SOLUTION: Max/Min Applications.
A farmer with 10,000 feet of fencing wants to enclose a rectangular field and then divide it into two (non-equal) plots by running a fence parallel to one
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Question 286354: Max/Min Applications.
A farmer with 10,000 feet of fencing wants to enclose a rectangular field and then divide it into two (non-equal) plots by running a fence parallel to one of its sides. What is the largest area that can be enclosed? What dimensions should he use to get the largest area?
2L+2W=10,000 is all i've gotten so far, im stuck.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
I think what you really want is 
because of the extra partition. So then 
The area is then 
so make the substitution and write a function for the area in terms of the width.
\ =\ \frac{10000w\ -\ 3w^2}{2})
A little rearranging:
\ =\ -\frac{3}{2}w^2\ +\ 5000w)
You should recognize that this will graph to a concave down parabola, so the vertex, located at
would give the maximum area. So if the width is 833 and one third, and there are three pieces of fence this size, that leaves 7500 for the two length pieces, or 3750 for the length. And finally, the area is then:
You can do that last bit of arithmetic.
John
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