Find feasible candidates
Where can the largest box be?
A square of side \(x\) is cut from each corner of a 16 cm by 10 cm sheet. The sides fold up into an open box, and the graph shows its volume \(V\).
A box needs \(10 - 2x \ge 0\), so \(0 \le x \le 5\). At \(x = 0\) and \(x = 5\) the box is flat and its volume is 0. A flat box is allowed, so both ends belong to the domain.
Drag \(x\). The orange line is the tangent to the graph, and its slope is \(V'(x)\). The largest volume can be where the tangent is flat, or at an end of the domain.
The tangent is flat at \(x = 2\) and at \(x = \tfrac{20}{3} \approx 6.67\). The second place is on the faint part of the graph, where the point is red because no box exists.
A value of \(x\) is feasible when the problem allows it. Of the two flat points, only \(x = 2\) is feasible, so discard \(x = \tfrac{20}{3}\). The two ends are feasible too, so the values to test are 0, 2 and 5.
The list of candidates
The figure shows an unnamed objective \(f\) on an interval \([a, b]\). A candidate is an input where the largest or smallest value of \(f\) can occur, and a small dot marks each one.
Drag the Tilt slider. The highest and lowest points of the curve move between candidates. They are always at an endpoint, at a flat point or at a corner.
At a flat point, \(f' = 0\). At a corner, \(f'\) does not exist. An endpoint is a candidate only if the interval includes it.
Notation
Find where the derivative is 0 or does not exist.
\[\begin{aligned} f'(x) &= 0 \\ \text{or} \quad f'(x) &\ \text{does not exist} \end{aligned}\]A solution inside the domain is a critical point, so keep it. A solution outside the domain, such as \(c_3\), is not a critical point, so discard it.
If the domain is a closed interval, add its endpoints.
\[x = a, \qquad x = b\]An endpoint is not a critical point, but it can give the absolute maximum or minimum.
The critical points and the endpoints are the candidates.
When it fails
The highest point of the curve is the endpoint \(b\). If the end is open, the interval does not include \(b\).
The values come close to the height at the open end but never reach it, so there is no absolute maximum.
Example 1: Profit with a capacity limit
A factory makes \(x\) hundred items a week, with \(0 \le x \le 7\). Its profit in thousands of dollars is \(P(x) = -x^3 + 6x^2 + 15x\). Find the candidates for the largest profit.
Differentiate. Factor the derivative and set it equal to 0.
\[\begin{aligned} P'(x) &= -3x^2 + 12x + 15 \\ &= -3(x - 5)(x + 1) = 0 \end{aligned}\]The solutions are \(x = 5\) and \(x = -1\). The value \(-1\) is outside the domain \([0, 7]\), so discard it.
\[x = 5\]The domain is closed, so add the endpoints. The candidates are 0, 5 and 7.
\[x = 0, \qquad x = 5, \qquad x = 7\]
Example 2: A plot with a limited width
A rectangular plot has area 450 m². Its width \(x\) must be between 10 m and 18 m. The fence around it has length \(P(x) = 2x + \dfrac{900}{x}\). Find the candidates for the shortest fence.
Differentiate and solve \(P'(x) = 0\).
\[\begin{aligned} P'(x) &= 2 - \frac{900}{x^2} = 0 \\ x^2 &= 450, \quad x = \pm 15\sqrt2 \end{aligned}\]A tempting wrong move: The critical point \(x = 15\sqrt2\) is the candidate.
\[x = 15\sqrt2 \approx 21.2\]The plot is at most 18 m wide, so no plot has the width 21.2. The point lies on the faint part of the curve, beyond the end of the domain, and its row in the table is red.
Neither solution lies in the domain, so discard both. The candidates are the two endpoints.
\[P(10) = 110, \qquad P(18) = 86\]
Practice
A poster has area 600 cm². The margins are 2 cm at each side and 3 cm at the top and bottom. Let \(x\) be the width of the poster. The printed area is \(P(x) = (x - 4)\left(\dfrac{600}{x} - 6\right)\) for \(4 \le x \le 100\). Find the candidates for the largest printed area.
Expand the product first. Keep only the solutions on the solid part of the curve. Add the endpoints.
Solution
Expand the product, so that the derivative is easy to find.
\[P(x) = 624 - 6x - \frac{2400}{x}\]Differentiate and solve \(P'(x) = 0\).
\[\begin{aligned} P'(x) &= -6 + \frac{2400}{x^2} = 0 \\ x^2 &= 400, \quad x = \pm 20 \end{aligned}\]The value \(-20\) is outside the domain, so discard it. The domain is closed, so add its endpoints, even though the printed area there is 0. The candidates are 4, 20 and 100.
\[x = 4, \qquad x = 20, \qquad x = 100\]
Questions
Questions 1 and 2 use the objective \(f(x) = 2x^3 - 9x^2 + 12x\), drawn as the chalk curve.
The domain is \([0, 3]\). Which values of \(x\) are candidates for the absolute extrema?
- 0, 1, 2 and 3
- 1 and 2
- 0 and 3
- 0, 1 and 2
Answer
0, 1, 2 and 3
The domain changes to \([1.5, 3]\). Which values of \(x\) are candidates now?
- 1.5, 2 and 3
- 1, 2 and 3
- 2 only
- 1, 1.5, 2 and 3
Answer
1.5, 2 and 3
Find the candidates for the absolute extrema of \(f(x) = x^3 - 12x\) on \([-1, 5]\).
- \(-1\), 2 and 5
- \(-2\) and 2
- \(-2\), \(-1\), 2 and 5
- 2 and 5
Answer
\(-1\), 2 and 5
Summary
You find where the derivative is 0 or does not exist, keep the points in the domain, and add the endpoints of a closed interval.
This works because a continuous function on a closed interval has its absolute extrema only at these points.
Use it to list the candidates before you compare any values.
This is one of four skills in topic 5.11 of Which & Why: AP Calculus.