Start by identifying the variables.

- \(h\) is the horizontal width of the rectangle
- \(v\) is the vertical length of the rectangle
- \(C\) is the cost of making the rectangle

Once we have identified our variables, we need to find a formula for the cost because that is what we want to minimize. Let \(p\) be the unit cost for a vertical cut of 1 cm so that \(2p\) is the cost of a horizontal cut of 1 cm. The rectangle involves two horizontal cuts and two vertical cuts. So the total cost is given by

\begin{equation*}
C = 2h \cdot 2p + 2v \cdot p = (4h+2v) p.
\end{equation*}

The value of \(p\) is just the unit cost, so we really want to minimize \(C = 4h+2v\) (measured in \(p\) units).

Our objective function involves two independent variables. This means there must be an additional constraint. Reviewing the problem, we recall that the total area needs to be \(500 \; cm^2\text{.}\) The area is computed by \(A=h \cdot v = 500\) so that we can treat \(v\) as another dependent variable,

\begin{equation*}
v = \frac{500}{h}.
\end{equation*}

Substituting this formula into our objective function, we can rewrite it involving only a single independent variable \(h\text{:}\)

\begin{equation*}
C = 4h + 2 \cdot \frac{500}{h} = 4h + \frac{1000}{h}.
\end{equation*}

Next, we find the derivative:

\begin{equation*}
C'(h) = 4 - \frac{1000}{h^2}.
\end{equation*}

We want to see where the derivative is positive and negative, so we find a common denominator and then factor:

\begin{equation*}
C'(h) = \frac{4h^2-1000}{h^2} = \frac{4(h^2-250)}{h^2}.
\end{equation*}

As a mathematical function, there is a vertical asymptote at \(h=0\) and intercepts at \(h= \pm \sqrt{250}\text{.}\) Geometrically, only \(h \gt 0\) makes physical sense, so we do sign analysis on \((0,\sqrt{250})\) and \((\sqrt{250},\infty)\text{.}\)

The sign analysis of \(C'(h)\) informs us that \(C(h)\) is decreasing on \((0,\sqrt{250})\) and increasing on \((\sqrt{250},\infty)\text{.}\) The first derivative test then guarantees that \(C(h)\) has a local minimum at \(h=\sqrt{250}\) over the interval \((0,\infty)\) which is the entire physically relevant domain. A graph of the cost as a function of \(h\) illustrates our result.

We finish by interpreting our mathematics. The question was how to cut the rectangle. Our analysis gave us a value for \(h=\sqrt{250} \approx 15.81 \; cm\text{.}\) We also need \(v\text{,}\) which was another dependent variable:

\begin{equation*}
v=\frac{500}{h} = \frac{500}{\sqrt{250}} = 2 \sqrt{250} \approx 31.62 \; cm.
\end{equation*}

The minimal cost rectangle would have a horizontal cut of 15.81 cm and a vertical cut of 31.62 cm.