At x = 0 the tangent is flat, and the curve is lower on both sides near 0. Which statement about f(0) = 0 on [-3, 3] is true?

It is a local maximum, but not the absolute maximum.

It is the absolute maximum, because f' changes from positive to negative at 0.

It is the absolute maximum, because f''(0) < 0.

It is not a local maximum, because f(2) is higher.

At x = 2 the value is f(2) = -16. Which reason shows that it is the absolute minimum on [-3, 3]?

The candidates give 9, -16, 0, -16 and 9, and -16 is the smallest.

f' changes from negative to positive at 2.

For g(x) = x + \dfrac{16}{x} on x > 0, which reason shows that g(4) = 8 is the absolute minimum?

g''(x) = \dfrac{32}{x^3} > 0 for every x > 0, so the graph is concave up on the whole domain.

Topic 5.11

Justify the best candidate

Builds on listing the candidates for an optimization problem. 7 minutes.

Deliveries to a shop

A shop pays 3600 dollars for each delivery, and stock held longer costs more to store. For deliveries every \(x\) days, the average cost per day is \(c(x) = \tfrac{3600}{x} + 4x\).

Drag \(x\). The curve is red where the cost falls and green where it rises. The tangent is flat at \(x = 30\), where \(c(30) = 240\).

A flat tangent alone does not make 240 the least cost. The chart under the graph shows the sign of \(c'\): negative for every \(x\) below 30 and positive for every \(x\) above.

The cost falls until 30 and rises after it, so \(x = 30\) gives the least cost.

A window on the whole domain

The figure shows a function \(f\) with no formula and a flat point at \(c\). Two faint edges mark a window around \(c\), and the orange line is at the height \(f(c)\).

The bands between the curve and the orange line are green where \(f\) is higher than \(f(c)\) and red where \(f\) is lower.

Make the window small. If it shows no red, nothing near \(c\) is lower, so \(f(c)\) is a local minimum. Widen it to the whole domain. If it still shows no red, nothing on the domain is lower, so \(f(c)\) is the absolute minimum.

Notation

On a closed interval, compare the values of \(f\) at all the candidates. The largest is the absolute maximum and the smallest is the absolute minimum.
\[f(a), \quad f(c), \quad f(b)\]
A continuous function on a closed interval has both.
On any domain, use the sign of \(f'\). If \(f' < 0\) at every \(x\) before \(c\) and \(f' > 0\) at every \(x\) after \(c\), then \(f\) falls until \(c\) and rises after it. No value is lower, so \(f(c)\) is the absolute minimum.
\[\begin{aligned} f' < 0 \ &\text{for } x < c \\ f' > 0 \ &\text{for } x > c \end{aligned}\]
For a maximum, reverse the signs.
Or use \(f''\). If \(f'(c) = 0\) and \(f'' > 0\) on the whole domain, the graph is concave up everywhere. It stays above the flat tangent at \(c\), so \(f(c)\) is the absolute minimum.
\[\begin{aligned} f'(c) &= 0 \\ f'' > 0 \ &\text{on the domain} \end{aligned}\]

When it fails

Lower the left side. Near \(c\), \(f'\) still changes from negative to positive, so \(f(c)\) is a local minimum. But \(f'\) is not negative at every \(x\) before \(c\), because the curve rises after the dip.

The window across the whole domain shows a red band where the curve is below \(f(c)\). A sign change near one point proves only a local minimum.

Example 1: A can with least material

A closed can holds \(54\pi\) cm³. With radius \(r\) cm, its surface area is \(S(r) = 2\pi r^2 + \dfrac{108\pi}{r}\), for \(r > 0\). Show that the least material is at \(r = 3\).

Differentiate and solve \(S'(r) = 0\).
\[\begin{aligned} S'(r) &= 4\pi r - \frac{108\pi}{r^2} = 0 \\ r^3 &= 27, \quad r = 3 \end{aligned}\]
The domain \(r > 0\) has no endpoints to compare, so use the second derivative on the whole domain.
\[S''(r) = 4\pi + \frac{216\pi}{r^3}\]
Both terms are positive for every \(r > 0\), so the graph is concave up on the whole domain and stays above the tangent. Its flat point is the lowest point.
\[S''(r) > 0 \ \text{for all } r > 0\]
The least area is at \(r = 3\).
\[S(3) = 18\pi + 36\pi = 54\pi\]

Example 2: A heated tank

A reaction heats a tank. The temperature rise in degrees Celsius after \(t\) minutes is \(T(t) = t^3 - 9t^2 + 24t\), for \(0 \le t \le 6\). When is the temperature rise greatest?

