Question 311400: A lifeguard uses 700m of marker buoys to rope off a rectangular swimming area in a lake. One side of the swimming area is a sandy beach. Determine the dimensions that will result in the maximum swimming area. What is the maximum swimming area?
This is how I interpreted the question, but I'm not positive that it's correct.
Let x represent the width and y the length. A=xy
2x+2y=700
2y=700-2x
y=(700-2x)/2
y=350-x
A(x)=x(350-x)
0=x(350-x)
x=0 or x=350
the mid value of 0 and 350 is x=175
when x=175, y=350-175
y=175
A(x)=x(350-x) is A(x)=350x-x^2
therefore the dimensions are 175 by 175 and max area is 30625. I don't believe my answer is right due to the fact I didn't take that 1 side of the beach is sandy and it doesn't make sense that the dimensions are 175 by 175 when the swimming area is suppose to be a rectangle. I would really appreciate the help. Thank you in advance.
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
You missed it by that much.
Your analysis was generally in the right direction, but you made a small error at the very beginning. In order to account for the beach being one side of the rectangle, you had to alter the perimeter formula.
Let  represent the width of the rectangle and  represent the length, just as you stated.
The area of the rectangle is indeed \ =\ xy) , but the perimeter of that portion of the rectangle bounded by the rope is not  but rather  because only one length is bounded by the rope.
Given that, proceed.
hence
\ =\ 700x\ -\ 2x^2)
Now, depending on whether you know how to take a derivative of a polynomial function or not you will either recognize this as a concave down parabola with a vertex with an -coordinate at:
}\ =\ 175)
or you will just take the first derivative of the function and set it equal to zero:
\ =\ 700\ -\ 4x)
and then submitting this value for to the modified perimeter formula,
\ +\ y\ =\ 700\ \ \Rightarrow\ \ y\ =\ 350)
For a maximum area of 
So you actually had the correct answer for the width dimension, but you only came up with half of the correct maximum area. What you did was instructive however. If you think about it, what you did solve was for the dimensions of the maximum area rectangle where you have to consider all four sides in the perimeter. And that is, in general, a square with sides measuring one-fourth of the perimeter. If you are interested, write back and I'll send you the general proof. And by the way, the fact that you derived a square when the problem asked for a rectangle is not, in and of itself, sufficient reason to discard your answer. The fact is, a square is a rectangle -- just a very special one where all 4 sides are the same measure.
John
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