Variables, Parameters, and Building a Model

Example For the rectangular pen with 24 meters of fencing, a model for the area in terms of the width $x$ in meters is

The assumptions are that all 24 meters are used, that the pen is a genuine rectangle, and that the fence has no gate or gap. The rest of this page explains where each piece comes from.

Example In the pen, the width $x$, the length $y$ and the area $A$ are variables, since different pens have different values. The total fencing $24$ and the number $2$ in the perimeter formula $2x + 2y$ are constants.

Example Let the pen use $P$ meters of fencing. The same reasoning as before gives

Here $x$ is the variable and $P$ is a parameter. Setting $P = 24$ gives the earlier model, and setting $P = 40$ gives $A(x) = x(20 - x)$ on $0 < x < 20$. Each value of $P$ gives one member of a family of models, and within one calculation $P$ does not change. The letter does not decide the role. The $24$ was a constant in the first model and plays the part of a parameter here, and if you fixed the width and let the fencing vary, $P$ would become the variable.

Example If you choose the width of the pen, the area follows. Width $x$ is independent and area $A$ is dependent.

You could instead choose the area and ask for the width, but then two widths give the same area of $20$, so width would not be a function of area.

Example The pen has width $x$ and length $y$, and the fencing gives the constraint

Solving for the length gives $y = 12 - x$. The area $A = xy$ has two variables, and substituting the constraint leaves one,

This is the step calculus needs, because the tools of single-variable calculus work with functions of one input.

Example The formula $x(12 - x)$ is defined for every real $x$. A genuine rectangle has both sides strictly positive, so $x > 0$ and $12 - x > 0$. Together these give

At $x = 0$ or $x = 12$ the formula gives area $0$, but the pen has collapsed into a segment, so those endpoints are excluded under the assumption of a nondegenerate rectangle. At $x = 13$ the formula gives $-13$, a negative area, which is a clear sign the input is outside the model.

Example With $x$ in meters, the length $12 - x$ is in meters because $12$ is a length in meters. The area $x(12 - x)$ is meters times meters, square meters. For a delivery charge of $4$ dollars plus $1.50$ dollars per kilometer for $d$ kilometers,

The kilometers cancel, so both terms are in dollars and the sum makes sense.

Example A courier charges a fixed initial fee of $4$ dollars, then $1.50$ dollars per kilometer for the first $10$ kilometers and $1$ dollar per kilometer for each kilometer beyond that. Assume the distance $d$ is any nonnegative real number of kilometers, the $4$ dollar fee is charged once on every delivery, and within each band the charge grows in proportion to the distance, with no rounding up to whole kilometers. The cost in dollars is

The second piece starts from $19$ because the first $10$ kilometers cost $4 + 1.5 \cdot 10 = 19$ dollars. Both pieces give $19$ at $d = 10$, so the cost has no jump there. A ride of $6$ kilometers costs $C(6) = 13$ dollars and a ride of $14$ kilometers costs $C(14) = 23$ dollars.

A common slip is to write the second piece as $4 + 1 \cdot d$, which charges the lower rate for the whole trip and makes the cost drop suddenly from $19$ to about $14$ just past $10$ kilometers. Checking the value at the breakpoint catches it.