A piece of wire of length P is to be bent into a rectangle.?

Question

What is the maximum area that can be enclosed by the rectangle?

Answer 1

Perimeter = P; Area = Q
2 sides = x and y
Q = Px - x^2 is the maximum area that can be enclosed.
How to derive this, refer: http://www.math.tamu.edu/~don.allen/masters/egypt_babylon/area/babylon_area_perimeter.htm

Answer 2

The maximum area is P^2/16.
The calculations are as follows
Let one side of the rectangle be 'x'
Then the other side will be (P-2x)/2
The area will be
A = x((P/2)-x) =x(P/2) - x^2
Differentiate A with respect to x
A' = P/2 - 2x
and the second derivative will be
A" = -2
For maximum A
A' = 0 and A"<0
thus P/2 - 2x = 0
2x = P/2
x = P/4
Other side = (P - 2x)/2 = (P - 2P/4)/2 = (P/2)/2 = P/4
Thus maximum area = P/4 * P/4 = P^2/16

Answer 3

P^2 / 16

A=LW
Where L is the chosen length for a side of the rectangle, and w is the width

P is the length of the wire, P is also the perimeter of the rectangle formed. Since we are forming a rectangle we know that:
P=2L+2W

Therefore,
2L = P-2W
L = (P/2) - W

By Substituting into our area function.
A= W((P/2) - W)
A = W(P/2) -W^2
or
A= W^2-W(P/2)

Which is a concave down quadratic, therefore the vertex is the coordinate of the maximum of this function. x=-b/2a is the x (in this case w) coordinate of the vertex of a quadratic, therefore:

W=P/[2(2)]
W = P/4

And since,
L = (P/2) - W
L=(P/2) - P/4
L = P/4

A=LW
A=(P/4)(P/4)
A = P^2 /16

Answer 4

To create maximum area in this situation you would need to create a square. All sides would be equal. P/4 gives you the length of each side... Area equals base x height... So... (P/4)x(P/4) or (P/4)^2 is the answer.

Answer 5

In case you are not familiar with calculus (and because the method fascinates me)
Let P = 2(L + W)
A = LW
L = (P/2) - W
A = W((P/2) - W)
A = - W^2 + (P/2)W
A = - (W^2 - (P/2)W + (P/4)^2) + (P/4)^2
A = - (W^2 - (P/4))^2 + (P/4)^2
which is a parabola opening down with vertex at
((p/4), (P/4)^2), so the maximum area occurs when W = P/4 (= L), and is P^2/16

Answer 6

A square of side P/4 will have area (P^2)/16

Answer 7

If P is bent in four equal pieces you obtain a square having sides the lenght of 1/4P. Area would then be 1/4P x 1/4P, area would be 1/16Psquares....

Answer 8

I think it is P squared but I'm not so sure