Question

Find the maximum Area of a rectangle whose perimeter is 320 ft. and find the dimensions of the one with the largest area.

Answer 1

Ok here is how I see it.

Imagine a square with sides in length of 'a':
\[
\text{square_side} = a \\
\text{square_area} = a^2
\]
Let's say we wanna make a rectangle with same area are but different sides.
I know it's not what in the question, but it's easier for me to go that way.
Let's say there is a factor 'x' that you choose any value you wish to modify one of the sides.
In order to keep the same area:
\[
\text{For rectangle with side } (x \cdot a) \text{ and b}: \\
\text{rectangle_area} = (x \cdot a) \cdot b \\\;\\
\text{rectangle_area} = \text{square_area} \quad \text{(by definition of the rectangle)} \\
x \cdot a \cdot b = a^2 \; // \div (x\cdot a) \\
b = \frac{a}{x}
\]
In short, that means that if we take a square and multiply one side by X, in order for the new rectangle to have same area, other side has to be multiplied by 1/X.
That also means that if one side grows, other has to shrink in order for area to be the same (trivial).

Now we wanna make an inequality between the modification of the sides. In order to make this inequality right we'll say x > 1. We can say that because x < 1 means the same, just sides swapped..
\[
x > 1 \\
\text{growth_value} = \text{rectangle_ax} - \text{square_side} = a \cdot x -a \\
\text{reduction_value} = \text{square_side} - \text{rectangle_b} = a - \frac{a}{x} \\
\text{reduction_value} < \text{growth_value} \quad\quad \text{ (A claim to prove)}\\
a - \frac{a}{x} < a \cdot x - a \\
\frac{a \cdot x - a}{x} < a \cdot x - a \;// \div (a \cdot x - a) \\
\frac{1}{x} < 1
\]
That is true since we defined x > 1...
That looks a bit messy, but what that basically means is that in order to make any rectangle with same area of the square, we have to make one side grow more than we reduce the other one.
That means that if we sum the rectangle's sides up it is bigger than the square's sides. (And so is the perimeter..)

Since the square shape has the smallest possible perimeter, it means that if we want to modify one of the rectangles to have the same perimeter as the square, we would have to shrink them in some matter, and that would make their area smaller.

Square therefore has the largest possible area for the perimeter, so
\[
\text{square_perimeter} = 320_{ft} \\
\text{square_side} = a = \frac{\text{square_perimeter} }{4} = \frac{320_{ft}}{4} = 80_{ft} \\
\text{square_area} = a^2 = (80_{ft})^2 = 6400_{ft^2}
\]
So, a square with sides of 80ft has the largest possible area for the perimeter which is 6400ft^2.

Sorry it's so long =|
I'm sure there is a much simpler way to show it.. but i'm too tired to think of it