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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.
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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