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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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प्रश्न

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.

बेरीज
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उत्तर

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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पाठ 2: Applications of Derivatives - Exercise 2.4 [पृष्ठ ९०]

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