Optimization

What dimensions give the biggest garden for a fixed amount of fencing? What shape of can uses the least metal? Which route for a cable costs the least? These are optimization problems: finding the best value of something. You’ve done a few with quadratics, but derivatives let you optimize almost any function. Most of the work is in the setup; the calculus part uses the tests you already know.

Key ideas

Setting up

1. Draw a picture and label it with variables.
2. Identify the objective: the quantity to maximize or minimize (area, volume, cost, distance). Write it as a formula.
3. Write the constraint: the fixed condition (total fencing, fixed volume, a curve the point must lie on).
4. Use the constraint to get one variable. Solve it for one variable and substitute, so the objective depends on a single variable.
5. State the domain: which values of that variable make physical sense (lengths positive, cuts no bigger than the sheet, and so on).

Solving and justifying

6. Differentiate and find the critical points in the domain.
7. Justify that you’ve found the absolute max or min. Use one of these:
   • Candidates test: if the domain is a closed interval, compare the objective at the critical points and the endpoints.
   • Only one critical point: if there’s just one critical point in the domain and the first derivative test (sign of $f'$ changes from $+$ to $-$) or the second derivative test ($f'' \lt 0$) shows it’s a relative maximum, it’s the absolute maximum. Same idea for a minimum.
8. Answer the question asked, with units. Read carefully: does it want the dimensions, the maximum value, or both?

AP free-response questions award a point for the justification, so don’t skip step 7.

A shortcut for distance

To minimize a distance $D = \sqrt{\dots}$, minimize $D^2$ instead. The square root is an increasing function, so $D$ and $D^2$ are smallest at the same point, and $D^2$ has no square root to differentiate.

Worked examples

Example 1: Fencing a divided field

A farmer has $600$ m of fencing to enclose a rectangular field and divide it into three pens with two fences parallel to one side. What dimensions give the largest total area?

Solution.

Variables: let $y$ be the length of each fence parallel to the dividers (the two ends of the field and the two dividers), and $x$ the length of the other two sides. So there are $4$ fences of length $y$ and $2$ of length $x$.

Constraint: $2x + 4y = 600$, so $y = 150 - \dfrac{x}{2}$.

Objective:

$$A(x) = xy = x\left(150 - \frac{x}{2}\right) = 150x - \frac{x^2}{2}$$

Domain: $0 \lt x \lt 300$ (both $x$ and $y$ must be positive).

Critical point: $A'(x) = 150 - x = 0$ gives $x = 150$.

Justify: $A''(x) = -1 \lt 0$, so $x = 150$ is a relative maximum, and it’s the only critical point in the domain, so it’s the absolute maximum.

Answer: $x = 150$ m and $y = 150 - 75 = 75$ m, for a maximum area of $150 \times 75 = 11\,250$ m².

Example 2: The open box

An open-top box is made from a $30$ cm by $48$ cm sheet of cardboard by cutting equal squares from the corners and folding up the sides. What size of square gives the largest volume?

Cutting squares of side $x$ leaves a base of $(48 - 2x)$ by $(30 - 2x)$ and a height of $x$.

Solution. Let $x$ be the side of each square, in cm. The box has height $x$ and a base of $(48 - 2x)$ by $(30 - 2x)$.

$$V(x) = x(48 - 2x)(30 - 2x) = 4x^3 - 156x^2 + 1440x$$

Domain: $0 \le x \le 15$ (at $x = 15$ the short side is used up). Including the endpoints is fine, since they just give $V = 0$, and it lets you use the candidates test.

$$V'(x) = 12x^2 - 312x + 1440 = 12(x^2 - 26x + 120) = 12(x - 6)(x - 20)$$

Only $x = 6$ is in the domain. Candidates:

$x$	$0$	$6$	$15$
$V(x)$	$0$	$6 \cdot 36 \cdot 18 = 3888$	$0$

Cut $6$ cm squares. The box is $36$ cm by $18$ cm by $6$ cm, with a maximum volume of $3888$ cm³.

