Question 409607
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There are 2 long sides and then 7 pieces of fence that equal the short side measure.  Draw yourself a picture to check my numbers.


Hence the 1200 feet of fencing must be used in the following way:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2l\ +\ 7w\ =\ 1200]


Solving for *[tex \Large l] we get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ l\ =\ 600\ -\ \frac{7}{2}w]


Since *[tex \Large A(l,w)\ =\ lw]


By substitution we get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(w)\ =\ -\frac{7}{2}w^2\ +\ 600w]


Since the Area function is continous and twice differentiable over its domain, take the first derivative:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dA}{dw}\ =\ -7w\ +\ 600]


set it equal to zero


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -7w\ +\ 600\ =\ 0]


Then solve for the value of the independent variable that gives an extreme point:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w_{ext}\ =\ \frac{600}{7}]


Take the second derivative:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d^2A}{dw^2}\ =\ -7\ <\ 0\ \forall\ w\ \in\ \text{Dom}(A)]  therefore *[tex \LARGE \left(w_{ext},A\left(w_{ext}\right)\right)] is a maximum.


Substitute *[tex \Large w_{ext}\ =\ \frac{600}{7}] back into *[tex \Large 2l\ +\ 7w\ =\ 1200] and solve for *[tex \Large l] to get the long dimension.


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