Given a particular perimeter, what rectangle dimentions gives the largest possible area?

Question

the perimeter of a rectangle is 300ft, and you want the most space possible. what should the dimensions be, and how did you get it? what formula should i use?

the perimeter of a rectangle is 300ft, and you want the most space possible. what should the dimensions be, and how did you get it? what formula should i use?

btw, i know its a square. i just dont know how to incorporate a formula

Answer 1

The largest possible area would be given by a square with side length of 75 feet. This can be verified using differentiation (calculus):

x and y = lengths of sides of rectangle
area = x * y
2x + 2y = 300 formula for perimeter
so x + y = 150
area = x * (150 - x) = 150x - x^2

need to find the value of x that will maximize the area. This is determined by finding the first derivative of the above equation, and setting it to 0. On a graph, this is represented by a parabola pointing down, and you're trying to find the point on the parabola that is the largest amount, i.e. the top of the curve, where the slope of the line is 0.

a(x) = 150x - x^2
a'(x) = 150 -2x = 0
2x = 150
x = 75
y = 75
(x and y are equal, so all the sides are of equal length, so you end up with a square)

Answer 2

Where it comes to rectangles, the square always gives the maximum area (75ft x 75ft).

You can prove this using calculus (the maximum point when plotting area against width is 5626ft area for 75ft width, so length must also be 75ft by subtraction and division of perimeter) like so:

Perimeter P = 2 * Length L + 2 * Width W
P = 2L + 2W
2L = P - 2W
L = P/2 - W
L = 150 - W

Therefore

Area A = Length L * Width W
A = LW
A = W (150 - W)
A = -W^2 + 150W

The maximum point on this curve may be found by taking the first derivative:

dA/dW = -2W + 150

This is zero when

150 = 2W
75 = W

We know this must be a maximum point because the second derivative is -2 throughout, so there are no minimum points.

Answer 3

It should be very quickly obvious that the area is maximized when you're dealing with a square, but if you need to go through all the gory details of proving it, here's how you can do it.

The perimeter, or (length + length + width + width), is 300 feet.
Hence, (length + width) is 150 feet.
Let x be the length of the rectangle (in feet).
Then the width must be (150-x).

What you're trying to do is maximize area, or (length * width).

(length * width) = x * (150 - x), or -x^2 + 150x. This is a downward-opening parabola, so the maximum area will be at the tip of the parabola.

Obviously, x * (150-x) equals zero when x=0 or x=150. Since parabolas are symmetrical, that means the tip of the parabola is where x=75. At this point, (150-x) is also 75.

Hence, the area is maximized when your rectangle is a square with 75-ft sides.

Hopefully that helps!

Answer 4

Simply divide the perimeter by 4 . You will get the dimension of the length of side of square which is the largest possible area you can get from the given perimeter

Answer 5

A^2=P^2/16

where A is the sides dimension and P is the perimeter

Answer 6

A square has the greatest area of any rectangle with a given perimeter. For example, a 75x75 square has p=300 and
A=5625 square units. a 100x50 rectangle has p=300 but A=5000 sq units sq units. A 10x280 rectangle has p=300, A=2800 sq units The longer and skinnier the rectangle becomes, the smaller the area given equal perimeters.

Answer 7

if one side of the required rectangle is 'x'
then the other side is (300-2x)/2 i.e. 150-x
so area is x*(150-x)
now area will be maxm. if d/dx of x*(150-x) is 0 and d^2/dx^2 is less than 0.
now d/dx of x*(150-x) = 150-2x : if 150-2x is zero then x=75
so other side is also 75.
now d^2/dx^2 of 150-2x is -2 which is less than 0. so answer is correct.

Answer 8

It's a square.

Answer 9

the largest possible area is hillary clintons thighs and her fat butt