First, let’s draw a generic rectangle of perimeter $12$.

The area of the rectangle is $$A = x\cdot y.$$ In order to be able to use calculus techniques, we have to express the area as a function of a single variable.

To this end, we have to express the variable, say $y$, in terms of $x$. How can this be achieved?

Recall, we are given the constraint: “…rectangles of perimeter $12$…”

If a rectangle has perimeter $12$, then $2x+2y=12$, and we can solve for $y$ in terms of $x$, $$y=\answer [given]{6-x}.$$

Hence the area of a rectangle of perimeter $12$ can be expressed as a function of a single variable, $x$, $$A(x) = x\left (\answer [given]{6-x}\right ).$$ However, for the side lengths to be physically relevant, we must assume that $x>0$ and $6-x>0$.

Therefore, the domain of the function $A$ is the interval $\left (\answer [given]{0},\answer [given]{6}\right )$.

So, to maximize the area of the rectangle, we have to find the global maximum of the function $A$ on the interval $(0,6)$. Since the domain of $A$ is an open interval, the global maximum (if it exists!) has to be a local maximum and, therefore, it occurs at a critical point. So, we have to find all critical points of $A$ on the interval $(0,6)$. Since $$A'(x) = \answer [given]{6-2x},$$ in order to find the critical points, we have to solve the equation $$\answer [given]{6-2x} = 0.$$ It follows that the only critical point of $A$ is at $x=\answer [given]{3}$.

We can make a sign chart for $A'(x)$ on the interval $(0,6)$:

Since $A'(x)$ is positive on $(0,3)$ and negative on $(3,6)$, the function $A$ has a global maximum at $x=3$. This is exactly when the rectangle is a square!

First we draw a picture. Here is a rectangle with an area of $100$.

To this end, we have to express the variable, say $y$, in terms of $x$, using the constraint, $A=x\cdot y=100$, $$y=\answer [given]{\frac {100}{x}}.$$

Hence the perimeter of a rectangle with area $100$ can be expressed as a function of a single variable, $x$, $$P(x) = 2x+2\left (\answer [given]{\frac {100}{x}}\right ).$$ Again, for the side lengths to be physically relevant, we must assume that $x>0$.

Therefore, the domain of the function $P$ is the interval $(\answer [given]{0},\answer [given]{\infty })$.

So, to minimize the perimeter of the rectangle, we have to find the global minimum of the function $P$ on the interval $\left (0,\infty \right )$.

Since the domain of $P$ is an open interval, the global minimum (if it exists!) has to be a local minimum and, therefore, it occurs at a critical point. So, we have to find all critical points of $P$ on the interval $(0,\infty )$. Since $$P'(x) = \answer [given]{2-\frac {200}{x^2}},$$ in order to find the critical points, we have to solve the equation $$\answer [given]{2-\frac {200}{x^2}} = 0.$$ Solving for $x$ gives us $x=\pm \answer [given]{10}$. Only the value $x=\answer [given]{10}$ is in the domain of $P$. Since $P'(x)$ is defined everywhere on the interval $(0,\infty )$, there are no more critical values, and there are no endpoints. Is there a local maximum, minimum, or neither at $x=10$? The second derivative is $$P''(x)=\answer [given]{400/x^3},$$ and $P''(10)>0$, so there is a local minimum. Since there is only one critical point, this is also the global minimum.

So the rectangle with smallest perimeter is the $10\times 10$ square.

We can illustrate our result with the graph of $P$.

The picture below shows three such rectangles. There are infinitely many such rectangles, one for each choice of $x$ in $(0,\sqrt {a})$.

We want to maximize $A$, the area of the rectangle. The lower right corner of the rectangle is at $(x,x^2)$, and once this is chosen the rectangle is completely determined. Let $h$ denote the height and let $w$ denote the width of the rectangle.

Then $$A=w\cdot h,$$ where $$w=2x \text { and } h=\answer [given]{a-x^2}.$$ Therefore, we can express the area $A$ as a function of $x$ $$A(x)=2x\left (\answer [given]{a-x^2}\right ).$$ We want the maximum value of $A(x)$ when $x$ is in $[0,\sqrt {a}]$. You might object to allowing $x=0$ or $x=\sqrt {a}$, since then the “rectangle” has either no width or no height, so is not “really” a rectangle. But the problem is somewhat easier if we simply allow such rectangles, which have zero area as we may then apply the Extreme Value Theorem and see that we indeed have a maximum and minimum value.

Setting $0=A'(x)=\answer [given]{-6x^2+2a}$ we find $x=\answer [given]{\sqrt {a/3}}$ as the only critical point. Testing this and the two endpoints (as the maximum could also be there), we have $A(0)=A(\sqrt {a})=\answer [given]{0}$ and $A(\sqrt {a/3})=\answer [given]{(4/9)\sqrt {3}a^{3/2}}$. Hence, the maximum area occurs when the rectangle has dimensions $2\sqrt {a/3}\times (2/3)a$.