APPLICATION OF DERIVATIVES

Chapter 6

Application of Derivatives

Big idea: the derivative measures how fast something changes. Once you can measure change, you can ask the most useful questions: when is it rising, when is it falling, and when is it at its highest or lowest?

In Chapter 5 you learned to find the derivative. In this chapter you will use it. A derivative tells you the slope of a graph at a point, and a slope tells you a lot:

• A positive slope means the value is going up. A negative slope means it is going down.
• A slope of zero means the graph is level for a moment. That is where a hill top or a valley bottom can be.
• A large slope means a fast change. A small slope means a slow change.

Look at the curve below. Moving from left to right, it rises, then falls, then rises again. The two turning points are marked.

The curve is $y = x^{3} - 6 x^{2} + 9 x$. Its derivative is $y' = 3 x^{2} - 12 x + 9$, which is zero at $x = 1$ and $x = 3$, exactly where the peak and the valley are.

A car travels 120 km in 2 hours. Its average speed is $\frac{120}{2} = 60$ km/h. But the speedometer does not show an average. It shows the speed at this moment. That is an instant rate of change.

The derivative gives exactly this instant rate. For a quantity $y$ that depends on $x$, the number $\frac{dy}{dx}$ is the rate of change of $y$ with respect to $x$. Let us see how an average rate turns into an instant rate. Take $y = x^2$ and look at the average rate between $x = 3$ and $x = 3 + h$ for smaller and smaller $h$:

The average rates move closer and closer to $6$. And indeed $\frac{d}{dx}x^2 = 2x$, which is $6$ at $x = 3$. The instant rate is the number that the average rates approach.

Often two or more quantities change with time $t$ at the same time, and they are linked by a formula. The chain rule links their rates:

$$\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$$

How to solve a rate problem:

1. Draw a picture and name each changing quantity with a letter.
2. Write a formula that links the quantities.
3. Differentiate both sides with respect to time $t$. Use the chain rule.
4. Only now put in the numbers for the moment you are asked about.
5. Write the answer with its sign and its units.

A square metal plate expands when heated. Its side $x$ grows at $0.2$ cm/s. How fast is its area growing when the side is $10$ cm?

The area is $A = x^2$. Differentiate with respect to $t$: $\frac{dA}{dt} = 2x \cdot \frac{dx}{dt}$. Now put in $x = 10$ and $\frac{dx}{dt} = 0.2$:

$$\frac{dA}{dt} = 2 \cdot 10 \cdot 0.2 = 4 \text{ cm}^2/\text{s}$$

The radius $r$ of a spherical soap bubble grows at $0.5$ mm/s. How fast is its volume growing when $r = 6$ mm?

The volume is $V = \frac{4}{3}\pi r^3$. Differentiating gives $\frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}$. At $r = 6$:

$$\frac{dV}{dt} = 4\pi \cdot 36 \cdot 0.5 = 72\pi \approx 226.19 \text{ mm}^3/\text{s}$$

A ladder $5$ m long leans on a wall. The bottom slides away from the wall at $0.3$ m/s. How fast does the top slide down when the bottom is $3$ m from the wall?

Let $x$ be the distance of the bottom from the wall and $y$ the height of the top. The ladder, wall and floor form a right triangle, so $x^2 + y^2 = 25$. Differentiate with respect to $t$: $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$, so

$$\frac{dy}{dt} = -\frac{x}{y}\cdot\frac{dx}{dt}$$

When $x = 3$, we have $y = 4$ (because $9 + 16 = 25$). So $\frac{dy}{dt} = -\frac{3}{4} \cdot 0.3 = -0.225$ m/s. The minus sign means the top is moving down, at $0.225$ m/s.

The cost, in cost units, of making $x$ items is $C(x) = 0.05 x^{2} + 4 x + 120$. The marginal cost is the rate of change of cost with respect to the number of items: $C'(x) = 0.1 x + 4$.

At $x = 20$ the marginal cost is $C'(20) = 6$. Compare this with the true extra cost of making the 21st item: $C(21) - C(20) = 226.05 - 220 = 6.05$. The marginal cost ($6$) is a very good estimate of the cost of one more item ($6.05$).

