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Question
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 – l2
∴ 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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f(x) = `square`
∴ f'(x) = `square`
∴ f''(x) = `square`
For extreme value, f'(x) = 0, we get
x = `square`
∴ f''`(square)` = – 2 < 0
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|
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