Question 1181411
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 *[illustration PigPenFence.jpg].

Length of Fence: *[tex \Large 2x\ +\ y\ =\ 32]


Length of Pen as a function of width:  *[tex \Large y(x)\ =\ 32\ -\ 2x]


Area of pen:  *[tex \Large A\ =\ xy]


Area of pen as a function of width:  *[tex \Large A(x)\ =\ 32x\ -\ 2x^2]


Take the first derivative:


*[tex \Large \frac{dA}{dx}\ =\ 32\ -\ 4x]


Set the derivative equal to zero to find an extremum:


*[tex \Large 32\ -\ 4x\ =\ 0]


*[tex \Large x\ =\ 8]


Take the second derivative:


*[tex \Large \frac{d^2A}{dx^2}\ =\ -4]


Therefore *[tex \Large \frac{d^2A}{dx^2}\ <\ 0\ \forall\,x] in the domain of A, and therefore the one extremum is guaranteed to be a maximum.


If the maximum area is obtained when *[tex \Large x\ =\ 8], then calculate *[tex \Large y(8)] to get the other dimension.


																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[illustration darwinfish.jpg]

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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