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Homework Help: Maximize and Minimize Area of 2 rectangles

  1. Dec 20, 2009 #1
    A man has 340 yds of fencing for enclosing two separate fields, one of which is to be rectangular twice as long as it is wide and the other a square. The square field must contain at least 100 sq. yds. And the rectangular one must contain at least 800 sq. yds. Find the maximum and the minimum values of the width x of the rectangular field. What is the greatest number of square yards that can be enclosed in the two fields?

    What I have so far: (not too much)

    A= bh

    One rectangle is bh (area 100 sq. yards or more), the other us 2b x 2h (area 800 sq. yards or more).

    bh+ (2b)(2h) = 340 (???)

    I am really having trouble coming up with an equation and solving the problem in general. Any help would be greatly appreciated.
  2. jcsd
  3. Dec 20, 2009 #2


    Staff: Mentor

    You are confusing the two fields. The square field is the one that has to contain at least 100 sq. yds. The rectangular one has to contain at least 800 sq. yds. Also, your dimensions are wrong for the rectangular field.
    Draw a sketch of the two fields, and label the sides. The sketch doesn't have to be very accurate, since all you need are variables to represent the dimensions.

    Identify your variables with descriptions of what they represent, like this:
    Let w = the width of the rectangular field.
    Let 2w = the length of the rectangular field.
    Let s = the length of any side of the square field.

    Now, write an equation that represents the total length of fencing used in the two fields.
    Then, write an equation that represents the total area. This equation will involve w and s.

    Solve the first equation for one variable, say w.
    Substitute for w in the second equation to get the total area as a function of a single variable. The new equation will be that of a parabola. Use the constraints on the minimum sizes of the two fields to find the maximum area of both fields and the minimum area.
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