Question 286354
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I think what you really want is *[tex \Large 2l\ +\ 3w\ =\ 10000] because of the extra partition.  So then *[tex \Large l\ =\ \frac{10000\ -\ 3w}{2}]


The area is then *[tex \Large A\ =\ lw] so make the substitution and write a function for the area in terms of the width.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(w)\ =\ \frac{10000w\ -\ 3w^2}{2}]


A little rearranging:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(w)\ =\ -\frac{3}{2}w^2\ +\ 5000w]


You should recognize that this will graph to a concave down parabola, so the vertex, located at


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w\ =\ \frac{-5000}{-6}\ =\  833\frac{1}{3}]


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:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 833\frac{1}{3}\ \times\ 3750]


You  can do that last bit of arithmetic.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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