Optimizing.?

Question

Find the dimensions of a rectangle with area 1000 m whos perimeter is as small as possible. I try 1000 = xy and p = 2(x+y) so 1000 = (x-p/2)(y-p/2). what am i doing wrong?

Answer 1

1000 = xy
p = 2(x + y)


To minimize perimeter, take the derivative but first express the perimeter in terms of one variable:

y = 1000/x

p = 2( x + (1000/x) )
p = 2x + 2000/x

Now, the derivative and set equal to zero to find the minimum:

p' = 2 - (2000/x^2) = 0

(2000/x^2) = 2
2000 = 2x^2
x = sqrt(1000); which is the length of the side that minimizes the perimeter. Plug into "p" to find minimum perimeter:

p = 2[sqrt(1000) + 1000/(sqrt(1000) ) ]
= 2[31.6 + 1000/31.6]
= 126.5 m

Answer 2

In order to solve this you must eliminate one of the variables. Use y = 1000 / x to rewrite the formula for p:

p = 2(x + 1000/x)

To find the minimum, take the derivative and equate it to zero:

dp/dx = 2 - 2000/x^2 = 0

It follows that x^2 = 1000, so x = sqrt 1000 ~ 31.6; and y = 1000 / x = sqrt 1000 as well. (Not strange... square is rectangle with smallest perimeter!)


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Or try it this way: if I add a small amount dx to x, I must subtract something from y in order to keep the area 1000; we want to keep

(x + dx) * (y + dy) = x * y
x * dy + y * dx = 0

The perimeter will increase by 2(dx + dy). At the minimum, this has to be zero, so dy = -dx. We find

x * (-dx) + y * dx = 0
y - x = 0
x = y

so the rectangle must be a square. (See... you don't even need to know the area!)

Answer 3

you have to optimise p
so eliminate x or y

y = 1000/x

p = 2{ x + (1000/x)}

dp/dx = 2{ 1 - 1000/(x^2)}

second derivative is 2000/(x^3)

dp/dx = 0 implies 2{ 1 - 1000/(x^2)} = 0

implies x = √1000

and second derivative is positive at x = √1000

so that p is minimum for x = √1000

and y = x/1000 = √1000

for minimum p, x = y ,that is the rectangle has to be a square