Example 3: The cheapest can

A closed cylindrical can must hold $500$ cm³. What radius and height use the least material (smallest surface area)? Give answers to three decimal places.

Solution. Let $r$ be the radius and $h$ the height, in cm.

Objective (top, bottom, and side): $S = 2\pi r^2 + 2\pi rh$.

Constraint: $\pi r^2 h = 500$, so $h = \dfrac{500}{\pi r^2}$. Substitute:

$$S(r) = 2\pi r^2 + 2\pi r \cdot \frac{500}{\pi r^2} = 2\pi r^2 + \frac{1000}{r}, \qquad r \gt 0$$ $$S'(r) = 4\pi r - \frac{1000}{r^2} = 0 \quad\Rightarrow\quad r^3 = \frac{250}{\pi} \quad\Rightarrow\quad r = \sqrt[3]{\frac{250}{\pi}} \approx 4.301$$

Justify: $S''(r) = 4\pi + \dfrac{2000}{r^3} \gt 0$ for all $r \gt 0$, so this is a relative minimum, and it’s the only critical point, so it’s the absolute minimum.

$h = \dfrac{500}{\pi r^2} \approx 8.603$ cm. The best can has radius about $4.301$ cm and height about $8.603$ cm, using about $348.734$ cm² of material.

Notice $h = 2r$: the height equals the diameter. (You can show this exactly: $r^3 = \frac{250}{\pi}$ gives $h = \frac{500}{\pi r^2} = \frac{2 r^3}{r^2} = 2r$.)

Example 4: Closest point on a curve

Find the point on the curve $y = \sqrt{x}$ that is closest to the point $(3, 0)$.

Solution. A point on the curve is $(x, \sqrt{x})$, with $x \ge 0$. Minimize the squared distance:

$$D^2 = (x - 3)^2 + (\sqrt{x} - 0)^2 = x^2 - 6x + 9 + x = x^2 - 5x + 9$$ $$\frac{d}{dx}(D^2) = 2x - 5 = 0 \quad\Rightarrow\quad x = \frac{5}{2}$$

The derivative changes from negative to positive at $x = \frac{5}{2}$, and it’s the only critical point on $[0, \infty)$, so it gives the absolute minimum. (At the endpoint $x = 0$, $D^2 = 9$, which is larger.)

The closest point is $\left(\dfrac{5}{2}, \sqrt{\dfrac{5}{2}}\right) \approx (2.5, 1.581)$, at a distance of $\sqrt{\frac{25}{4} - \frac{25}{2} + 9} = \dfrac{\sqrt{11}}{2} \approx 1.658$.

Common mistakes

Differentiating before you have one variable. $A = xy$ can’t be optimized until you use the constraint to replace $y$. Get the objective in terms of a single variable first.

Skipping the domain. The domain tells you which critical points count and gives the endpoints for the candidates test. In Example 2, $x = 20$ solves $V'(x) = 0$ but makes no sense: you can’t cut $20$ cm squares from a $30$ cm side.

No justification. Finding a critical point isn’t enough. Say why it’s the absolute max or min: a candidates table, or “the only critical point, and $f'$ changes from positive to negative there.”

Answering the wrong question. If the question asks for the dimensions, give both dimensions with units. If it asks for the maximum volume, give the volume. In Example 2, "$x = 6$" alone isn’t the volume.

Mixing up the constraint and the objective. The objective is what you’re making as large or small as possible. The constraint is what’s fixed. In Example 3, the volume is fixed ($500$ cm³) and the surface area is minimized.

Forgetting a side or a face. Check your formula against the picture. Is the box open or closed? Is one side of the pen against a wall? Do the dividers count?

Practice

1. (Warm-up) Two positive numbers have a product of $64$. What is the smallest possible value of their sum?

Solution

Let the numbers be $x$ and $\dfrac{64}{x}$, with $x \gt 0$. The sum is $S(x) = x + \dfrac{64}{x}$.

$S'(x) = 1 - \dfrac{64}{x^2} = 0$ gives $x = 8$. $S''(x) = \dfrac{128}{x^3} \gt 0$, so it’s a minimum, and it’s the only critical point.

