Find when an accumulated amount is greatest or least
People in a store
The top curve is a store's net rate, how fast the number of people inside changes, in people per hour, t hours after opening. The lower graph shows the number of people, 40 at t = 0.
Move the End slider to the right. Where the curve is above the axis the region is green and the number rises, because the store gains people. Where it is below, the region is red and the number falls.
The curve crosses the axis at t = 1 and at t = 4, where the number has a peak and then a low point.
The greatest and least numbers occur at these two turns or at the ends, t = 0 and t = 5. At those four times the numbers are 40, 51, 24 and 35, so the greatest is 51 and the least is 24.
Any net rate
The top curve is any net rate r, and the lower graph is the amount A. Where r is above the axis A rises, and where it is below, A falls.
A peak of A is where r changes from positive to negative. A low point is where r changes from negative to positive.
The lowest point of the rate curve is not a low point of the amount, which turns only where the rate crosses the axis.
Notation
The slope of the amount is the net rate, so the amount rises where \(r\) is positive and falls where \(r\) is negative. The orange tangent shows this slope.
\[A'(t) = \cy{r(t)}\]A candidate is a time where the amount might be greatest or least. The candidates are the two ends and the times where \(r\) changes sign, because the amount turns only there.
\[\begin{aligned} &t = a, \quad t = b, \\ &r(t) = 0 \text{ with a sign change} \end{aligned}\]At each candidate \(c\), the amount is the starting amount plus the integral of the rate from \(a\) to \(c\).
\[A(c) = A(a) + \cb{\int_a^c} \cy{r(t)}\,dt\]Compare the amounts. The largest is the absolute maximum, and the smallest is the absolute minimum. Here the peak is only a local maximum, because the right end is higher.
When it fails
Raise the curve until its low point touches the axis. There \(r\) is 0 but positive on both sides, so there is no sign change.
The amount flattens for an instant and then rises again, so the peak and the low point merge and disappear. A zero without a sign change is not a candidate.
Example 1: a reservoir
A reservoir holds 100 million gallons at \(t = 0\) days. Its net rate is \(r(t) = 12 - 4t\) million gallons per day for \(0 \le t \le 5\). Find when the reservoir holds the most water and the least water.
The net rate is 0 at \(t = 3\). It is positive before 3 and negative after, so the amount rises and then falls.
\[12 - 4t = 0 \;\Rightarrow\; t = 3\]The candidates are \(t = 0, 3, 5\). The amount is \(100 + 12t - 2t^2\), the starting 100 plus the integral of the rate.
\[\begin{aligned} A(0) &= 100 \\ A(3) &= 118 \\ A(5) &= 110 \end{aligned}\]The greatest amount is 118 at \(t = 3\). The least amount is 100 at \(t = 0\), an end of the interval.
\[\cg{118}, \qquad \cr{100}\]
Example 2: bacteria in a dish
A dish holds 10 thousand bacteria at \(t = 0\) hours. The net rate is \(r(t) = 3(t - 2)(t - 4)\) thousand per hour for \(0 \le t \le 6\). Find when the dish holds the most bacteria.
The rate is positive before \(t = 2\), negative from 2 to 4 and positive after 4. The amount is 10 plus the integral of the rate.
\[A(t) = 10 + t^3 - 9t^2 + 24t\]A tempting wrong move: The rate changes from positive to negative at \(t = 2\), so the greatest amount is there.
\[A(2) = 30\]The amount rises again after t = 4 and ends higher than at t = 2, so a sign change is only a candidate.
Compare every candidate: both ends and both sign changes.
\[\begin{aligned} A(0) &= 10 \\ A(2) &= 30 \\ A(4) &= 26 \\ A(6) &= 46 \end{aligned}\]The greatest amount is 46 thousand at t = 6, an end of the interval.
\[\cg{46 \text{ thousand}}\]
Practice
A water tower holds 80 kiloliters at \(t = 0\) days. Its net rate is \(r(t) = -6t^2 + 42t - 60\) kiloliters per day for \(0 \le t \le 6\). Find when the tower holds the least water and the most water.
Move the end across the two times where the curve crosses the axis. Which way does the amount turn?
Solution
The rate is negative before \(t = 2\), positive from 2 to 5 and negative after 5, so the amount falls, rises and falls again.
\[r(t) = -6(t - 2)(t - 5)\]The candidates are \(t = 0, 2, 5, 6\). The amount is \(80 - 2t^3 + 21t^2 - 60t\).
\[\begin{aligned} A(0) &= 80 \\ A(2) &= 28 \\ A(5) &= 55 \\ A(6) &= 44 \end{aligned}\]The least amount is 28 kiloliters at t = 2. The greatest is 80 kiloliters at t = 0.
\[28 \text{ and } 80\]
Questions
The graph shows the net rate of change of the water in a pond, in liters per minute, for the first 8 minutes.
On which interval is the amount of water decreasing?
- \(2 < t < 6\)
- \(0 < t < 2\)
- Only at \(t = 4\)
Answer
\(2 < t < 6\)
Which times do you compare to find the greatest amount for \(0 \le t \le 8\)?
- \(t = 0, 2, 6\) and \(8\)
- \(t = 2\) only
- \(t = 4\)
- \(t = 0\) and \(8\) only
Answer
\(t = 0, 2, 6\) and \(8\)
A tank holds 30 liters at \(t = 0\). Its net rate is \(2t - 10\) liters per minute for \(0 \le t \le 8\). What is the least amount of water in the tank?
- 5 liters
- 14 liters
- −25 liters
Answer
5 liters
Summary
You find the times where the net rate changes sign, and compare the amount at those times and at both ends.
This works because the amount rises where the rate is positive and falls where it is negative, so it turns only at a sign change.
You use it when a problem asks when a quantity is greatest or least and gives its rate of change.
This is one of five skills in topic 8.3 of Which & Why: AP Calculus.