Calculus 1

Optimization

A sheet of metal, a fixed volume, a speed limit. A situation asks for the most or the least of something but hands us no function. How do we build the function, decide which inputs make sense, and know for certain that the value we find is the best one?

You need

The first and second derivative testsCalculus 1 Extrema on a closed intervalCalculus 1 Maxima, minima and critical pointsGrade 11

Where you are 11 in Applications of derivatives

Calculus8

Applications of derivatives11

▴ 6 above The mean value theorem Related rates Curve sketching and connecting f, f' and f''L'Hopital's rule

Optimization

When you reach for it

A problem asks for the largest or smallest value of some quantity (a volume, a cost, a time, a distance, a profit) and leaves a measurement up to us. Extrema on a closed interval finds the absolute extremes of a continuous function on [a, b], and The first and second derivative tests shows when a single critical point gives the absolute extreme on any interval. Optimization is the work that comes first: turning a description into one function of one variable, on a domain that makes physical sense, and then choosing the right tool. The general vocabulary of maxima and minima is in Maxima, minima and critical points.

When the function turns out to be a quadratic, the vertex answers the question without calculus, as in Maximum, minimum and projectile problems; a linear quantity over a region cut out by inequalities is Linear programming. Every problem below leads to a function that is not a quadratic, so the derivative is needed.

The procedure

Day 1: from words to a function on an interval

Begin by naming each quantity that can change, with its units, and draw a picture when there is geometry. Then write the objective, the quantity to be made largest or smallest, as a formula. It usually involves two or more variables. The situation also ties them together by a constraint: a fixed amount of material, a set distance, a required volume. Solving the constraint for one variable and substituting leaves the objective as a function of a single variable.

Next comes the sensible domain, the inputs for which the situation exists. Lengths are not negative, a cut cannot be wider than the sheet, a speed must lie between the legal limits. Often the domain is a closed interval [a, b]. Then, when the objective is continuous, the The extreme value theorem guarantees that an absolute maximum and an absolute minimum exist, and Extrema on a closed interval finds them: the candidates are the Critical points and local versus absolute extrema inside (a, b) and the two endpoints, and the largest and smallest values among them are the answers. An endpoint may describe a degenerate object, such as a box of zero height. Including it costs one easy evaluation, and it is what makes the interval closed. An endpoint can also be the answer, when the best value the formula would like lies outside what the situation allows (Example 1).

Last, answer the question that was asked. It may want the best value, the input that gives it, or the dimensions of the object.

Day 2: geometry

In geometric problems the constraint comes from a formula: the volume of a cylinder, similar triangles inside a cone, the equation of a curve. Two habits help. To find the point of a curve nearest a given point, minimize the square of the distance. Since \displaystyle \sqrt{u} is increasing for u \ge 0, the distance and its square are smallest at the same input, and the square has no root to differentiate. And choose whichever variable makes the algebra simplest; it need not be the quantity the question names.

Day 3: open domains and justification

Sometimes no endpoint makes sense: a radius may be any positive number, a workshop may produce any amount. On an open or unbounded interval the extreme value theorem promises nothing, and an extreme may fail to exist. Comparing values is then replaced by an argument, and The first and second derivative tests supplies three.

The sign of the derivative. If f' < 0 on the whole part of the interval left of c and f' > 0 on the whole part right of c, then f(c) is the absolute minimum on the interval (reverse the signs for a maximum).
The only critical point. If c is the only critical point of a continuous f inside the interval and f has a local minimum at c, then f(c) is the absolute minimum there; likewise for a maximum.
The second derivative. A value f''(c) > 0 with f'(c) = 0 gives a local minimum, which becomes absolute when c is the only critical point, or when f'' > 0 on the whole interval.

For a business making q units, profit is revenue minus cost, P(q) = R(q) - C(q). Where P is differentiable, a critical point has P'(q) = 0, that is,

R'(q) = C'(q):

marginal revenue equals marginal cost, in the language of The derivative as a rate of change. The equation only finds candidates. It also holds where profit is smallest nearby, so every solution needs a check (Example 6).

