Comments on: A farmer wants to fence an area of 37.5 million square feet in a rectangular field? http://www.thearealist.com/area-5/a-farmer-wants-to-fence-an-area-of-37-5-million-square-feet-in-a-rectangular-field Fri, 03 Feb 2012 13:07:59 -0500 http://wordpress.org/?v=2.8.4 hourly 1 By: Victor http://www.thearealist.com/area-5/a-farmer-wants-to-fence-an-area-of-37-5-million-square-feet-in-a-rectangular-field/comment-page-1#comment-1831 Victor Tue, 09 Feb 2010 20:06:59 +0000 http://www.thearealist.com/area-5/a-farmer-wants-to-fence-an-area-of-37-5-million-square-feet-in-a-rectangular-field#comment-1831 Let one side be length A and the other B Total area = A.B = k where k is the constant value 37.5 million Then B = k/A...................(i) The total length of the sides plus the division fencing will be L = 2A + 2B + A if we let the division fence be parallel to A: it doesn't matter either way. So L = 2A + 2k/A + A = 3A + 2k/A ................. (ii) Differentiate L with respect to A, to get L' = 3 - 2k/(A^2) = 0 for max or min This gives A^2 = 2k/3 and so A = sqr root of 2k/3 Substitute in (i) to get B = k/[sqr root of 2k/3] = sqr root of 3k/2 by division and rearrangement. Now put the value of k in place as 37.5 million and the length of side A should be 5000 ft and the length of side B should be 7500 ft and the division fence should be parallel to side A and measure 5000 ft. Check the second derivative of L, to get L'' = 4k/(A^3) which is positive since k and A are both positive, so we have a minimum. The total length of wire will be 30,000 feet by substituting values in (ii) and this is the minimum quantity and therefore minimum cost. OK?<br><b>References : </b><br> Let one side be length A and the other B
Total area = A.B = k where k is the constant value 37.5 million

Then B = k/A……………….(i)

The total length of the sides plus the division fencing will be

L = 2A + 2B + A if we let the division fence be parallel to A: it doesn’t matter either way.

So L = 2A + 2k/A + A = 3A + 2k/A …………….. (ii)

Differentiate L with respect to A, to get

L’ = 3 – 2k/(A^2) = 0 for max or min

This gives A^2 = 2k/3 and so A = sqr root of 2k/3

Substitute in (i) to get

B = k/[sqr root of 2k/3] = sqr root of 3k/2 by division and rearrangement.

Now put the value of k in place as 37.5 million and
the length of side A should be 5000 ft and
the length of side B should be 7500 ft and the division fence should be parallel to side A and measure 5000 ft.

Check the second derivative of L, to get

L” = 4k/(A^3) which is positive since k and A are both positive, so we have a minimum.

The total length of wire will be 30,000 feet by substituting values in (ii) and this is the minimum quantity and therefore minimum cost.

OK?
References :

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