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प्रश्न
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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उत्तर
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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The rectangle has area of 50 cm2. Complete the following activity to find its dimensions for least perimeter.
Solution: Let x cm and y cm be the length and breadth of a rectangle.
Then its area is xy = 50
∴ `y =50/x`
Perimeter of rectangle `=2(x+y)=2(x+50/x)`
Let f(x) `=2(x+50/x)`
Then f'(x) = `square` and f''(x) = `square`
Now,f'(x) = 0, if x = `square`
But x is not negative.
∴ `x = root(5)(2) "and" f^('')(root(5)(2))=square>0`
∴ by the second derivative test f is minimum at x = `root(5)(2)`
When x = `root(5)(2),y=50/root(5)(2)=root(5)(2)`
∴ `x=root(5)(2) "cm" , y = root(5)(2) "cm"`
Hence, rectangle is a square of side `root(5)(2) "cm"`
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