1. Find the rate of change of the area of a circle with respect to its radius $r$ when $r = 4$ cm.
2. The side of a square grows at $0.5$ cm/s. How fast is the area growing when the side is $6$ cm?
3. The edge of a cube grows at $2$ cm/s. How fast is the volume growing when the edge is $3$ cm?
4. A stone dropped in still water makes a circular ripple whose radius grows at $4$ cm/s. How fast is the enclosed area growing when the radius is $10$ cm?
5. Air is pumped into a spherical balloon at $100$ cm³/s. How fast is the radius growing when $r = 5$ cm?
6. A ladder $13$ m long leans on a wall. Its bottom slides away at $0.2$ m/s. How fast does the top slide down when the bottom is $5$ m from the wall?
7. The revenue from selling $x$ items is $R(x) = 3x^2 + 36x + 5$ (in revenue units). Find the marginal revenue when $x = 15$.
8. A point moves on the curve $y = x^3 + 2$. Find the points where the $y$-coordinate changes $3$ times as fast as the $x$-coordinate.

Answers

1. $\frac{dA}{dr} = 2\pi r$. At $r = 4$ it equals $8\pi \approx 25.1327$ cm²/cm.
2. $\frac{dA}{dt} = 2x\frac{dx}{dt} = 2(6)(0.5) = 6$ cm²/s.
3. $\frac{dV}{dt} = 3x^2 \frac{dx}{dt} = 3(9)(2) = 54$ cm³/s.
4. $\frac{dA}{dt} = 2\pi r \frac{dr}{dt} = 2\pi(10)(4) = 80\pi \approx 251.33$ cm²/s.
5. From $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$: $\frac{dr}{dt} = \frac{100}{4\pi \cdot 25} = \frac{1}{\pi} \approx 0.3183$ cm/s.
6. $x^2 + y^2 = 169$, so $\frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt}$. With $x = 5$, $y = 12$: $-\frac{5}{12}(0.2) \approx -0.0833$ m/s. The top moves down at about $0.0833$ m/s.
7. $R'(x) = 6x + 36$. At $x = 15$ it equals $126$ revenue units per item.
8. We need $\frac{dy}{dt} = 3\frac{dx}{dt}$, and $\frac{dy}{dt} = 3x^2\frac{dx}{dt}$. So $3x^2 = 3$, which gives $x = \pm 1$. The points are $(1, 3)$ and $(-1, 1)$.

Imagine walking from left to right along a graph. If you are going uphill, the function is increasing. If you are going downhill, it is decreasing. In exact words, on an interval:

• A function is increasing if a larger input always gives a larger output: $x_1 < x_2$ means $f(x_1) < f(x_2)$.
• A function is decreasing if a larger input always gives a smaller output: $x_1 < x_2$ means $f(x_1) > f(x_2)$.

You do not need to draw the graph. The sign of the slope tells you:

If $f'(x)$ is zero only at a few separate points and has the same sign everywhere else, the function is still increasing (or decreasing) through those points. For example, $y = x^3$ has slope $0$ at $x = 0$ but still climbs the whole time.

1. Find $f'(x)$.
2. Solve $f'(x) = 0$. These points cut the number line into intervals.
3. Choose one test point inside each interval and find the sign of $f'$ there.
4. Write down where $f$ is increasing and where it is decreasing.

Find where $f(x) = 2 x^{3} - 9 x^{2} + 12 x$ is increasing and decreasing.

Step 1. $f'(x) = 6 x^{2} - 18 x + 12 = 6(x - 1)(x - 2)$.

Step 2. $f'(x) = 0$ at $x = 1$ and $x = 2$. These split the line into three intervals.

So $f$ is increasing on $(-\infty, 1]$ and on $[2, \infty)$, and decreasing on $[1, 2]$. The values at the turning points are $f(1) = 5$ and $f(2) = 4$.

Find where $f(x) = x^{4} - 8 x^{2}$ is increasing and decreasing.

$f'(x) = 4 x^{3} - 16 x = 4x(x - 2)(x + 2)$, which is zero at $x = -2, 0, 2$. Test points: at $x = -3$ the slope is $-60$ (negative); at $x = -1$ it is $12$ (positive); at $x = 1$ it is $-12$ (negative); at $x = 3$ it is $60$ (positive).

For $f(x) = e^{-x}$ we get $f'(x) = -e^{-x}$. An exponential $e^{(\cdot)}$ is always positive, so $f'(x) < 0$ for every $x$. The function is decreasing on the whole real line. This is the pattern of a battery that loses a fixed fraction of its charge every hour.

