Perimeter-Area Inequality

Let the side lengths of the rectangle be $x$ and $y$ .

We have that $A=xy$ and $P=2x+2y$ .

By AM-GM, $\dfrac{x+y}{2}\ge \sqrt{xy}$ , or $\dfrac{P}{4}\ge \sqrt{A}$ .

Squaring both sides gives $\dfrac{P^2}{16}\ge A$ , and rearranging gives $\dfrac{P^2}{A}\ge \boxed{16}$

I used a simple logic for this. I don't know whether this logic is fully correct or not.

The minimum value of $\frac{P^2}{A}$ will be achieved when $A$ is largest and $P$ is smallest. For all rectangles, square has the largest area for the least perimeter. Let side of square be $x$ units.

So, $\frac{P^2}{A} = \frac{(4x)^2}{x^2} = \frac{16x^2}{x^2} = 16$

Let the side lengths be $x$ and $y$ . So we have: $\dfrac{P^2}{A}=\dfrac{4(x^2+2xy+y^2)}{xy}=4\left(\dfrac x y + \dfrac y x + 2\right)$ Since $(x,y)\in\mathbb{R_+}\times\mathbb{R_+}$ , by AM-GM inequality $\dfrac x y +\dfrac y x\geq 2$ Hence: $\dfrac{P^2}{A}=4\left(\dfrac x y + \dfrac y x + 2\right)\geq 4(2+2)=\fbox{16}$

The area of a variable rectangle (with fixed perimeter) is minimum when it's a square. If the perimeter is P, then each side of the square is P/4. So its area is: $A = \frac{{P}^{2}}{16}$ Therefore, $\frac{{P}^{2}}{A} = 16$

A very simple solution. We all know that a square is also a rectangle and using its formula will improve accuracy and increase efficiency. Taking one side of the square as a simply 1 unit, P^2/A= 4side/side.side = (4side)^2/1.1 16Side/1side

=16.

My solution is pretty short and easy to understand. (it looks a little lengthier because I have explained each and every step)

This is for all rectangles, so make a square with a side of 1. Perimeter squared = 16. Area = 1.

By AM-GM inequality x/y + y/x>2 hence, P^2/A>=16

Let the sides be x and y. Then A=xy, and P=2(x+y). P^2=4(x+y)(x+y)>= 4(2(sqroot(xy))(2(sqroot(xy)) = 16xy (by AM GM). If we divide both sides of the inequality by A, we get P^2/A>=16xy/xy=16.

I have no any idea about mathematic equations to solve.

I just have a conceptual idea about rectangle. One of the most basic rectangle is 'square'. It has the minimum area (i'm not sure but it is one that i have in my mind) when compared with others of the same shortest length.

So, it is a reason why 16 is an answer.

Square is also a rectangle. Use the smallest number which is 1 as the mesurement of one side. Then solve for P2/A. P=1x4 = 4. 4^2 = 16. A=1x1. 16/1 = 16.

O menor retângulo possível é um quadrado, portanto, o perímetro é 4 lado (l) e a área é lado lado(l²). (4l)²/l²=16

minimum value is possible when length=breadth it becomes a square so perimeter is 4s where is side of the square and P^2 =16s^2 and area is s^2 so on dividing p^2 by A we get 16

as we know, p = 2(x+y)..................................eq2 a = x*y.......................................eq 1 putting x = a/y, p = 2(a/y+y) p^ = 4(a/y+y)^ diff w.r .t.x we get 2pdp/dy = 8(a/y+y)(-a/y^+1)....................................................eq 3 for max or minima putting dp/dy = 0 a/y+y = 0 (not possible) -a/y^+1 = 0 y^ = a putting y in eq 1 x^ = a putting x , y in eq 2 we get p^/a = 16 diff eq 3 w.r.tx , 2pd^p/dp^+2(dp/dy)^ = +ve so the func. will be minima. ]

The rectangle has minimum area when it is a square.
Area of the sq=a^2.
Perimeter=4a.
On dividing, we get 16.

P*P/A=4(X+Y)^2/XY Use condition for maxima and minima for minimum value first order derivative wrt x=0 second order derivative wrt x>0

Let the side lengths of the rectangle be a and b. We have that A=ab and P=2(a+b). from this problem we can get {2(a+b)}^2/ab. now the lowest value of a is 1 and b is 1.1 because it is a rectangle.so {2(a+b)}^2/ab=16.something.From this we can say that the lowest value of P^2/A=16.

Smallest perimeter=1+1+1+1=4 4 4=16 Area of the same square=1 1=1 So16/1=16