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A rod of 108 m long is bent to form a rectangle. Find it’s dimensions when it’s area is maximum.

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#### Solution

Let the length and breadth of a rectangle be l and b respectively.

∴ Perimeter of rectangle = 2(l + b) = 108 m

∴ l + b = 54

∴ b = 54 – l ...(i)

Area of rectangle = l × b

= l(54 – l) ...[From (i)]

Let f(l) = l(54 – l)

= 54l – l^{2}

∴ f'(l) = 54 – 2l

∴ f"(l) = – 2

Consider, f'(l) = 0

∴ 54 – 2l = 0

∴ 54 = 2l

∴ l = 27

For l = 27,

f"(27) = – 2 < 0

∴ f(l) i.e., area is maximum at l = 27

and b = 54 – 27 ...[From (i)]

= 27

∴ The dimensions of rectangle are 27 m × 27 m.

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