Worked examples

Example 1

Day 1: Nivi’s tanker truck

Nivi drives a tanker truck on a 300-mile run. She is paid 24 dollars per hour, and the truck’s fuel and running costs are 6 + \dfrac{v^2}{120} dollars per hour at a steady v miles per hour. (a) On a road with a minimum speed of 40 mph and a limit of 55 mph, what speed makes the trip cheapest? (b) What if the limit is 70 mph?

Solution

The quantities that change are the speed v (mph), the time t (hours) and the cost C (dollars). The objective is the cost, dollars per hour times hours. Since 24 + 6 = 30,

C = \left(30 + \frac{v^2}{120}\right)t.

The constraint is distance equals speed times time, vt = 300, so t = \dfrac{300}{v} and

\begin{aligned} C(v) &= \frac{300}{v}\left(30 + \frac{v^2}{120}\right) \\ &= \frac{9000}{v} + \frac{5v}{2}. \end{aligned}

Its derivative is

C'(v) = -\frac{9000}{v^2} + \frac{5}{2} = \frac{5(v^2 - 3600)}{2v^2},

which exists for every v > 0 and is zero only at v = 60 (the root -60 is not a speed).

(a) The sensible domain is [40, 55], and C is continuous there. The critical point 60 is outside it, so the only candidates are the endpoints: C(40) = 225 + 100 = 325 and \displaystyle C(55) = \frac{6625}{22} \approx 301.14. The cheapest trip, about 301.14 dollars, comes from driving at the limit. The derivative says why: C'(v) < 0 on the whole interval, so every increase in speed saves money there.

(b) Now the domain is [40, 70] and 60 lies inside. The candidates are C(40) = 325, C(60) = 150 + 150 = 300 and \displaystyle C(70) = \frac{2125}{7} \approx 303.57. The cheapest speed is 60 mph, for 300 dollars; above it, the extra fuel costs more than the wages saved.

Example 2

Day 1: Aputsiaq’s pewter tray

Aputsiaq has a sheet of pewter 21 inches by 16 inches. She cuts a square of side x inches from each corner and folds up the four flaps to make an open-top tray (Figure 1). What size of square gives the tray of largest volume, and what is that volume?

Figure 1The sheet of Example 2. Cutting an x by x square from each corner and folding along the dashed lines gives a tray x deep on a base 21 - 2x by 16 - 2x.

Solution

After folding, the tray is x inches deep, and its base is 21 - 2x by 16 - 2x inches, since a square comes off both ends of each side. The objective is

\begin{aligned} V(x) &= x(21 - 2x)(16 - 2x) \\ &= 4x^3 - 74x^2 + 336x. \end{aligned}

The sheet’s size is already built into the formula, so no separate constraint is needed. The sensible domain needs x \ge 0 and a base width 16 - 2x \ge 0, so it is [0, 8]. Both ends give flat sheets, V(0) = V(8) = 0, and keeping them makes the interval closed, so the extreme value theorem promises a largest volume.

V is a polynomial, differentiable everywhere, with

\begin{aligned} V'(x) &= 12x^2 - 148x + 336 \\ &= 4(x - 3)(3x - 28). \end{aligned}

The zeros are 3 and \displaystyle \frac{28}{3} \approx 9.33, and only 3 lies in (0, 8). The candidates give V(0) = 0, V(3) = 3 \cdot 15 \cdot 10 = 450 and V(8) = 0. The largest tray comes from 3-inch squares: it measures 15 by 10 by 3 inches and holds 450 cubic inches.

Example 3

Day 2: Nivi’s birdbath

Nivi shapes a birdbath basin as an open-top cylinder of sheet metal holding 64\pi cubic inches. To sit in its stand, the radius must be at least 2 and at most 6 inches. Which radius uses the least metal, and which the most?

