Solving a Simple Math Equation for the Smallest Perimeter of a Rectangle

  • Thread starter JayDub
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In summary, the smallest perimeter of a rectangle with an area of 64 cm^2 can be found by graphing x(64/x) and finding the minimum value, which would be 2(x) + 2(64 / x). Similarly, for a rectangle with a perimeter of 100 cm, the largest area can be found by graphing x(50-x) and finding the maximum value. This is because in both cases, we are trying to either minimize or maximize the given value.
  • #1
JayDub
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I am trying to remember how I would go about solving an equation like so.

What is the smallest perimiter of a rectangle with area of 64 cm^2.

I know I can use the graphing calculator.

So I know there are two sides, x and 64 / x.

If I try and graph x(64/x) I just get a straight line at 64. So how does one go about solving a question like this?

Thank you.
 
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  • #2
If I try and graph x(64/x) I just get a straight line at 64.
Which is good -- you're graphing the area of your rectangle. It had better be a constant value!
 
  • #3
Lol, I understand why it is giving me that (because the x's cancel) which I am sure you know.

If I had the inverse of this question, something like. The perimiter of a rectangle is 100 cm. What is the largest area it can have?

I would go

Side 1: x
Side 2: 50 - x

Graphing: x(50-x) thus at the maximum, x = 25.
Therby making the other side 25 as well, for a maximum area of 625 cm.
 
  • #4
Well if you're trying to minimize perimeter in your original problem, what should you be graphing?
 
  • #5
When looking for the rectangle with the largest area, why do you do those steps?
 
  • #6
wait a minute... If

Side 1: x
Side 2: 64 / x

To find the smallest area couldn't I graph 2(x) + 2(64 / x) and find the smallest perimiter?
 
  • #7
Thus graphing, gives me an x value of 8 and a perimiter of 32.
 
  • #8
That sounds good.
 

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