State the feasible domain
A box from a square sheet
A square sheet of cardboard is 18 cm on each side. A square of side \(x\) is cut from each corner, and the sides fold up to make an open box.
Drag \(x\) to the right. The corner squares grow, and the base shrinks to \(18 - 2x\). At \(x = 9\), the corner squares touch, and the base has width 0.
Past \(x = 9\), the squares overlap and turn red. No box exists there, even though the formula \(x(18 - 2x)^2\) still gives a number.
The thick part of the x-axis marks the values of \(x\) that make a box, from 0 to 9.
Any sheet
Take a sheet of any size \(S\). The cut \(x\) cannot be negative, and it cannot be more than half of \(S\), because the two cuts along one side must fit.
These limits come from the picture, not from the formula.
Notation
The feasible domain is the set of inputs for which the situation exists.
Write each requirement as an inequality. The cut \(x\) and the base \(18 - 2x\) are lengths, so neither can be negative.
\[\begin{aligned} &x \ge 0, \qquad 18 - 2x \ge 0 \\ &\Rightarrow\quad 0 \le x \le 9 \end{aligned}\]Decide each endpoint. Keep it if the setting still gives a value there, even a flat box of volume 0. Drop it if no shape exists there or the formula is undefined.
Write the domain beside the formula.
\[\begin{aligned} V(x) &= x(18 - 2x)^2 \\ &\cq{0 \le x \le 9} \end{aligned}\]
When it fails
Some domains have an end that does not belong. Take an open tank that holds 32 m³, with a square base of side \(x\). At \(x = 0\) there is no tank, and its surface area \(x^2 + \tfrac{128}{x}\) divides by 0, so the left end is open.
The tank can be as wide as you like, so the domain has no right end.
Example 1: A pen against a short wall
A farmer has 100 m of fence for a rectangular pen against a wall. The wall is 30 m long, so the side along the wall is at most 30 m. The sides next to the wall are \(x\) long, and the side along the wall is \(100 - 2x\). Find the feasible domain of \(x\).
The side along the wall cannot be negative. This sets the upper limit of \(x\).
\[100 - 2x \ge 0 \quad\Rightarrow\quad x \le 50\]The wall is 30 m long, so the side along it is at most 30 m. The red gap is the part of a longer side past the wall. A longer side means a smaller \(x\), so this sets the lower limit.
\[100 - 2x \le 30 \quad\Rightarrow\quad x \ge 35\]Both endpoints are real pens, so both stay in the domain. At \(x = 35\), the pen uses the whole wall. At \(x = 50\), the side along the wall is 0 and the area is 0.
\[\cq{35 \le x \le 50}\]
Example 2: A fruit stand
A stand sells \(x\) kg of apples a day at a price of \(12 - \tfrac{x}{5}\) dollars per kg. It has only 25 kg to sell. The income is \(R(x) = x\left(12 - \tfrac{x}{5}\right)\). Find the feasible domain of \(x\).
The stand cannot sell a negative amount, so \(x \ge 0\). The upper limit can come from the price or from the supply.
A tempting wrong move: The price is not negative when \(12 - \tfrac{x}{5} \ge 0\). This gives \(x \le 60\), so the domain is \(0 \le x \le 60\).
\[0 \le x \le 60\]The stand has only 25 kg, so the curve to the right of 25 shows sales that cannot happen. The highest point of the whole curve, at \(x = 30\), is not a real sale.
The supply limit of 25 kg is the stricter one, so it sets the upper end.
\[\cq{0 \le x \le 25}\]Both endpoints are real sales, so both stay in the domain. At \(x = 25\), the price is 7 dollars and \(R = 25 \cdot 7 = 175\). At \(x = 0\), nothing is sold and \(R = 0\).
Practice
A cable crosses a river and then follows the bank to a factory. The factory is 4 km down the bank from the point straight across. The cable lands \(x\) km down the bank from that point. Find the feasible domain of \(x\).
Picture the two extreme cables. One lands straight across, and one goes straight to the factory.
Solution
The cable lands between the point straight across and the factory, because a landing outside this stretch makes both parts of the cable longer.
\[0 \le x \le 4\]Both endpoints are real cables, so both stay in the domain. At \(x = 0\), the cable crosses straight across and follows the bank for 4 km. At \(x = 4\), it goes straight to the factory.
\[\cq{0 \le x \le 4}\]
Questions
The figure shows the box cut from the 18 cm square sheet, with volume \(V(x) = x(18 - 2x)^2\).
For which values of \(x\) does the figure show a box?
- \(0 \le x \le 9\)
- \(x \ge 0\)
- \(0 < x < 9\)
- \(0 \le x \le 18\)
Answer
\(0 \le x \le 9\)
What does the figure show at \(x = 9\)?
- A limiting box of volume 0
- A box of volume 324
- No value, because the formula is undefined
- A box that cannot exist, because the squares overlap
Answer
A limiting box of volume 0
A farmer has 60 m of fence for a pen beside a cliff, with no fence along the cliff. The sides next to the cliff are \(x\) long, and the side along the cliff is \(60 - 2x\). What is the feasible domain of \(x\)?
- \(0 \le x \le 30\)
- \(0 \le x \le 60\)
- \(0 < x < 30\)
- \(x \ge 0\)
Answer
\(0 \le x \le 30\)
Summary
You write each requirement of the situation as an inequality, solve for the allowed inputs, and decide each endpoint.
The formula gives a number even where no shape exists, so only the situation can set the domain.
You find the domain before you look for critical points, because its endpoints can also hold the largest or smallest value.
This is one of four skills in topic 5.10 of Which & Why: AP Calculus.