Show that $f(x) = x - \sin x$ is increasing on the whole real line. We have $f'(x) = 1 - \cos x$. Since $\cos x \le 1$, we get $f'(x) \ge 0$ always. It equals $0$ only at $x = 0, \pm 2\pi, \dots$, which are separate points. So $f$ is increasing everywhere.

For $f(x) = x^{2} - 6 x + 5$ we get $f'(x) = 2x - 6$. This is negative when $x < 3$ and positive when $x > 3$. So $f$ is decreasing on $(-\infty, 3]$ and increasing on $[3, \infty)$. The lowest point of the parabola is at $x = 3$.

1. Find the intervals where $f(x) = x^{2} + 4 x - 1$ is increasing and decreasing.
2. Find the intervals where $f(x) = x^{3} - 12 x$ is increasing and decreasing.
3. Find the intervals where $f(x) = x^{4} - 2 x^{2}$ is increasing and decreasing.
4. Show that $f(x) = e^{2 x}$ is increasing on the whole real line.
5. Find the intervals where $f(x) = x e^{-x}$ is increasing and decreasing.
6. For $x > 0$, find where $f(x) = x - 2 \ln x$ is increasing and decreasing.
7. Show that $f(x) = x^{3} - 6 x^{2} + 15 x + 3$ is increasing on the whole real line.
8. Show that $f(x) = \cos x$ is decreasing on the interval $(0, \pi)$.
9. Find all values of $k$ for which $f(x) = x^2 + kx + 3$ is increasing on $[2, \infty)$.

Answers

1. $f'(x) = 2 x + 4$. Decreasing on $(-\infty, -2]$; increasing on $[-2, \infty)$.
2. $f'(x) = 3(x-2)(x+2)$. Increasing on $(-\infty, -2]$ and $[2, \infty)$; decreasing on $[-2, 2]$.
3. $f'(x) = 4x(x-1)(x+1)$. Decreasing on $(-\infty, -1]$ and $[0, 1]$; increasing on $[-1, 0]$ and $[1, \infty)$.
4. $f'(x) = 2 e^{2 x}$, which is positive for every $x$ because $e^{2x} > 0$.
5. $f'(x) = (1 - x)e^{-x}$ and $e^{-x} > 0$. So $f'(x) > 0$ for $x < 1$ and $f'(x) < 0$ for $x > 1$. Increasing on $(-\infty, 1]$; decreasing on $[1, \infty)$.
6. $f'(x) = 1 - \frac{2}{x} = \frac{x - 2}{x}$. Decreasing on $(0, 2]$; increasing on $[2, \infty)$.
7. $f'(x) = 3x^2 - 12x + 15 = 3\left((x-2)^2 + 1\right)$. Because $(x-2)^2 \ge 0$, we get $f'(x) \ge 3 > 0$ for every $x$.
8. $f'(x) = -\sin x$. On $(0, \pi)$ we have $\sin x > 0$, so $f'(x) < 0$.
9. $f'(x) = 2x + k$, which is smallest at $x = 2$, where it equals $4 + k$. We need $4 + k \ge 0$, so $k \ge -4$. Check: for $k = -4$, $f(x) = x^{2} - 4 x + 3$ is increasing on $[2, \infty)$. For $k = -5$ the slope at $x = 2.1$ is $2(2.1) - 5 = -0.8 < 0$, so it fails.

A local maximum is the top of a small hill on the graph: nearby points are all lower. A local minimum is the bottom of a small valley: nearby points are all higher. The word extremum means either of them. "Local" means we compare only with nearby points, so a local maximum need not be the highest point of the whole graph.

A critical point of $f$ is a point $x = c$ inside the domain where either $f'(c) = 0$ or $f$ is not differentiable. Peaks and valleys can only happen at critical points.

Move from left to right through a critical point $c$ and watch the sign of $f'$:

Suppose $f'(c) = 0$ and the second derivative exists at $c$. The second derivative tells how the slope itself is changing:

Find the local maximum and minimum values of $f(x) = x^{3} - 3 x^{2} - 9 x + 5$.

$f'(x) = 3 x^{2} - 6 x - 9 = 3(x - 3)(x + 1)$, so the critical points are $x = -1$ and $x = 3$. The second derivative is $f''(x) = 6 x - 6$.

Remember that these are local values. The curve goes on rising to very large values for large $x$, so the local maximum ($10$) is not the largest value of the function. A local value is only the best one nearby.