Solution

With radius r and height h in inches, the metal is the bottom plus the side, A = \pi r^2 + 2\pi r h. The constraint is the volume, \pi r^2 h = 64\pi, so h = \dfrac{64}{r^2} and

A(r) = \pi r^2 + \frac{128\pi}{r}

on 2 \le r \le 6. Differentiating,

A'(r) = 2\pi r - \frac{128\pi}{r^2} = \frac{2\pi(r^3 - 64)}{r^2},

which is zero only at r = 4, inside (2, 6). The candidates are A(2) = 4\pi + 64\pi = 68\pi, A(4) = 16\pi + 32\pi = 48\pi and \displaystyle A(6) = 36\pi + \frac{64\pi}{3} = \frac{172\pi}{3}. The least metal, 48\pi \approx 150.8 square inches, goes into the basin of radius 4, where \displaystyle h = \frac{64}{16} = 4: as tall as its radius. The most, 68\pi \approx 213.6 square inches, goes into the narrowest basin allowed, with r = 2 and h = 16.

Example 4

Day 2: Seren’s nearest point

Seren wants the point of the parabola y = x^2 nearest the point (10, 2), and its distance (Figure 2).

Figure 2Example 4. The point of y = x^2 nearest (10, 2) is (2, 4), at distance \displaystyle 2\sqrt{17}.

Solution

A point of the parabola is (x, x^2), at distance \displaystyle \sqrt{(x - 10)^2 + (x^2 - 2)^2} from (10, 2). Minimize the square instead,

g(x) = (x - 10)^2 + (x^2 - 2)^2.

Then

\begin{aligned} g'(x) &= 2(x - 10) + 4x(x^2 - 2) \\ &= 4x^3 - 6x - 20 \\ &= 2(x - 2)(2x^2 + 4x + 5). \end{aligned}

The quadratic factor has discriminant 16 - 40 < 0 and a positive leading coefficient, so it is positive for every x, and g' has the sign of x - 2. Every real x is allowed, so there are no endpoints to compare. Instead, g' < 0 for all x < 2 and g' > 0 for all x > 2, so g decreases up to 2 and increases after it, and g(2) is the absolute minimum. Since g(2) = 64 + 4 = 68, the nearest point is (2, 4), at distance \displaystyle \sqrt{68} = 2\sqrt{17} \approx 8.25.

Example 5

Day 2: Seren’s soapstone carving

Seren has a cone of soapstone 15 centimeters tall on a base of radius 6 centimeters. She will carve from it a cylinder standing on the cone’s base and sharing its axis. Find the radius and height of the largest such cylinder, and the fraction of the stone it keeps.

Solution

Slice through the axis (Figure 3). The cone becomes a triangle with base 12 and height 15, and a cylinder of radius r and height h becomes a rectangle. For the largest cylinder of a given radius, its top corners touch the slanted sides. The objective is V = \pi r^2 h. The constraint comes from similar triangles: above the cylinder sits a small cone of radius r and height 15 - h, similar to the whole cone, so

\frac{r}{15 - h} = \frac{6}{15}, \quad\text{so}\quad h = 15 - \frac{5r}{2}.

With h already solved for, r is the simpler variable:

\begin{aligned} V(r) &= \pi r^2\left(15 - \frac{5r}{2}\right) \\ &= 15\pi r^2 - \frac{5\pi r^3}{2} \end{aligned}

on 0 \le r \le 6. The ends are degenerate, V(0) = V(6) = 0. Next,

V'(r) = 30\pi r - \frac{15\pi r^2}{2} = \frac{15\pi r}{2}(4 - r),

whose only zero inside (0, 6) is r = 4. There h = 15 - 10 = 5 and V(4) = \pi \cdot 16 \cdot 5 = 80\pi \approx 251.3 cubic centimeters, larger than both endpoint values. The cone holds \displaystyle \frac{1}{3}\pi \cdot 36 \cdot 15 = 180\pi, so the cylinder keeps \displaystyle \frac{80\pi}{180\pi} = \frac{4}{9} of the stone. Exercise 12 shows the fraction is \displaystyle \frac{4}{9} for every cone.

