Find extrema of the accumulation

A reservoir that turns around

A reservoir holds water to a height of 10 meters at time \(t = 0\) months. The upper graph shows the rate \(f(t)\) at which the height changes, in meters per month. The lower graph shows the height \(H(t)\), and the shaded region is red below the axis and teal above it.

Drag t to the right. The lower graph is red where the height falls and green where it rises.

At \(t = 4\), the rate crosses the axis from below to above. The red region has area 8, so the height is \(10 - 8 = 2\) meters. After \(t = 4\) the height rises, so 2 is a low point.

At \(t = 8\), the rate crosses the axis from above to below. The teal region from 4 to 8 has area 12, so the height is \(2 + 12 = 14\) meters. After \(t = 8\) the height falls, so 14 is a high point.

Where the rate crosses the axis

Drag \(x\) along the graph. Each time the upper curve crosses the axis, the lower graph turns around.

If the curve crosses from above to below, the lower graph changes from rising to falling. This point is a relative maximum of the accumulation.

If the curve crosses from below to above, the lower graph changes from falling to rising. This point is a relative minimum.

The rate gives the place of an extremum, and the starting value plus the signed area gives its height.

Show the ends of the lower graph. The right end is above the relative maximum, so the largest value is at an end.

Notation

1. The lower graph is the accumulation \(F\), whose slope is \(f\). A critical point of \(F\) is where its slope is zero, so the critical points are the zeros of \(f\).

   \[F'(x) = \cy{f(x)} = 0\]

2. A relative maximum is larger than the values near it. \(F\) has one at a critical point \(c\) if \(f\) changes from positive to negative there.

   \[\cg{f > 0} \text{ before } c, \quad \cr{f < 0} \text{ after } c\]

3. A relative minimum is smaller than the values near it. \(F\) has one if \(f\) changes from negative to positive at \(c\).

   \[\cr{f < 0} \text{ before } c, \quad \cg{f > 0} \text{ after } c\]

4. The value at an extremum is the starting value \(F(a)\) plus the signed area from \(a\) to \(c\).

   \[F(c) = F(a) + \cb{\int_a^c f(t)\,dt}\]

5. The absolute maximum and minimum are the largest and smallest values on \([a, b]\). Compare \(F\) at each critical point and at both endpoints.

   An endpoint is never a relative extremum, because that needs points on both sides. It can still be an absolute extremum.

When it fails

A zero of \(f\) alone is not enough. Lifted, the curve touches the axis at one point but stays above it on both sides.

The slope of \(F\) is zero there and positive on both sides, so the lower graph rises through it with no extremum.

Example 1: two turns

\(F(x) = 3 + \displaystyle\int_0^x \cos t\,dt\) for \(0 \le x \le 2\pi\). Find the relative extrema of \(F\) and their values.

1. The slope of \(F\) is \(\cos x\), which is zero at \(x = \tfrac{\pi}{2}\) and \(x = \tfrac{3\pi}{2}\).

   \[F'(x) = \cos x = 0\]

2. The cosine is positive before \(\tfrac{\pi}{2}\) and negative after it, so \(F\) has a relative maximum there.

   \[\cg{\cos x > 0} \;\to\; \cr{\cos x < 0}\]

3. The cosine is negative before \(\tfrac{3\pi}{2}\) and positive after it, so \(F\) has a relative minimum there.

   \[\cr{\cos x < 0} \;\to\; \cg{\cos x > 0}\]

4. An antiderivative of \(\cos t\) is \(\sin t\), and \(\sin 0 = 0\), so each value is 3 plus \(\sin x\).

   \[\begin{aligned} F\!\left(\tfrac{\pi}{2}\right) &= 3 + \sin\tfrac{\pi}{2} = \cg{4} \\ F\!\left(\tfrac{3\pi}{2}\right) &= 3 + \sin\tfrac{3\pi}{2} = \cr{2} \end{aligned}\]

Example 2: the largest value on an interval

\(G(x) = 10 + \displaystyle\int_0^x (3t^2 - 12t)\,dt\) for \(0 \le x \le 5\). Find the absolute maximum and the absolute minimum of \(G\).

1. The slope of \(G\) is \(3x^2 - 12x = 3x(x - 4)\). It is zero at \(x = 0\) and \(x = 4\), and only \(x = 4\) lies inside the interval, because \(x = 0\) is its left endpoint.

   \[G'(x) = 3x(x - 4)\]

2. The slope is negative before 4 and positive after 4, so \(G\) has a relative minimum at \(x = 4\). It has no relative maximum.

   \[\begin{aligned} G(4) &= 10 + \left[t^3 - 6t^2\right]_0^4 \\ &= 10 + (64 - 96) = \cr{-22} \end{aligned}\]

3. A tempting wrong move: There is no relative maximum, so \(G\) has no maximum.

   \[\begin{aligned} &\text{no relative maximum} \\ &\Rightarrow\; \text{no maximum} \end{aligned}\]

   The lower graph is highest at its left end, at height 10. An endpoint is never a relative maximum, but it can be the absolute maximum.

4. Compare \(G\) at the critical point and at both endpoints. The absolute maximum is 10, at \(x = 0\). The absolute minimum is \(-22\), at \(x = 4\).

   \[\begin{aligned} G(0) &= 10 \\ G(4) &= -22 \\ G(5) &= 10 + (125 - 150) = -15 \end{aligned}\]

Practice

\(H(x) = 4 + \displaystyle\int_0^x 3(t - 3)(t + 1)\,dt\) for \(0 \le x \le 5\). Find the absolute maximum and the absolute minimum of \(H\).

The lower graph turns where the integrand crosses the axis. Compare H there and at both endpoints.

Questions

The upper graph shows a rate \(f\) on \([0, 8]\), a straight-line path through \((0, -2)\), \((1, 0)\), \((3, 4)\), \((5, 0)\) and \((8, -3)\). The amount is \(F(x) = 5 + \int_0^x f(t)\,dt\).

1. Where does \(F\) have a relative maximum?

   • \(x = 5\)
   • \(x = 3\)
   • \(x = 8\)
   Answer

   \(x = 5\)

2. What is the absolute minimum value of \(F\) on \([0, 8]\)?

   • 4
   • 5
   • \(-1\)
   Answer

   4

3. \(K(x) = 10 + \displaystyle\int_0^x (2t - 6)\,dt\). \(K\) has a relative minimum at \(x = 3\). What is \(K(3)\)?

   • 1
   • 10
   • \(-9\)
   Answer

   1

Summary

You find an extremum of the accumulation where the rate changes sign, and compare the endpoints for absolute extrema.

This works because the accumulation turns only where its slope changes sign, and its height there is the starting value plus the signed area.

You use it to find the largest or smallest accumulated amount on an interval.

This is one of five skills in topic 6.5 of Which & Why: AP Calculus.