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Question
A metal wire of 36 cm long is bent to form a rectangle. By completing the following activity, find it’s dimensions when it’s area is maximum.
Solution: Let the dimensions of the rectangle be x cm and y cm.
∴ 2x + 2y = 36
Let f(x) be the area of rectangle in terms of x, then
f(x) = `square`
∴ f'(x) = `square`
∴ f''(x) = `square`
For extreme value, f'(x) = 0, we get
x = `square`
∴ f''`(square)` = – 2 < 0
∴ Area is maximum when x = `square`, y = `square`
∴ Dimensions of rectangle are `square`
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Solution
Let the dimensions of the rectangle be x cm and y cm.
∴ 2x + 2y = 36
∴ x + y = 18
∴ y = 18 – x
Let f(x) be the area of rectangle in terms of x, then
f(x) = = xy = x(18 – x) = 18x – x2
∴ f'(x) = 18 – 2x
∴ f''(x) = – 2
For extreme value, f'(x) = 0, we get
18 – 2x = 0
∴ 18 = 2x
∴ x = 9
∴ f''(9) = – 2 < 0
∴ Area is maximum when x = 9, y = 18 – 9 = 9
∴ Dimensions of rectangle are 9 cm × 9 cm.
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