Figure 3The cross-section of Example 5. The cylinder’s rim touches the cone, so its radius r and height h are tied by similar triangles. The thin lines mark the radius 6 of the cone and the radius r of the cylinder.

Example 6

Day 3: Aputsiaq’s accordions

Aputsiaq’s workshop builds accordions. Making q of them a month costs

C(q) = q^3 - 60q^2 + 1500q + 4000

dollars (a made-up cost formula, chosen so the numbers come out whole), and each sells for 975 dollars. How many a month bring the largest profit?

Solution

Revenue is R(q) = 975q, so the profit is

\begin{aligned} P(q) &= R(q) - C(q) \\ &= -q^3 + 60q^2 - 525q - 4000. \end{aligned}

The domain is [0, \infty), with no upper end, so the closed interval method does not apply. (A count is a whole number; we treat q as continuous and check that the answer is whole.) Marginal revenue is R'(q) = 975 and marginal cost is C'(q) = 3q^2 - 120q + 1500. Setting them equal,

\begin{aligned} 3q^2 - 120q + 525 &= 0, \\ (q - 5)(q - 35) &= 0. \end{aligned}

Marginal revenue equals marginal cost at two outputs, and they mean opposite things. Since P''(q) = -6q + 120, P''(5) = 90 > 0 gives a local minimum of profit at 5, while P''(35) = -90 < 0 gives a local maximum at 35.

With two critical points, neither is yet known to be absolute. Factoring, P'(q) = -3(q - 5)(q - 35): negative on [0, 5), positive on (5, 35) and negative for q > 35. So P falls from q = 0 to 5, rises to 35, and falls forever after, and the largest value is P(0) = -4000 or P(35). Since

\begin{aligned} P(35) &= -42{,}875 + 73{,}500 - 18{,}375 - 4000 \\ &= 8250, \end{aligned}

the largest profit is 8250 dollars a month, from 35 accordions. At 5 the profit is P(5) = -5250 dollars, the worst output nearby. In marginal terms, below 5 each extra accordion costs more than 975 dollars to make, and between 5 and 35 it costs less.

Example 7

Day 3: Seren’s ride to the farmhouse

Seren starts at a point A on a straight road. A farmhouse F is 3 miles from the road, and the point B of the road nearest F is 8 miles from A. She bikes at 15 mph on the road and 9 mph across the field. Where should she leave the road to reach the farmhouse soonest?

Solution

Let her leave the road at P, x miles short of B (Figure 4). She rides 8 - x miles of road and, by the Pythagorean theorem, \displaystyle \sqrt{x^2 + 9} miles of field. Time is distance over speed, so the objective, in hours, is

T(x) = \frac{8 - x}{15} + \frac{\sqrt{x^2 + 9}}{9}

on 0 \le x \le 8. Here x = 0 means riding all the way to B and x = 8 means cutting across from the start.

Figure 4Example 7. Seren rides 8 - x miles of road from A to P, then \displaystyle \sqrt{x^2 + 9} miles of field from P to F.

The derivative is

T'(x) = -\frac{1}{15} + \frac{x}{9\sqrt{x^2 + 9}},

and T'(x) = 0 when \displaystyle \dfrac{x}{\sqrt{x^2 + 9}} = \dfrac{9}{15} = \dfrac{3}{5}. Squaring, 25x^2 = 9x^2 + 81, so \displaystyle x^2 = \frac{81}{16} and \displaystyle x = \frac{9}{4} (the left side is positive, so the root is positive). There \displaystyle \sqrt{x^2 + 9} = \frac{15}{4} and

\begin{aligned} T\left(\tfrac{9}{4}\right) &= \frac{23/4}{15} + \frac{15/4}{9} \\ &= \frac{23}{60} + \frac{25}{60} = \frac{4}{5}. \end{aligned}

The endpoints give \displaystyle T(0) = \frac{8}{15} + \frac{1}{3} = \frac{13}{15} and \displaystyle T(8) = \frac{\sqrt{73}}{9} \approx 0.949. The least time is \displaystyle \frac{4}{5} hour, 48 minutes, leaving the road \displaystyle 8 - \frac{9}{4} = \frac{23}{4} miles from A; riding to B takes 52 minutes and cutting straight across about 57.

