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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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Question

A metal wire of  36 cm length is bent to form a rectangle. Find its dimensions when its area is maximum.

Sum
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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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Chapter 4: Applications of Derivatives - Exercise 4.3 [Page 109]

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Balbharati Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
Chapter 4 Applications of Derivatives
Exercise 4.3 | Q 3 | Page 109

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