Question 476600
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The area of a rectangle is given by:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ lw]

so we can say:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ l\ =\ \frac{A}{w}]

The perimeter is given by:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P\ =\ 2l\ +\ 2w]

Substituting we can write a function for the perimeter in terms of *[tex \Large w]:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(w)\ =\ \frac{2A}{w}\ +\ 2w]

Extremum where the first derivative is equal to zero:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dP}{dx}\ =\ 2\ -\ \frac{2A}{w^2}]

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

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

But discard the negative root, hence:

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

The extremum is a minimum if the second derivative is positive at the extreme point.

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d^2P}{dx^2}\ =\ \frac{4A}{w^3}]

Since *[tex \Large w] and *[tex \Large A] are both positive quantities, the second derivative is positive for all values of *[tex \Large w] and therefore the extremum is a minimum.

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