A second argument confirms it without the endpoints. The ratio \displaystyle \dfrac{x}{\sqrt{x^2 + 9}} increases with x, so T' changes from negative to positive at \displaystyle \frac{9}{4}, and that only critical point is the absolute minimum on any interval containing it. This argument never uses the 8 miles, which is why the turn-off does not depend on them as long as B is at least \displaystyle \frac{9}{4} miles from A (Exercises 17 and 19).

Boundaries

The closed interval method needs a closed, bounded interval and a continuous objective. On (0, \infty), or across a break, values may be compared only after an argument that the extreme exists: the sign of f', the only critical point, or f'' of one sign throughout.
An extreme may not exist. The basin of Example 3, with any positive radius allowed, has no largest area, since A(r) \to \infty as r \to 0^+. In Exercise 4 the smallest value occurs only at degenerate endpoints, so no actual rectangle attains it (Exercise 5).
A critical point outside the sensible domain is discarded, and the answer may then sit at an endpoint (Example 1(a)).
R'(q) = C'(q) is necessary, not sufficient. It needs P differentiable, it also marks profit minimums (Example 6), and a domain end such as q = 0 must still be compared.
f''(c) < 0 alone gives a local maximum, not an absolute one, unless c is the only critical point in the interval or f'' < 0 throughout it.
The answer is only as good as the model. The optimum is the best value of the formula, and it describes the situation only where the formula does (the truck’s running cost, the workshop’s cost).

Where this goes next

When the equation f'(x) = 0 cannot be solved exactly, Newton's method closes in on its root, and the closed interval method then runs on the approximation. Quantities linked by a constraint that change in time are the subject of Related rates, which differentiates the same kind of geometric relation used here. In multivariable calculus the same plan (an objective, a constraint, a domain) returns for functions of several variables.

Exercises

Day 1: setting up, and the closed interval method

Two nonnegative numbers add to 18. Find the largest and smallest values of the first times the square of the second.
A barge makes a 120-mile run. The crew costs 288 dollars per hour, and fuel costs 4.5v^2 dollars per hour at v mph. (a) If the barge must travel between 5 and 10 mph, find the cheapest and the most expensive speeds and their costs. (b) Find the cheapest speed if the top speed is 7 mph.
A sheet 16 inches by 6 inches has a square of side x cut from each corner, and the flaps fold up into an open-top box. (a) Find the largest volume. (b) Find the largest volume if the box must be at least 2 inches deep.
A rectangle has a diagonal of 10 centimeters. Find the largest perimeter it can have, and the shape that gives it.

Day 1: reasoning

In Exercise 4 the closed interval method gives a smallest value of 20, at both endpoints. Is there a rectangle with diagonal 10 and perimeter 20? What does this say about the smallest perimeter?
Let C(v) = \dfrac{A}{v} + Bv with A, B > 0, on [a, b] with 0 < a < b. Show that the least value is at \displaystyle \sqrt{A/B} if that number lies in [a, b] and otherwise at the endpoint nearer to it, and that the largest value is always at an endpoint.

Day 2: geometry

Find the point of the curve y = \dfrac{4}{x}, x > 0, nearest the origin, and its distance.
Find the points of y = x^2 nearest the point (0, 3), and their distance. What happens at x = 0?
A right circular cylinder is inscribed in a sphere of radius 6 inches. Using the cylinder’s height h as the variable, find its largest volume.
A rectangle has its base on the x-axis and its two upper corners on the parabola y = 12 - 3x^2. Find the largest area and the dimensions.
A closed cylinder of sheet metal must hold 54\pi cubic feet, and its radius must be between 2 and 4 feet. Find the radii giving the least and the most surface area, with those areas.

