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Question
Solve the following:
A wire of length l is cut into two parts. One part is bent into a circle and the other into a square. Show that the sum of the areas of the circle and the square is the least, if the radius of the circle is half of the side of the square.
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Solution
Let r be the radius of the circle and x be the length of the side of the square.
Then,
Total length of the wire = circumference of the circle + perimeter of the square = l
∴ 2πr + 4x = l
∴ r = `(l - 4x)/(2pi)`
A = area of the circle + area of the square
= πr2 + x2
= `pi((l - 4x)/(2pi))^2 + x^2`
= `x^2+ (1)/(4pi) (l - 4x)^2`
= f(x) ...(Say)
Then f'(x) = `2x + (1)/(4pi) xx 2(l - 4x)(- 4)`
= `2x - (2)/pi(l - 4x)`
and
f"(x) = `2 - (2)/pi( - 4)`
= `2 + (8)/pi`
Now, f'(x) = 0 when `2x - (2)/pi (l - 4x)` = 0
i.e. when 2πx – 2l + 8x = 0
i.e when 2(π + 4)x = 2l
i.e. when x = `l/(pi + 4)`
and
f"`(l/(pi + 4)) = 2 + (8)/pi > 0`
∴ By the second derivative test, f has a minimum,
When x = `l/(pi + 4)`.
For this value of x,
r = `(l - 4(l/(pi + 4)))/(2pi)`
= `(pil + 4l - 4l)/(2pi(pi + 4)`
= `l/(2(pi + 4)`
= `x/(2)`
This shows that the sum of the areas of circle and square is least, when radius of the circle = `(1/2)` side of the square.
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