# PreCalc (still confused)

posted by .

Hello and Happy Halloween! Trying to get this problem and could use some help.

Find the dimensions of the rectangle with minimum perimeter if its area is 400 square meters. Find the least perimeter.

I know that a rectangle has two sides that are equal and two other sides that are equal but I am unsure about what type of equation to use. I thought maybe I would try guess and check but I would rather have an equation.

PreCalc - jim, Saturday, October 31, 2009 at 2:16pm
"two sides that are equal and two other sides that are equal" is a good start. Call the length of one type of side x, and the other y.

Then the area is x * y

And the perimeter is 2x + 2y

Check this on a small 5 * 3 rectangle, to be sure you understand it.

We are told the area is 400. Does this give you enough to go further?

***************************************
Ok, so I understand the equations but I'm not sure how to go about solving them. If I am supposed to solve a system of equations, both equations should be equal to something. so X*Y=400, but what does 2X+2Y equal?

• PreCalc (still confused) -

Good question!

2X+2Y could be equal to lots of things: your question is to find the _minimum_ they can be equal to.

For example, we know

xy=400

so x=5, y = 80 works for that.

Then 2x + 2y = 10 + 160 = 170. That is one answer.

x=10, y=40 also gives us xy=400.

Then 2x + 2y = 20 + 80 = 100. That is another answer, and a better one than 170, since it's smaller, and therefore closer to the minimum.

A lot of calculus is about finding the minimum or maximum value, which is often the "best" or "worst" case in real life, but we can't use calc here, so can you find another set of values for x and y so that

x * y = 400

and 2x + 2y is as small as possible?

• PreCalc (still confused) -

Let me phrase it as a real-life-ish word puzzle:

You need to build a pool 400 sq. m. in area. Building curved or slanted sides is too complicated and expensive, so the sides have to be straight, so you're going to make it a rectangle.

The sides cost you \$10 per metre to put up.

What are the lengths of the sides of the cheapest pool you can build?

• PreCalc (still confused) -

So basically, I have to guess and check.

20 X 20 would be the smallest but is a square.

How about 16 X 25 = 82?

I would much rather have a mathematical way to go about it.

• PreCalc (still confused) -

First thing to remember is that a square is a rectangle; it's just a special case of a rectangle.

Getting back to a "mathematical way to go about it", there are two I can think of, and one involves calculus. Maybe this is set to make you appreciate calculus when you get to it. :-)

Anyway consider that since we know:

xy = 400

we know that

y = 400 / x

which means that we can restate

2x + 2y

as

2x + 2(400 / x).

So we can restate the problme as finding the minimum value of

x + 400 / x

Does that prompt any new ideas?

• PreCalc (still confused) -

Ok, well now I'm really confused. I thought about substituting 400/x into y but I still don't have anything to put it equal to.

Can you please try to explain it.

• PreCalc (still confused) -

Actually, putting 400/x into y will give you something, here's a geometrical approach of finding the minimum.

Let x stands for one side of the "rectangle", not forgetting that a square is also a rectangle, as Jim has pointed out. The adjacent side will be 400/x, thus assuring that the product of the two sides will give an area of 400.

The perimeter, P is therefore given by
P = 2(x+400/x)

For x to be the value for which the perimeter is the minimum, then any perturbation, h, in the value of x will result in a bigger perimeter, i.e.

2((x+h)+400/(x+h))≥2(x+400/x)
for any value of h, i.e. h≥0 or h≤0.

Expanding and simplifying, we obtain
h(1-400/(x(x+h)))≥0
if h≥0, x(x+h)≥400
if h≤0, x(x+h)≤400
The only value of h that satisfies both inequalities is h=0, or x²=400, hence h=20.

If you would like to have some practice of the method, you can try to find the minimum perimeter of three sides of a rectangle of a given area, A. The problem arises when we have to fence an area against a wall. You will find that the two opposites sides are √(A/2), and the side opposite the wall is 2√(A/2).

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