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Question
A wire of length 36 metres is bent in the form of a rectangle. Find its dimensions if the area of the rectangle is maximum.
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Solution
Let x metres and y metres be the length and breadth of the rectangle.
Then its perimeter is 2(x + y) = 36
∴ x + y = 18
∴ y = 18 – x
Area of the rectangle = xy = x(18 – x)
Let f'(x) = x(18 – x) = 18x – x2
∴ f'(x) = `d/(dx)(18x - x^2)` = 18 – 2x
and f'(x) = `d/(dx)(18 - 2x)` = 0 – 2 × 1 = –2
Now, f'(x) = 0, if 18 – 2x = 0
i.e. if x = 9
and f'(9) = – 2 < 0
∴ By the second derivative test, f has maximum value at x = 9.
When x = 9, y = 18 – 9 = 9
∴ x = 9 cm, y = 9 cm
∴ Rectangle is a square of side 9 metres.
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