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Calculus! Max. and Min. Question

  1. Apr 17, 2005 #1
    The lifeguard at a public beach has 400m of rope available to lay out a rectangular restricted swimming area using the straight shoreline as one side of the triangle.
    a.) If she wants to maximize the swimming area, what will the dimensions of the rectangle be?
    b.)To ensure the safety of the swimmers, she decides that nobody should be more than 50m from shore. What should the dimensions of the swimming area be with this added restriction?....

    What do i doo?

    A is supposed to be 100m by 200m..and b is 50m by 300m...i can't get those at all though!
     
  2. jcsd
  3. Apr 17, 2005 #2

    OlderDan

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    First, repair the statement of your problem. The area is in the shape of a rectangle, not a triangle

    a) Three sides of the rectangle are formed by the rope. Let x = the length of the side parallel to the shore. Figure out how to express the other two sides in terms of x. Express the area of the rectangle in terms of x. Maximize that area. You will need the derivative of the area set equal to zero.

    b) This is a simple calculation based on your expression of the area in terms of x. What is the value of x in this case?
     
    Last edited: Apr 17, 2005
  4. Apr 17, 2005 #3
    oops yah ur right, the last word of the problem should have been rectangle..
     
  5. Apr 17, 2005 #4

    dextercioby

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    The rectangle has sides a,a,b,b.Okay?From the text of the problem u're given that

    [tex] 2a+b=400 [/tex] and you're asked about the values of "a" and "b" to maximize the area,i.e.the product [itex] a\cdot b[/itex].

    From the constraint,u find that [itex] b=400-2a [/itex] and so the area function which needs to be maximized becomes

    [tex] S=a(400-2a) [/tex]

    Find "a" for which the function S is maximum...

    Daniel.
     
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