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Re: Attempt this question for your practice

by ldoolitt » Wed Oct 24, 2007 8:08 am
pahwa wrote:John plans to fence rectangular garden with 128feet of wire. How can he get maximum area?
Please write the reasoning as well.

Hint: This has something to do with my Tip 1 from Tip of the day thread.
Solve this not using calculus? I assume that was what you were asking.
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by samirpandeyit62 » Wed Oct 24, 2007 8:27 am
128 feet = perimeter

so sum of sides = 64 feet

area = l*b

so to maximise area we need to maximise the product of l * b which would be highest when l=b = 32

area = 32^2
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by jangojess » Wed Oct 24, 2007 2:02 pm
ya square will have maximum area.....so max area = (128/4)^2 = 1024...

The same with li'll twist....John has a wire with 128 ft long. wht will be the max area obtained???
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by pahwa » Wed Oct 24, 2007 9:20 pm
Yes. That is correct.

Please ensure that you remember this. To maximise the area, Square has to be formed!!!
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by jangojess » Wed Oct 24, 2007 9:45 pm
pahwa....u r partially correct...in ur Q u've mentioned that a rectangular fence needs to be fenced...square will have max area in a quadrilaterals for a fixed perimeter but when it comes to a 2d figure ALWAYS a circle wins......... :)
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