Day 2: reasoning

A cylinder stands inside a cone of radius R and height H, on its base and sharing its axis. Show that the largest such cylinder has radius \displaystyle \frac{2R}{3} and height \displaystyle \frac{H}{3} and holds \displaystyle \frac{4}{9} of the cone’s volume.
Let (x_0, f(x_0)), with x_0 an interior point of the domain of a differentiable f, be the point of the graph nearest a point (a, b) off the graph. Show that the segment from (a, b) to that point is perpendicular to the tangent there, and check it on Example 4.

Day 3: open domains, cost and revenue, travel time

An open-top box with a square base must hold 96 cubic feet. Base material costs 3 dollars per square foot and side material 1 dollar per square foot. Find the cheapest dimensions and cost, and justify that the cost is an absolute minimum.
A shop sells birdhouses at 60 dollars each, and making q of them a month costs C(q) = 0.01q^3 - 0.6q^2 + 24q + 500 dollars (a made-up formula). Use marginal revenue and marginal cost to find the output with the largest profit, and justify it.
A dune buggy starts on a straight paved track. A camp is 5 miles off the track, and the point of the track nearest the camp is 12 miles ahead. The buggy goes 50 mph on the track and 40 mph on sand. Where should it leave the track, and what is the least time?
In Example 7, suppose B is only 2 miles from A, all else the same. Where should Seren leave the road, and how long does the trip take?
Show that the basin of Example 3, with any positive radius allowed, has its absolute minimum area at r = 4 and has no absolute maximum.

Day 3: reasoning

Generalize Example 7: road speed u, field speed w with w < u, farmhouse d miles from the road, and B at distance L from A. Show that the best turn-off is \displaystyle x = \dfrac{dw}{\sqrt{u^2 - w^2}} short of B when this is at most L. What happens if w \ge u?
Suppose R and C are twice differentiable, R'(q_0) = C'(q_0) and R''(q_0) < C''(q_0). Prove that profit has a local maximum at q_0. Which condition fails at q = 5 in Example 6?

