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?
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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) - jim, Sunday, November 1, 2009 at 1:25pm
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) - jim, Sunday, November 1, 2009 at 1:30pm
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) - Muffy, Sunday, November 1, 2009 at 1:44pm
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) - jim, Sunday, November 1, 2009 at 2:20pm
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) - MUFFY, Sunday, November 1, 2009 at 3:01pm
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) - MathMate, Sunday, November 1, 2009 at 4:34pm
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))¡Ã2(x+400/x)
for any value of h, i.e. h¡Ã0 or h¡0.
Expanding and simplifying, we obtain
h(1-400/(x(x+h)))¡Ã0
if h¡Ã0, x(x+h)¡Ã400
if h¡0, x(x+h)¡400
The only value of h that satisfies both inequalities is h=0, or x©÷=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 ¡î(A/2), and the side opposite the wall is 2¡î(A/2).
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I feel like this is so complicated. I don't understand where h came from, I don't understand how to solve it, I just don't get it.
First, grab your graphing calculator.
Area= lw
P=2L+2W
or
P=2L+2(400/l)
Graph that and see if you get a minimum P for as a function of L.
Now, analytically, I don't think I can do better than that. When you get into Calc, and realize how to take a derivitave and what it means, this problem will be trivial.
When I graphed it, I got a reciprocal function not a parabola. I don't know how to find a minumum that's not a parabola or a cubic.
Should I just go with the guess and check and call it 20 X 20?
If P = 2x + 2y and xy=400, then the perimeter is P(x)=2x + 800/x. In a pre-calc course you could only be expected to find the minimum value of P(x) by inspection of a graph. See the "graph of p = 2x + 800/x from 0 to 40" at wolfram alpha.
OR, by creativity ...
The minimum value of the perimeter exists, so let it be K. Then the equation
2x + 800/x = K
can have only one solution. Multiply by x and get the equation in standard form to get
2x^2 - Kx + 800 = 0
Since there is only one solution, the discriminant of the quadratic formula equals 0. So it follows that
(-K)^2 -4(2)(800) = 0 , or K=80
Since K=80, x = -(-80)/[2(2)] = 20, by the quadratic formula.
Hope you are still tuned in, I enjoyed solving the problem.
Well, it seems like finding the minimum perimeter of a rectangle given its area can be quite challenging. But let me try to explain it in a way that might be a bit more fun and entertaining!
Imagine you are a clown architect tasked with building the cheapest rectangular pool possible. The area of the pool is 400 square meters, and the sides of the pool will cost you $10 per meter. Your goal is to minimize the amount of money you spend on the sides.
Now, let's think about the shape of the pool. The sides of the pool can be represented by two equal lengths, which we'll call x, and two equal widths, which we'll call y.
If the sides of the pool were a square, the minimum perimeter would be achieved. However, since the pool must have an area of 400 square meters, we can't make it a square.
To find the minimum perimeter, we need to find an x and y that give us the minimum cost. Since the cost is determined by the perimeter, which is 2x + 2y, we want to minimize that expression.
One way to approach this is to recognize that the product of x and y is equal to the area, which is 400. So, we have xy = 400.
To minimize 2x + 2y, we can rewrite it as 2(x + y). Since xy = 400, we can substitute y = 400/x into the equation to get 2(x + 400/x).
Now, let's have some fun! If you think about it, this expression looks like the formula for a clown's hat. You have a big hat (400/x) on top of a smaller hat (x). Our goal is to find the smallest possible hat size that minimizes the total hat cost.
To find this minimum, we can take the derivative of the expression, set it equal to zero, and solve for x. But since we don't want to use calculus, let's use our clown intuition.
Think about it this way - the smaller the hat size (x), the bigger the big hat (400/x) has to be to maintain the same area of 400 square meters. But as the big hat gets bigger, the cost goes up. So, the smallest possible hat size (x) will result in the smallest cost.
In this case, the smallest possible hat size (x) is 20 meters (remember h = 20 from the previous explanation). This means the other side (y) will also be 20 meters to maintain an area of 400 square meters.
So, the dimensions of the rectangle with the minimum perimeter are 20 meters by 20 meters. And the least perimeter is 2(20) + 2(20) = 80 meters.
Congratulations! You've just built the cheapest clown pool with the minimum perimeter. Now you can enjoy some clown-shaped water balloons and entertain your friends!
Finding the dimensions of the rectangle with the minimum perimeter can be done using a mathematical approach rather than guess and check. Here's how you can solve this problem:
1. Start by considering a rectangle with two sides that are equal, let's call them x. The other two sides are also equal, let's call them y. So, the dimensions of the rectangle are x and y.
2. The area of the rectangle is given as 400 square meters, so we have the equation xy = 400.
3. The perimeter of a rectangle is given by the equation 2x + 2y.
4. We want to find the minimum perimeter, so we need to minimize the expression 2x + 2y.
5. We can rewrite the equation xy = 400 as y = 400/x.
6. Substituting y = 400/x into the expression for the perimeter, we get 2x + 2(400/x).
7. We want to find the minimum value of 2x + 2(400/x). We can do this by finding the derivative of this expression with respect to x and setting it equal to zero.
8. Differentiating 2x + 2(400/x) with respect to x, we get 2 - 800/x^2.
9. Setting 2 - 800/x^2 = 0, we can solve for x.
10. Solving the equation, we get x^2 = 400. Taking the square root of both sides, we get x = 20.
11. Plugging x = 20 into the equation y = 400/x, we get y = 400/20 = 20.
12. So, the dimensions of the rectangle with the minimum perimeter are 20 meters by 20 meters.
13. The minimum perimeter can be found by plugging x = 20 and y = 20 into the expression 2x + 2y. We get 2(20) + 2(20) = 80 meters.
Therefore, the dimensions of the rectangle with the minimum perimeter are 20 meters by 20 meters, and the minimum perimeter is 80 meters.