Look at $f(x) = x^4$. Here $f'(x) = 4x^3$ is zero at $x = 0$, and $f''(x) = 12x^2$ is also $0$ at $x = 0$. The second derivative test cannot decide. Use the first derivative test: for $x < 0$ the slope $4x^3$ is negative, and for $x > 0$ it is positive. The slope changes from negative to positive, so $x = 0$ is a local minimum.

Now compare $f(x) = x^3$. Again $f'(0) = 0$ and $f''(0) = 0$. But $f'(x) = 3x^2$ is positive on both sides of $0$. The sign does not change, so $x = 0$ is neither a maximum nor a minimum.

Find the local extrema of $f(x) = \sin x + \cos x$ on $(0, 2\pi)$.

$f'(x) = \cos x - \sin x = 0$ when $\tan x = 1$, so $x = \frac{\pi}{4}$ or $x = \frac{5\pi}{4}$. The second derivative is $f''(x) = -\sin x - \cos x$.

• At $x = \frac{\pi}{4}$: $f'' = -1.4142$ (negative), so there is a local maximum of $f = \sqrt{2} \approx 1.4142$.
• At $x = \frac{5\pi}{4}$: $f'' = 1.4142$ (positive), so there is a local minimum of $f = -\sqrt{2} \approx -1.4142$.

Let $f(x) = |x - 2| + 1$. At $x = 2$ the graph has a sharp corner, so $f$ is not differentiable there, and $x = 2$ is a critical point for that reason. For $x < 2$ the slope is $-1$ and for $x > 2$ it is $+1$. The slope changes from negative to positive, so $x = 2$ is a local minimum, with value $f(2) = 1$. The second derivative test could not be used here at all.

A local maximum is only the best value nearby. Often we want the best value over a whole stretch of the graph, for example the highest temperature during one day. This is the absolute maximum. The lowest value over the stretch is the absolute minimum.

A closed interval $[a, b]$ includes both of its end points. Here is an important fact:

Where can they be? Only at two kinds of places:

• at a critical point inside the interval (where $f' = 0$ or $f$ is not differentiable), or
• at an end point of the interval.

That is why the method is simple: list these few candidate points, compute $f$ at each one, and compare.

Take $f(x) = x^{3} - 3 x^{2} - 9 x + 5$ from the last section. Its critical points are $x = -1$ and $x = 3$. The values are $f(-1) = 10$ and $f(3) = -22$. The answer depends on the interval you are given:

If the interval is not closed, the best value may not exist. For $f(x) = \frac{1}{x}$ on $(0, 1]$, the function keeps growing as $x$ gets closer to $0$, so there is no absolute maximum. Also, $f(x) = x$ on the open interval $(0, 1)$ has no minimum and no maximum, because the end points $0$ and $1$ are not allowed.

To find the absolute maximum and minimum of a continuous function $f$ on a closed interval $[a, b]$:

1. Find the critical points of $f$ inside $(a, b)$: the points where $f'(x) = 0$, and the points where $f$ is not differentiable.
2. Add the end points $a$ and $b$ to the list.
3. Calculate $f$ at every point in the list.
4. Compare. The largest value is the absolute maximum. The smallest value is the absolute minimum.

Find the absolute maximum and minimum of $f(x) = 2 x^{3} - 15 x^{2} + 36 x + 1$ on $[1, 5]$.

Step 1. $f'(x) = 6 x^{2} - 30 x + 36 = 6(x - 2)(x - 3)$. The critical points are $x = 2$ and $x = 3$. Both lie inside $[1, 5]$.

Step 2. The end points are $x = 1$ and $x = 5$.

Step 3. Calculate the values:

Step 4. The largest value is $56$, so the absolute maximum is $56$ at $x = 5$. The smallest value is $24$, so the absolute minimum is $24$ at $x = 1$. Both are at end points. The local maximum $29$ at $x = 2$ is only a local one.

Find the absolute maximum and minimum of $f(x) = |x^2 - 4|$ on $[-3, 3]$.

The graph of $x^2 - 4$ dips below the $x$-axis between $-2$ and $2$. The absolute value flips that part upward, so $f$ has sharp corners at $x = -2$ and $x = 2$. These are not differentiable, so they are critical points. Between $-2$ and $2$ we have $f(x) = 4 - x^2$, with $f'(x) = -2x$, which is zero at $x = 0$.

The absolute maximum is $5$, reached at $x = -3$ and at $x = 3$. The absolute minimum is $0$, reached at the two corners $x = -2$ and $x = 2$. The small hill at $x = 0$ (value $4$) is only a local maximum.