Answers

P(x) = x(18 - x)^2 on [0, 18], with P'(x) = (18 - x)(18 - 3x), zero at 6 and 18. P(0) = P(18) = 0 and P(6) = 864. Largest 864 (the numbers 6 and 12), smallest 0.
C(v) = \dfrac{120}{v}(288 + 4.5v^2) = \dfrac{34{,}560}{v} + 540v, and C'(v) = 540 - \dfrac{34{,}560}{v^2} is zero at v = 8. (a) C(5) = 9612, C(8) = 8640, C(10) = 8856: cheapest at 8 mph for 8640 dollars, most expensive at 5 mph for 9612 dollars. (b) C' < 0 on [5, 7], so the cheapest is 7 mph: C(7) = \dfrac{61{,}020}{7} \approx 8717.14 dollars.
V(x) = x(16 - 2x)(6 - 2x) = 4x^3 - 44x^2 + 96x on [0, 3], with V'(x) = 4(3x - 4)(x - 6). (a) \displaystyle V\left(\frac{4}{3}\right) = \frac{4}{3} \cdot \frac{40}{3} \cdot \frac{10}{3} = \frac{1600}{27} \approx 59.26 cubic inches; the endpoints give 0. (b) V' < 0 on [2, 3], so the largest is V(2) = 2 \cdot 12 \cdot 2 = 48 cubic inches.
With sides x and \displaystyle \sqrt{100 - x^2}, \displaystyle P(x) = 2x + 2\sqrt{100 - x^2} on [0, 10]. \displaystyle P'(x) = 2 - \dfrac{2x}{\sqrt{100 - x^2}} is zero when \displaystyle x = 5\sqrt{2}. \displaystyle P(5\sqrt{2}) = 20\sqrt{2} \approx 28.28 and P(0) = P(10) = 20. Largest \displaystyle 20\sqrt{2} centimeters, for the square of side \displaystyle 5\sqrt{2}.
No. For 0 < x < 10, with \displaystyle y = \sqrt{100 - x^2} > 0, (x + y)^2 = 100 + 2xy > 100, so the perimeter 2(x + y) exceeds 20. The endpoints are segments, not rectangles. Perimeters come as close to 20 as we like, but no rectangle has the smallest one.
C'(v) = B - \dfrac{A}{v^2} is negative for \displaystyle v < \sqrt{A/B} and positive for \displaystyle v > \sqrt{A/B}. If \displaystyle \sqrt{A/B} is in [a, b], C falls and then rises, so the minimum is there. If \displaystyle \sqrt{A/B} > b, C decreases on [a, b] and the minimum is at b; if \displaystyle \sqrt{A/B} < a, it increases and the minimum is at a. In each case C only falls, only rises, or falls then rises, so no interior point beats both endpoints and the maximum is at a or b.
g(x) = x^2 + \dfrac{16}{x^2}, g'(x) = 2x - \dfrac{32}{x^3} = \dfrac{2(x^4 - 16)}{x^3}, zero at x = 2, negative before and positive after. The nearest point is (2, 2), at distance \displaystyle \sqrt{8} = 2\sqrt{2}.
g(x) = x^2 + (x^2 - 3)^2 = x^4 - 5x^2 + 9 and g'(x) = 2x(2x^2 - 5), zero at 0 and \displaystyle \pm\frac{\sqrt{10}}{2}, with signs -, +, -, +. So g falls to \displaystyle -\frac{\sqrt{10}}{2}, rises to 0, falls to \displaystyle \frac{\sqrt{10}}{2}, then rises, and its least value is \displaystyle g\left(\pm\frac{\sqrt{10}}{2}\right) = \frac{11}{4}. Nearest points \displaystyle \left(\pm\frac{\sqrt{10}}{2}, \frac{5}{2}\right), distance \displaystyle \frac{\sqrt{11}}{2} \approx 1.66. At 0, g has a local maximum: (0, 0), at distance 3, is farther than the points beside it.
With radius r, \displaystyle r^2 + \left(\frac{h}{2}\right)^2 = 36, so \displaystyle V(h) = \pi\left(36 - \frac{h^2}{4}\right)h on [0, 12]. \displaystyle V'(h) = \pi\left(36 - \frac{3h^2}{4}\right) is zero at \displaystyle h = 4\sqrt{3}, where r^2 = 24. The largest volume is \displaystyle 96\sqrt{3}\pi \approx 522.4 cubic inches; the endpoints give 0.
With corners (\pm x, 12 - 3x^2), A(x) = 2x(12 - 3x^2) = 24x - 6x^3 on [0, 2]. A'(x) = 24 - 18x^2 is zero at \displaystyle x = \frac{2\sqrt{3}}{3}. The largest area is \displaystyle \frac{32\sqrt{3}}{3} \approx 18.48, with width \displaystyle \frac{4\sqrt{3}}{3} and height 8.
