How? Find the dimensions of a rectangle with area 1000 sq-m whose perimeter is as small as possible?

Answer 1

Let x and y be the dimensions. Then the perimeter is 2x+2y, and the area is xy = 1000.

To minimize the perimeter, set y = 1000/x and obtain the perimeter in terms of x alone,

2x + 2000/x.

To minimize this, differentiate it, to get

2 - 2000/x^2

and set this equal to zero. You get x = sqrt(1000), so y=1000/sqrt(1000)=sqrt(1000) as well, so the rectangle is actually a square.

Note: to confirm that x=sqrt(1000) is a minimizer, not a maximizer, you can compute the second derivative, which is 4000/x^3, and see that it's positive, so this critical point is a minimizer.

Answer 2

Let z = one side of the rectangle, and y = the other side.

Thus, z x y =1000

You also know that the perimeter of this rectangle is also (2 x z) + (2 x y) Since the goal is to minimize this equation, you will need to take the derivitive of it.

Since we have 2 variables, solve fo z in terms of y --> z = 1000/y

Minimize 2 x 1000/y + 2y by taking the derivitive of y.

d/dy = (-2000/y^2) + 2

Seting d/dy = 0, you can solve for y to get y = 1000^(1/2) The square root of 1000 = 31.62277...

Thus the variable z = 1000/31.62277... also equals the square root of 1000. It appears as if your rectangle is actually a square with a side equaling the square root of 1000.

Think about this some more and you will find that this makes sense. This is a common problem, and you will find that by minimizing the sides you will have an equilateral polygon, no matter how many sides it has 3, 4, 5, 30, the answer will always be the same, a polygon with sides of equal length!

Kind of confusing, but think of it graphically and you will understand this to be true as well.

Answer 3

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let sides be a & b area = ab = 1000 p = perimeter = 2[a+b] p = 2a + 2000/a dp/da = 0 = minimum = 2 - 2000/a^2 = 0 ======================== d^p/da^2 = +ve =========================== a^2 = 1000 a = 10 root10 b = 1000/10 root10 = 100/root10 = 10 root10 a=b=10root10 = square

Answer 4

answer: a square whose side is sqrt(1'000)


how you get there:

(1) area A=x*y

(2) perimeter P=2(x+y)

From (1) you can extract, say, y=A/x
which you plug into (2)
P=2x+2A/x

You want a minimum so you're looking for a zero on the derivative
dP/dx=2-2A/x^2

set it equal to 0, rearrange, you find
x=sqrt(A)

for a minimum. Well, to be professional you'd need to check that this is really a minimum, not a maximum. To do this you'd need to take the second derivative, d^2P/dx^2, which you'd find to be 4A/x^3, which is indeed positive for x=sqrt(A), so it IS a minimum indeed.

so your minimum perimeter will be 4*sqrt(A)


hope this helps

Answer 5

What you want is a rectangle that is as close to a square (sides equal) as possible.

I think 25 by 40 is about as close as you can get. the perimeter would be 130m

Answer 6

A rectangle with smallest possible perimiter is always a square.

Side is approx. 31.62; perim. is appox. 126.5. No other rectangle has perim. as small.

Answer 7

It's a square with each side of length root1000m