Maximizing the Area of Rectangles with Fixed perimeter

[flan]

Isn't this type of stuff in the linear programming topic in decision maths?

Maximizing profit while keeping spending small (for a business plan).

I decision. We weren't allowed to do Decision 2, supposed to be too easy, Have to do fp2 instead

Edit: off topic, sorry.

 

[vcuber13]

anyway. Here's an example. For the rectangle/square problem, let \( x \) and \( y \) be the sides of the rectangle with a fixed perimeter \( P \). Of course, \( P=2(x+y) \). AM-GM states that \( \frac{x+y}{2} \ge \sqrt{xy} \), or \( \frac{P}{4} \ge \sqrt{xy} \). Since both sides are positive, we can square both sides: \( xy \le \frac{P^2}{16} \). \( xy \) is maximized when \( xy = \frac{P^2}{16} \), and AM-GM states that this equality is only true when \( x = y \). Therefore, the maximum area of a rectangle with a fixed perimeter occurs when the rectangle is a square.

thats quite interesting, but will that only work if it is an actual rectangle? like if p=3x+2y will it still work?

 

[einstein00]

thats quite interesting, but will that only work if it is an actual rectangle? like if p=3x+2y will it still work?

Sure. I'm guessing some people might want to try this on their own, so it's in the spoiler.

Spoiler

\( P = 3x+2y \)
\( A = xy \)


\( \frac{3x+2y}{2} \ge \sqrt{3x\times2y} \) (here, X (reference to my above post) = 3x and Y = 2y.)
\( \frac{P}{2} \ge \sqrt{6xy} \)
\( \frac{P^2}{4} \ge 6xy \)
\( A = xy \le \frac{P^2}{24} \)

We wish to maximize A, which happens at equality, meaning X = Y. (or 3x = 2y)

\( A = \frac{(600 \text{ meters})^2}{24} = \boxed{15000 \text{ meters}^2} \)

\( P = 3x+3x = 6x = 600 \text{ meters} \)
\( x = \boxed{100 \text{ meters}} \)
\( y = \boxed{150 \text{ meters}} \)

 

[vcuber13]

thats very interesting