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Question
A metal wire of 36 cm length is bent to form a rectangle. Find its dimensions when its area is maximum.
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Solution
Let the length and breadth of a rectangle be x and y, respectively.
∴ Perimeter of rectangle = 2 (x + y) = 36 cm
∴ x + y = 18
∴ y = 18 − x ....(i)
Area of rectangle = x × y
= x (18 − x) ...from eq (i)
Let f(x) = x(18 − x)
= 18x − x2
Then f'(x) = `d/dx (18 - x^2)`
= 18 × 1 − 2x
= 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 a maximum value at x = 9.
When x = 9, y = 18 − 9 = 9
Hence, the rectangle is a square of side 9 cm.
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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`
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