1. Find the local minimum value of $f(x) = x^2 - 4x + 7$.
2. Find the local maximum and local minimum values of $f(x) = x^3 - 12x$.
3. For $x > 0$, show that $f(x) = x + \frac{1}{x}$ has a local minimum at $x = 1$ and find its value.
4. Find the local extremum of $f(x) = xe^{-x}$ and say whether it is a maximum or a minimum.
5. Find the absolute maximum and minimum of $f(x) = 2x^3 - 3x^2 - 12x$ on $[-2, 3]$.
6. Find the absolute maximum and minimum of $f(x) = \sin 2x$ on $[0, \pi]$.
7. Find the absolute maximum and minimum of $f(x) = x\sqrt{4 - x^2}$ on $[0, 2]$.
8. Show that $f(x) = x^3$ has no local maximum and no local minimum.
9. Two positive numbers add up to $20$. Find them if their product is as large as possible.
10. A rectangle has area $100$ m². Find the sides for which its perimeter is as small as possible.

Answers

1. $f'(x) = 2x - 4 = 0$ gives $x = 2$. Since $f''(x) = 2 > 0$, there is a local minimum of $f(2) = 3$.
2. $f'(x) = 3x^2 - 12 = 0$ gives $x = \pm 2$. $f''(x) = 6x$ is negative at $x = -2$ and positive at $x = 2$. Local maximum: $f(-2) = 16$. Local minimum: $f(2) = -16$.
3. $f'(x) = 1 - \frac{1}{x^2} = 0$ gives $x = 1$ (since $x > 0$). $f''(x) = \frac{2}{x^3}$ is $2 > 0$ at $x = 1$. The local minimum value is $f(1) = 2$.
4. $f'(x) = (1 - x)e^{-x} = 0$ gives $x = 1$. $f''(x) = (x - 2)e^{-x}$, and $f''(1) = -0.3679$ is negative. So $x = 1$ is a local maximum, of value $f(1) = e^{-1} \approx 0.3679$.
5. $f'(x) = 6(x - 2)(x + 1)$, so the critical points are $x = -1$ and $x = 2$. Values: $f(-2) = -4$, $f(-1) = 7$, $f(2) = -20$, $f(3) = -9$. Absolute maximum: $7$ at $x = -1$. Absolute minimum: $-20$ at $x = 2$.
6. $f'(x) = 2\cos 2x = 0$ gives $2x = \frac{\pi}{2}$ or $\frac{3\pi}{2}$, so $x = \frac{\pi}{4}$ or $\frac{3\pi}{4}$. The values there are $1$ and $-1$. At the end points $f(0) = f(\pi) = 0$. Absolute maximum $1$ at $x = \frac{\pi}{4}$; absolute minimum $-1$ at $x = \frac{3\pi}{4}$.
7. $f'(x) = \frac{4 - 2x^2}{\sqrt{4 - x^2}} = 0$ gives $x = \sqrt{2}$. Values: $f(0) = 0$, $f(\sqrt{2}) = \sqrt{2} \cdot \sqrt{2} = 2$, $f(2) = 0$. Absolute maximum $2$ at $x = \sqrt{2}$; absolute minimum $0$ at $x = 0$ and $x = 2$.
8. $f'(x) = 3x^2$ is positive for every $x \neq 0$ and is $0$ only at $x = 0$. The sign of $f'$ never changes, so $f$ is increasing everywhere and has no peak and no valley.
9. Let the numbers be $x$ and $20 - x$, with product $P(x) = x(20 - x)$. $P'(x) = 20 - 2x = 0$ gives $x = 10$, and $P'' = -2 < 0$. Both numbers are $10$, and the largest product is $100$.
10. Let one side be $x$; the other is $\frac{100}{x}$. Perimeter $P(x) = 2\left(x + \frac{100}{x}\right)$. $P'(x) = 2\left(1 - \frac{100}{x^2}\right) = 0$ gives $x = 10$. Both sides are $10$ m (a square), and the least perimeter is $40$ m.

Many real "find the best" problems are solved the same way. Step 1: write the quantity to be made largest or smallest as a function of one variable. Step 2: decide the allowed interval for that variable. Step 3: apply the working rule.

A square sheet of card has sides of $24$ cm. A small square of side $x$ cm is cut from each corner and the sides are folded up to make a box with no lid. What value of $x$ gives the box of greatest volume?

