Understanding Rectangle Area and PerimeterDate: 11/08/2002 at 16:50:32 From: Mark Subject: Area Dear Dr.Math, I am a sophmore in high school and we are reviewing area and perimeter. My teacher gave us a problem and told us to prove it true or false. I know that the answer is false, but I need help understanding and explaining why it's wrong. Here is the problem: We have a given square and we see that if we increase the perimeter, area increases as well in our new rectangle. Is that always true? Give a thorough explanation and clear examples. Given Square: ------ 3ft | | | | | | ------ 3ft P = 12ft and A = 9 sq.ft New Rectangle: --------- | | 3ft | | | | --------- 4ft P = 14ft and A = 12 sq.ft I do see that the theory that "as perimeter increases, the area as well increases" is not always true, even though it works for the given problem. My example is that if our new rectangle looks like this, then our perimeter increased but our area decreased, so the theory is wrong or at least not always right. ---------------------- 1ft | | ---------------------- 6ft P = 14ft and A = 6ft I know how to do the math and figure out perimeter and area, but I don't know how to explain why it doesn't work if the one side of the rectangle is one, and why it did work for the given problem. Please help, Sincerely, Mark Date: 11/09/2002 at 17:22:49 From: Doctor Rick Subject: Re: Area Hi, Mark. You are correct that a rectangle with perimeter greater than 12 (the area of the given square) does not necessarily have an area greater than 9 (the area of the given square). When you are told to disprove a statement, all you need to do is to provide a counterexample - a case that satisfies the premises (rectangle with perimeter greater than 12) and does not fit the conclusion (area greater than 9). You have done this, so you have a thorough explanation with a clear example. You don't need to explain "why" the statement is not true in any deeper sense than "because here is an example in which it is not true." But I understand your desire to understand it on a deeper level. The fact is that if you require that the perimeter of a rectangle be THE SAME as that of the square, the area of the rectangle will ALWAYS be LESS than that of the square (unless the rectangle is the square itself). In other words, for a given perimeter, the rectangle that has the largest area is a square. Let's consider a rectangle with length L and width W. Its perimeter is 2(L+W) and its area is LW. If we require that the perimeter be some particular value P, then we can find what the width must be when we know the length of the rectangle: 2(L+W) = P L + W = P/2 W = P/2 - L Then the area will be LW = L(P/2 - L) A = (P/2)L - L^2 Thus the area of the rectangle is a quadratic function of the length L. Its graph is a parabola that opens downward, and the greatest area occurs at the vertex of the parabola. You can prove that the value of L at the vertex is P/4, so that the width is also P/4 and the rectangle with greatest area is a square. If you have further questions about this, please feel free to ask me. "Why" questions can lead in lots of different directions, and only you can tell me when we have hit on an explanation of the type that will satisfy you. Of course, a true mathematician will never be completely satisfied, there are always more directions to explore. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/