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# Calculus/calculus optimization problem

Question
The lifeguard at a public beach has 400 m of rope available to lay out a rectangular restricted swimming area using the straight shoreline as one side of the rectangle.
(a) If she wants to maximize the swimming area, what will the dimensions of the rectangle be?

(b)To ensure the safety of swimmers, she decides that nobody should be more than 50 m from shore.  What should the dimensions of the swimming area be with this added restriction?

(a) The rope would be used for the far side of length y and 2 shore to the far side of length x.
Thus, the length would be 2x + y = 400, so y = 400 - 2x.  The area would be xy, and
y = 400 - 2x, so the total area is x(400-2x) = 400x - 2x².

This is a downward sloping parabola, so the maximum is where the slope is 0.  The slope is the derivative, and that is 400 - 4x.  This is zero at x = 100.  Since the length is 2x + y = 400, that gives 200 + y = 400.  This means y = 200.  The dimensions will then be 100x200.

(b) If x is no more than 50, and 50 is the best, then x = 50.  Since 2x + y = 400,
that is the same, since x = 50, of 100 + y = 400, so y = 300.

Calculus

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#### Scotto

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