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Question
A wire of length 120 cm is bent in the form of a rectangle. Find its dimensions if the area of the rectangle is maximum
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Solution
Let the length and breadth of a rectangle be x cm and y cm
∴ Perimeter of rectangle = 2(x + y) = 120 cm
∴ x + y = 60 .......(i)
Let A be the area of the rectangle.
∴ A = xy
= x(60 − x) .......[From (i)]
= 60x − x2
Differentiating w. r. t. x, we get
`("dA")/("d"x)` = 60 − 2x
∴ `("d"^2"A")/("d"x^2)` = −2
For maximum area, `"dA"/("d"x)` = 0
∴ 60 − 2x = 0
∴ x = 30
For x = 30,
`(("d"^2"A")/("d"x^2))_(x = 30)` = − 2 < 0
When x = 30, area of the rectangle is maximum.
and y = 60 − 30 = 30 .......[From (i)]
∴ Area of the rectangle is maximum if length = breadth = 30 cm.
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