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Tamil Nadu Board of Secondary EducationHSC Science Class 12

A farmer plans to fence a rectangular pasture adjacent to a river. The pasture must contain 1,80,000 sq. mtrs in order to provide enough grass for herds. No fencing is needed along the river. - Mathematics

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Question

A farmer plans to fence a rectangular pasture adjacent to a river. The pasture must contain 1,80,000 sq. mtrs in order to provide enough grass for herds. No fencing is needed along the river. What is the length of the minimum needed fencing material?

Sum

Solution


Let the length of the pasture be ‘x’ m

Let the breadth of the pasture be ‘y’ m

Given Area = 1,80,000

xy = 1,80,000

y = 1,80,000

For fencing, we need 2y + x  .......(one side is River)

Let P = 2y + x

P = `2(180000/2) + x = (360000/x) + x`

`"dP"/("d"x) = - 360000/x^2 + 1`

For maximum or minimum,

`"dP"/("d"x)` = 0

⇒ – 360000 + x2 = 0

x2 = 360000

x = ± 600  ......[x = – 600 is not possible]

∴ x = 600

Now,  `("d"^2"P")/("d"x^2) = 720000/x^3`

At x = 600, `("d"^2"P")/("d"x^2) > 0`

∴ P is minimum when x = 600

y = `180000/600` = 300

∴ Length of the minimum needed fencing material = 2y + x

= 2(300) + 600

= 1200 m

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Applications in Optimization
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Chapter 7: Applications of Differential Calculus - Exercise 7.8 [Page 47]

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Samacheer Kalvi Mathematics - Volume 1 and 2 [English] Class 12 TN Board
Chapter 7 Applications of Differential Calculus
Exercise 7.8 | Q 6 | Page 47
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