need help on this..

[jacqueline]

A lifeguard marks a rectangular swimming area at a beach with a 200m rope. The width of the swimming area is x meters. The area enclosed is A squared meters, where A = x(200-2x). What is the greatest area that can be enclosed?

 

[soroban]

Hello, jacqueline!

A lifeguard marks a rectangular swimming area at a beach with a 200m rope.
The width of the swimming area is x meters.
The area enclosed is A squared meters, where A = x(200-2x).
What is the greatest area that can be enclosed?

If this is a Calculus course, take the derivative of \(\displaystyle A = 200x - 2x^2\) and maximize \(\displaystyle A\).

If this is PreCalc, note that the area function is a quadratic,
. . and its graph is a down-opening parabola.
Its maximum occurs at its <u>vertex</u> . . . got it now?

 

[jacqueline]

Thanks, I think I got it... is the answer 2500 squared meters?

 

[soroban]

Hello, jacqueline!

is the answer 2500 squared meters? . . . . sorry, no

From the nature of that area function they gave us
. . (and seeing many problems like this before),
I assume that only three sides of the swimming area are roped off.

           200 - 2x 
      * - - - - - - - - * 
      |                 |
     x|                 |x
      |                 |
    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
             beach

And that's how they got: .\(\displaystyle A\:=\:x(200\,-\,2x)\)

Your algebra/arithmetic is correct: .\(\displaystyle x\,=\,50\)

. . So the width of the swimming area is \(\displaystyle 50\) m.

. . But the length is: .\(\displaystyle 200\,-\,2x\:=\;200\,-\,2\cdot50\:=\:100\) m.

Therefore, the area is: .\(\displaystyle 50\,\times\,100\:=\:5000\) m\(\displaystyle ^2\).

 

[jacqueline]

Ooh, I forgot about that area function that was given. Thanks so much for your help!