Question 597839
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I hope you are taking Calculus, because that is the only way I know how to solve this one.


Minimize


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ F(w)\ =\ 2w\ +\ \frac{70}{w}]


Take the first derivative:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dF}{dw}\ =\ 2\ -\ \frac{70}{w^2}]


Set the first derivative equal to zero and solve for the value of the independent variable at a local extremum.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\ =\ \frac{70}{w^2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2w^2\ =\ 70]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w\ =\ \pm\sqrt{35}]


Discard the negative root; measurements of distance less than zero are absurd in this context.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w\ =\ \sqrt{35}]


Take the second derivative:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d^2F}{dw^2}\ =\ \frac{140}{w^3}]


Which is positive for all positive values of the independent variable.  Therefore the local extreme is a minimum.


Hence the minimum amount of fence is required when the width measures *[tex \LARGE \sqrt{35}].  You should be able to determine the length yourself using *[tex \LARGE l\ =\ \frac{70}{w}].  Using the calculator and rounding is also up to you.


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