Differentiate and solve \(T'(t) = 0\). The candidates are the flat points and the endpoints: 0, 2, 4 and 6.
\[\begin{aligned} T'(t) &= 3t^2 - 18t + 24 \\ &= 3(t - 2)(t - 4) \\ T'(t) &= 0 \ \text{at } t = 2, 4 \end{aligned}\]
A tempting wrong move: The derivative changes from positive to negative at \(t = 2\), so the greatest rise is \(T(2) = 20\).
\[T(2) = 20\]
The tangent is flat at \(t = 2\), but the graph rises again after \(t = 4\) to \(T(6) = 36\), above the dashed line at 20. A sign change near one point proves only a local maximum.
Compare the values at all four candidates.
\[\begin{aligned} T(0) &= 0, \quad T(2) = 20 \\ T(4) &= 16, \quad T(6) = 36 \end{aligned}\]
The greatest rise is 36, at the endpoint \(t = 6\).

Practice

A lamp hangs \(h\) meters above a table. At a point on the table 6 m from the spot under the lamp, the brightness is \(I(h) = \dfrac{h}{(h^2 + 36)^{3/2}}\), for \(h \ge 0\). Show that the brightness is greatest at \(h = 3\sqrt2\).

The curve rises to the flat point and falls after it. Show this with the sign of \(I'\) on each side.

Solution

Differentiate. The denominator is positive, so \(I'\) has the sign of \(36 - 2h^2\).
\[I'(h) = \frac{36 - 2h^2}{(h^2 + 36)^{5/2}}\]
Solve \(I'(h) = 0\). The negative root is outside the domain \(h \ge 0\).
\[h^2 = 18, \qquad h = 3\sqrt2\]
For \(0 \le h < 3\sqrt2\), the sign is positive, so \(I\) rises. For \(h > 3\sqrt2\), the sign is negative, so \(I\) falls.
The signs cover the whole domain, so \(I(3\sqrt2)\) is the absolute maximum.

Questions

Questions 1 and 2 use the objective \(f(x) = x^4 - 8x^2\) on \([-3, 3]\).

At \(x = 0\) the tangent is flat, and the curve is lower on both sides near 0. Which statement about \(f(0) = 0\) on \([-3, 3]\) is true?
- It is a local maximum, but not the absolute maximum.
- It is the absolute maximum, because \(f'\) changes from positive to negative at 0.
- It is the absolute maximum, because \(f''(0) < 0\).
- It is not a local maximum, because \(f(2)\) is higher.
Answer
It is a local maximum, but not the absolute maximum.
At \(x = 2\) the value is \(f(2) = -16\). Which reason shows that it is the absolute minimum on \([-3, 3]\)?
- The candidates give 9, \(-16\), 0, \(-16\) and 9, and \(-16\) is the smallest.
- \(f'\) changes from negative to positive at 2.
- \(f'(2) = 0\).
- \(f(2) < f(0)\).
Answer
The candidates give 9, \(-16\), 0, \(-16\) and 9, and \(-16\) is the smallest.
For \(g(x) = x + \dfrac{16}{x}\) on \(x > 0\), which reason shows that \(g(4) = 8\) is the absolute minimum?
- \(g''(x) = \dfrac{32}{x^3} > 0\) for every \(x > 0\), so the graph is concave up on the whole domain.
- \(g''(4) = \dfrac12 > 0\).
- \(g'(4) = 0\).
- \(g(4) = 8 < g(1) = 17\).
Answer
\(g''(x) = \dfrac{32}{x^3} > 0\) for every \(x > 0\), so the graph is concave up on the whole domain.

Summary

You compare the values at every candidate, or use the sign of \(f'\) or \(f''\) on the whole domain.

A sign change near one point proves only a local minimum or maximum, so the argument must cover the whole domain.

Use this after you list the candidates, to show which is best.

This is one of four skills in topic 5.11 of Which & Why: AP Calculus.