The numbers are $8$ and $8$, and the smallest sum is $16$.

2. (Warm-up) A rectangle has a perimeter of $40$ m. Write its area as a function of one side length $x$, state the domain, and find the largest possible area.

Solution

The other side is $20 - x$, so $A(x) = x(20 - x) = 20x - x^2$, with $0 \lt x \lt 20$.

$A'(x) = 20 - 2x = 0$ gives $x = 10$. $A''(x) = -2 \lt 0$, and it’s the only critical point, so it’s the absolute maximum.

The largest area is $10 \times 10 = 100$ m² (a square).

3. (Core) A rectangular garden of area $200$ m² is built against a house, so only three sides need fencing. What dimensions use the least fencing?

Solution

Let $x$ be each side perpendicular to the house and $y$ the side parallel to it. Constraint: $xy = 200$, so $y = \dfrac{200}{x}$.

$$L(x) = 2x + y = 2x + \frac{200}{x}, \qquad x \gt 0$$

$L'(x) = 2 - \dfrac{200}{x^2} = 0$ gives $x^2 = 100$, so $x = 10$. $L''(x) = \dfrac{400}{x^3} \gt 0$, so it’s a minimum, and it’s the only critical point.

The garden is $10$ m (out from the house) by $20$ m (along the house), using $40$ m of fencing.

4. (Core) A closed box with a square base must hold $2$ m³. The material for the top and bottom costs $10 per m², and the material for the sides costs $5 per m². Find the dimensions that minimize the cost, and the minimum cost.

Solution

Let the base be $x$ m by $x$ m and the height $h$ m. Constraint: $x^2 h = 2$, so $h = \dfrac{2}{x^2}$.

Top and bottom: $2x^2$ m² at $10. Four sides: $4xh$ m² at $5.

$$C(x) = 10(2x^2) + 5(4xh) = 20x^2 + 20x \cdot \frac{2}{x^2} = 20x^2 + \frac{40}{x}, \qquad x \gt 0$$

$C'(x) = 40x - \dfrac{40}{x^2} = 0$ gives $x^3 = 1$, so $x = 1$. $C''(x) = 40 + \dfrac{80}{x^3} \gt 0$, so it’s a minimum, and it’s the only critical point.

The box is $1$ m by $1$ m by $2$ m tall, and the minimum cost is $C(1) = 20 + 40 = 60$, or $60.

5. (Core) An open-top box is made from a $12$ cm by $12$ cm square sheet by cutting equal squares from the corners. Find the largest possible volume.

Solution

$V(x) = x(12 - 2x)^2$ for $0 \le x \le 6$.

$$\begin{aligned} V'(x) &= (12 - 2x)^2 + x \cdot 2(12 - 2x)(-2) \\ &= (12 - 2x)\big((12 - 2x) - 4x\big) \\ &= (12 - 2x)(12 - 6x) \end{aligned}$$

$V'(x) = 0$ at $x = 6$ and $x = 2$. Candidates:

$x$	$0$	$2$	$6$
$V(x)$	$0$	$2 \cdot 8^2 = 128$	$0$

The largest volume is $128$ cm³, with $2$ cm squares cut out.

6. (Core) Find the points on the parabola $y = x^2$ that are closest to the point $(0, 4)$, and the minimum distance.

Solution

A point on the parabola is $(x, x^2)$.

$$D^2 = x^2 + (x^2 - 4)^2 = x^4 - 7x^2 + 16$$

$\dfrac{d}{dx}(D^2) = 4x^3 - 14x = 2x(2x^2 - 7) = 0$ gives $x = 0$ or $x = \pm\sqrt{\dfrac{7}{2}}$.

