r be the radius of the circle formed. Then, `4a = x and 2pi r = 25 - x or a = (x)/(4) and r = (25 - x)/(2pi)` `:.` area of the square `= ((x^(2))/(16))` sq metres. Area of the circle `= pi ((25 - x)/(2pi))^(2) = ((25 -x)^(2))/(4pi)` Now, the combined area `A = (x^(2))/(16) + ((25 - x)^(2))/(4pi)` `:. (dA)/(dx) = (x)/(8) - ((25 -x))/(2pi) = ((pi + 4) x - 100)/((8pi) , and (d^(2)A)/(dx^(2)) = ((pi + 4))/(8pi)` For a maxima or minima, we have `(dA)/(dx) = 0` Now, `(dA)/(dx) = 0 rArr (pi + 4)x - 100 = 0 rArr x = (100)/((pi + 4))` And, `(d^(2)A)/(dx^(2)) = ((pi + 4))/(8pi) gt 0` for all values of x `:. x = (100)/((pi + 4))` is a point of minimum. `:.` lengths of the pieces are `((100)/(pi + 4))` metres and `((25pi)/(pi + 4))` metres. |
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