h = \dfrac{54}{r^2} and S(r) = 2\pi r^2 + \dfrac{108\pi}{r}; S'(r) = 4\pi r - \dfrac{108\pi}{r^2} is zero at r = 3. S(2) = 62\pi, S(3) = 54\pi, S(4) = 59\pi. Least 54\pi \approx 169.6 square feet at r = 3 (height 6); most 62\pi \approx 194.8 square feet at r = 2.
Similar triangles give \displaystyle h = H\left(1 - \frac{r}{R}\right), so \displaystyle V(r) = \pi H\left(r^2 - \frac{r^3}{R}\right) on [0, R], zero at both ends. \displaystyle V'(r) = \pi H\left(2r - \frac{3r^2}{R}\right) is zero at \displaystyle r = \frac{2R}{3}, where \displaystyle h = \frac{H}{3} and \displaystyle V = \frac{4\pi R^2 H}{27}. The cone holds \displaystyle \frac{\pi R^2 H}{3}, and the ratio is \displaystyle \frac{4}{9}.
g(x) = (x - a)^2 + (f(x) - b)^2 has a minimum at the interior point x_0, so g'(x_0) = 0: (x_0 - a) + (f(x_0) - b)f'(x_0) = 0. This is the dot product of (x_0 - a, f(x_0) - b) with the tangent direction (1, f'(x_0)), so the two are perpendicular. In Example 4, (2 - 10, 4 - 2) = (-8, 2) and (1, 4) give -8 + 8 = 0.
Base side x, height \dfrac{96}{x^2}, cost C(x) = 3x^2 + \dfrac{384}{x} on (0, \infty). C'(x) = 6x - \dfrac{384}{x^2} is zero at x = 4, and C''(x) = 6 + \dfrac{768}{x^3} > 0 for all x > 0, so C(4) is the absolute minimum. Base 4 by 4 feet, height 6 feet, cost 48 + 96 = 144 dollars.
C'(q) = 0.03q^2 - 1.2q + 24 = 60 gives q^2 - 40q - 1200 = 0, so (q - 60)(q + 20) = 0 and q = 60. P'(q) = 60 - C'(q) = -0.03(q - 60)(q + 20) is positive on (0, 60) and negative after, so P(60) is the absolute maximum on [0, \infty): P(60) = 3600 - 1940 = 1660 dollars. (P''(60) = -2.4 < 0 agrees.)
With x miles short of the nearest point, \displaystyle T(x) = \dfrac{12 - x}{50} + \dfrac{\sqrt{x^2 + 25}}{40} on [0, 12]. T' = 0 when \displaystyle \dfrac{x}{\sqrt{x^2 + 25}} = \dfrac{4}{5}, so \displaystyle x = \frac{20}{3}, and \displaystyle T\left(\frac{20}{3}\right) = \frac{8}{75} + \frac{5}{24} = \frac{63}{200} hour, 18.9 minutes. The endpoints give T(0) = 0.365 and T(12) = 0.325 hour. Leave the track \displaystyle \frac{16}{3} \approx 5.33 miles from the start.
\displaystyle T(x) = \dfrac{2 - x}{15} + \dfrac{\sqrt{x^2 + 9}}{9} on [0, 2]. The critical point \displaystyle \frac{9}{4} is outside, and T'(x) < 0 on [0, 2] (at x = 2, \displaystyle \frac{2}{9\sqrt{13}} \approx 0.062 < \frac{1}{15}). So x = 2: head straight across from A, taking \displaystyle T(2) = \frac{\sqrt{13}}{9} \approx 0.401 hour, about 24 minutes, against \displaystyle T(0) = \frac{7}{15} hour, 28 minutes.
A(r) = \pi r^2 + \dfrac{128\pi}{r} on (0, \infty), A'(r) = 0 only at r = 4, and A''(r) = 2\pi + \dfrac{256\pi}{r^3} > 0, so A(4) = 48\pi is a local minimum at the only critical point, hence the absolute minimum. Since A(r) \to \infty as r \to 0^+, there is no maximum.
\displaystyle T(x) = \dfrac{L - x}{u} + \dfrac{\sqrt{x^2 + d^2}}{w} and \displaystyle T'(x) = -\dfrac{1}{u} + \dfrac{x}{w\sqrt{x^2 + d^2}}. T' = 0 when \displaystyle \dfrac{x}{\sqrt{x^2 + d^2}} = \dfrac{w}{u}, so u^2x^2 = w^2(x^2 + d^2) and \displaystyle x = \dfrac{dw}{\sqrt{u^2 - w^2}}; the ratio increases with x, so T' changes from - to + there. Example 7 gives \displaystyle \frac{3 \cdot 9}{\sqrt{144}} = \frac{9}{4}. If w \ge u, then \displaystyle T'(x) < -\frac{1}{u} + \frac{1}{w} \le 0, so T decreases on [0, L] and the best route crosses straight from A.
P' = R' - C' is 0 at q_0 and P'' = R'' - C'' < 0 there, so the second derivative test gives a local maximum. At q = 5 in Example 6, R''(5) = 0 and C''(5) = 6 \cdot 5 - 120 = -90, so R''(5) > C''(5), and profit has a local minimum.