$x$	$0$	$\pm\sqrt{7/2}$
$D^2$	$16$	$\frac{49}{4} - \frac{49}{2} + 16 = \frac{15}{4}$

$D^2$ heads to infinity as $\lvert x \rvert$ grows, so the minimum is at $x = \pm\sqrt{\frac{7}{2}} = \pm\frac{\sqrt{14}}{2}$. The closest points are $\left(\pm\dfrac{\sqrt{14}}{2}, \dfrac{7}{2}\right)$, at distance $\dfrac{\sqrt{15}}{2} \approx 1.936$. (At $x = 0$, the point $(0, 0)$ is a relative maximum of the distance.)

7. (Core) A rectangle is inscribed in a semicircle of radius $10$ cm, with its base on the diameter. Find the dimensions that give the largest area.

Solution

Put the centre at the origin. If the top-right corner is $(x, y)$ on the circle, then $y = \sqrt{100 - x^2}$, the base is $2x$, and

$$A(x) = 2x\sqrt{100 - x^2}, \qquad 0 \le x \le 10$$$$A'(x) = 2\sqrt{100 - x^2} + 2x \cdot \frac{-x}{\sqrt{100 - x^2}} = \frac{2(100 - x^2) - 2x^2}{\sqrt{100 - x^2}} = \frac{200 - 4x^2}{\sqrt{100 - x^2}}$$

$A'(x) = 0$ when $x^2 = 50$, so $x = 5\sqrt{2}$. ($A'$ is undefined at $x = 10$, an endpoint.) Candidates: $A(0) = 0$, $A(10) = 0$, and

$$A(5\sqrt{2}) = 2(5\sqrt{2})\sqrt{50} = 10\sqrt{2} \cdot 5\sqrt{2} = 100$$

The rectangle is $10\sqrt{2} \approx 14.142$ cm wide and $5\sqrt{2} \approx 7.071$ cm tall, with area $100$ cm².

8. (Challenge) A power line runs from a station on one bank of a straight river $300$ m wide to a factory on the other bank, $1200$ m downstream. Cable costs $50 per metre under water and $30 per metre along the bank. The cable goes straight across the river to a point $x$ m downstream, then along the far bank. Find the cheapest route.

Solution

The underwater part has length $\sqrt{300^2 + x^2}$ and the land part $1200 - x$, for $0 \le x \le 1200$.

$$C(x) = 50\sqrt{90\,000 + x^2} + 30(1200 - x)$$$$C'(x) = \frac{50x}{\sqrt{90\,000 + x^2}} - 30 = 0$$$$50x = 30\sqrt{90\,000 + x^2} \;\Rightarrow\; 2500x^2 = 900(90\,000 + x^2) \;\Rightarrow\; 1600x^2 = 81\,000\,000 \;\Rightarrow\; x = 225$$

(Only the positive root counts.) Candidates:

$x$ (m)	$0$	$225$	$1200$
$C(x)$ ($)	$15\,000 + 36\,000 = 51\,000$	$50(375) + 30(975) = 48\,000$	$\approx 61\,847$

The cheapest route crosses the river to a point $225$ m downstream, then follows the bank. The minimum cost is $48 000.

9. (Challenge) Find the dimensions of the cylinder with the largest volume that fits inside a sphere of radius $6$ cm. Give the volume exactly and to three decimal places.

Solution

Let the cylinder have radius $r$ and height $h$. A cross-section through the centre shows a right triangle with legs $r$ and $\frac{h}{2}$ and hypotenuse $6$:

$$r^2 + \left(\frac{h}{2}\right)^2 = 36 \quad\Rightarrow\quad r^2 = 36 - \frac{h^2}{4}$$$$V(h) = \pi r^2 h = \pi\left(36h - \frac{h^3}{4}\right), \qquad 0 \le h \le 12$$

$V'(h) = \pi\left(36 - \dfrac{3h^2}{4}\right) = 0$ gives $h^2 = 48$, so $h = 4\sqrt{3} \approx 6.928$ cm.

$V(0) = V(12) = 0$, so this critical point gives the maximum. Then $r^2 = 36 - 12 = 24$, so $r = 2\sqrt{6} \approx 4.899$ cm.

$$V = \pi(24)(4\sqrt{3}) = 96\sqrt{3}\,\pi \approx 522.374 \text{ cm}^3$$