After cutting, the base is a square of side $24 - 2x$ and the height is $x$. So the volume is $V(x) = x \left(24 - 2 x\right)^{2}$, where $0 \le x \le 12$. Then $V'(x) = (24 - 2x)(24 - 6x)$, which is zero at $x = 12$ and $x = 4$. At $x = 12$ the volume is $0$. At $x = 4$ it is $V(4) = 4 \cdot 16^2 = 1024$. At $x = 0$ it is $0$ too.

The greatest volume is 1024 cm³, with $x = 4$ cm.

A farmer has $120$ m of fence and will fence three sides of a rectangular field. The fourth side is a long wall. Find the largest area.

Let each of the two equal sides be $x$ m. The side parallel to the wall is $120 - 2x$ m. The area is $A(x) = x \left(120 - 2 x\right)$ for $0 \le x \le 60$. $A'(x) = 120 - 4x = 0$ gives $x = 30$. At $x = 0$ and $x = 60$ the area is $0$. So the largest area is $A(30) = 1800$ m². The field is $30$ m by $60$ m: the side along the wall is twice the other sides.

A closed cylindrical can must hold $500$ cm³. Find the radius and height that use the least metal.

Let the radius be $r$ and the height $h$. The volume gives $\pi r^2 h = 500$, so $h = \frac{500}{\pi r^2}$. The surface area is $S = 2\pi r^2 + 2\pi r h = 2\pi r^2 + \frac{1000}{r}$. Now $S'(r) = 4\pi r - \frac{1000}{r^2} = 0$ gives $r^3 = \frac{250}{\pi}$, so $r = \sqrt[3]{250/\pi} \approx 4.3013$ cm.

Then $h = \frac{500}{\pi r^2} \approx 8.6025$ cm, which is exactly $2r$. The least surface area is about $348.73$ cm². Check: for very small $r$ the can is a tall thin tube and $S$ is large. For very large $r$ it is a flat disc and $S$ is also large. So the one critical point must give the minimum.

Which point of the curve $y = x^2$ is closest to the point $(0, 3)$?

The squared distance from $(x, x^2)$ to $(0, 3)$ is $D^2 = x^2 + (x^2 - 3)^2 = x^{4} - 5 x^{2} + 9$. Minimising $D^2$ is the same as minimising $D$ and avoids square roots. $\frac{d(D^2)}{dx} = 4x^3 - 10x = 2x(2x^2 - 5) = 0$ gives $x = 0$ or $x^2 = 2.5$. At $x = 0$, $D^2 = 9$, a local maximum. At $x^2 = 2.5$, $D^2 = 6.25 - 12.5 + 9 = 2.75$. So the closest points are $(\pm 1.5811, 2.5)$, and the least distance is $\sqrt{2.75} \approx 1.6583$.

A rectangle has a diagonal of length $10$. Find the largest possible area.

Let the diagonal make an angle $\theta$ with a side, where $0 < \theta < \frac{\pi}{2}$. The sides are $10\cos\theta$ and $10\sin\theta$, so the area is $A = 100\sin\theta\cos\theta = 50\sin 2\theta$. $A'(\theta) = 100\cos 2\theta = 0$ gives $2\theta = \frac{\pi}{2}$, that is $\theta = \frac{\pi}{4}$. The largest area is $50$. The sides are both $10\cos\frac{\pi}{4} \approx 7.0711$: the best rectangle is a square.

1. The radius of a spherical bubble grows at $0.1$ cm/s. How fast is its surface area growing when the radius is $3$ cm?
2. Grain is poured onto a flat floor at $4$ m³/min and forms a cone whose height always equals its base radius. How fast is the height growing when the height is $2$ m?
3. Show that $f(x) = x^3 - 3x^2 + 3x - 100$ is increasing on the whole real line.
4. Find the intervals where $f(x) = \frac{x^4}{4} - x^3$ is increasing and decreasing.
5. Find all local maximum and minimum values of $f(x) = x^2 e^{-x}$.
6. Find the absolute maximum and minimum of $f(x) = 3x^4 - 4x^3 - 12x^2 + 1$ on $[-2, 3]$.
7. A rectangle has area $64$ cm². Find its least possible perimeter.
8. A closed tank in the shape of a box with a square base and no lid must hold $32$ m³. Find the base side and height that use the least material for the base and four sides.
9. Show that the local maximum value of $f(x) = x + \frac{1}{x}$ is smaller than its local minimum value.
10. Find the value of $a$ for which $f(x) = x^2 + \frac{a}{x}$ has a local minimum at $x = 2$.