find the dimensions?

Question

find the dimensions of a rectangle a with the greatest are whose perimeter is 30 feet.

Answer 1

If you call one side x, then since the perimeter is 30ft,
the other side must be (30-2x)/2 ft.

The area as a function of x is:
a(x) = x(30-2x)/2 = 15x - x²

which is maximised when a'(x) = 0

a(x) = 15x - x²
a'(x) = 15 - 2x
=> a(x) is maximised when x=7.5ft

Then the dimensions for greatest area
are x=7.5ft and (30-2x)/2 = 7.5ft also.

(i.e. a square 7.5 x 7.5 ft as you might have guessed)

Answer 2

if the perimeter of a rectangle is given, the greatest area happens when it is square.

thus, a=30/4=7.5

area=7.5^2

Answer 3

Always remember that area is the greatest when the length and the breadth are the same (in case of quadrilaterals)

Now, given, 2(l + b) = 30
that is, l + b = 15

Since, l and b are the same to be of the greatest area l = b

so in the equation 2l = 15
l = b = 7.5 feet

thus, area is l*b = 7.5 * 7.5 = 56.25 sq. feet

Answer 4

30 = 2 * (L + W)
30 / 2 = L + W
15 = L + W
Spread 15 to equal measurement between length and width.

Answer:
Length = 7.5 feet and Width = 7.5 feet

Answer 5

A = LW
2(L + W) = 30
A/W + W = 15
A + W^2 = 15W
A = - (W^2 - 15W + (15/2)^2) + 225/4
A = - (W^2 - 15/2)^2 + 225/4
A = 225/4 = 56.25 ft^2
W = 15/2 = 7.5 ft.
L = 15/2 = 7.5 ft.

Answer 6

It will be a square that is 7.5 x 7.5