optimization problem?

Question

find the dimension of a rectangle with area 1000 m^2 whose perimeter is as small as possible

Answer 1

Let the sides of the rectangle be x and y.

We have xy = 1000.

The perimeter of the rectangle is 2x + 2y.

Using the area equations, solve for y in terms of x:

y = 1000/x

This gives a perimeter equation P(x):

P(x) = 2x + 2000/x.
P(x) = 2x + 2000x^(-1)

Take P'(x) and set it equal to 0:

P'(x) = 2 - 2000x^(-2) = 0

2 - 2000x^(-2) = 0
2 = 2000x^(-2)
2x^2 = 2000
x^2 = 1000
x = sqrt(1000)
x = 31.622 m.

The rectangle is a square having side length 31.622 m. The perimeter is 124.491 m

Answer 2

Area of a rectangle is
A = LW
The perimeter is
P = 2L + 2W
You know that A = 1000, so
LW = 1000
Solve for one of the variables. It doesn't matter which one.
L = 1000/W
Replace L with 100/W in the perimeter equation
P = 2L + 2W
P = 2(1000/W) + 2W
P = 2000/W + 2W
Take the derivative
P' = -2000/W^2 + 2
Set this equal to zero.
-2000/W^2 + 2 = 0
Get a common denominator.
(-2000 + 2W^2) / W^2 = 0
-2000 + 2W^2 = 0
2W^2 = 2000
W^2 = 1000
W = sqrt(1000)
W = 10*sqrt(10)
L = 1000 / W = 1000 / [10*sqrt(10)] = 10*sqrt(10)
It should come to no surprise to you that when you minimize the perimeter of a rectangle, it actually turns out to be a square.

Answer 3

Perimeter=2x+2y

Area=xy=1000
so
y=1000/x

Substitute this into the perimiter equation
2x+2(1000/x)
2x+2000/x
take the derivative
2-2000/x^2
set it equal to 0 ans solve
0=2-2000/x²
2000/x²=2
2000=2x²
1000=x²
x=± √1000
the negative can be ignored because you cant have a negative dimension

plug this number back into the equation to find y
y=1000/√1000

plug both of these numbers into the perimeter equation
P=2(√1000)+2(1000/√1000)
P=2√1000+2000/√1000
P≈126.49
hope this helps

Answer 4

10√10m x 10√10m
A square is the rectangle with least perimeter for a given area.

But if you need a proof:
Let Rectangle have dimensions x and y
x * y = 1000, or y = 1000/x
P = 2x + 2y = 2x + 2(1000/x) = 2x + 2000/x
take the derivative and set = 0 to find minimum

P' = 2 - 2000/x^2 = 0
x^2 = 1000
x =√1000 = 10√10
y = 1000/x = 10√10

Answer 5

Let the dimensions of the rectangle by x and y.
Area, A = xy = 1000
=> y = 1000/x

Perimeter, P = 2x + 2y
=> P = 2x + 2000/x
=> dP/dx = 2 - 2000/x^2

When dP/dx = 0,
2 - 2000/x^2 = 0
2 = 2000/x^2
x^2 = 1000
x = 31.6

d2P/dx2 when x=31.6 = 4000/(31.6)^3 > 0,
so x = 31.6 would result in a minimum value for P.

When x = 31.6, y = 1000/x = 1000/31.6 = 31.6

The dimensions of the rectangle is 31.6m by 31.6m; it is a square.

Answer 6

Just set it up:

Area = l*w... 1000 = x*y

Then perimeter equals 2x + 2y.

That should make it easier, just isolate the area equation for x or y, then plug it into the perimeter one, and derivate.

Answer 7

Let area(A) = length(l) x width(w) = 10^3
then l=A/w
perimeter(P) = 2l + 2w =2A/w + 2w
dP/dw = d/dw(2Aw^(-1) + 2w) = 2(-1)A/w^2 + 2 set= 0
simplifying, w^2 = A
w = sqr(A)
l= A/w = sqr(A)
sqr(A) = sqr(10^3) = 10sqr(10)