Practical maximum and minimum problems

AS · P1 · 16 min

How should a sheet of card be cut to make the biggest possible box? What shape of can holds $500\ \text{cm}^3$ using the least metal? These are optimisation problems: a quantity has to be made as large or as small as possible subject to a constraint. Calculus solves them by turning the situation into a function of one variable and finding its stationary points. Paper 1 usually sets one per paper, often with a "show that" first part that gives you the function, followed by the calculus.

The idea

Suppose you have $300$ m of fencing to enclose a rectangular field against a straight river, so only three sides need fencing. A long thin field has almost no area; so does a short wide one. Somewhere in between is the best shape. If the area is written as a function of one length, the best shape is where the graph of area against that length reaches its peak, and at the peak the gradient is zero.

y = 300x - 2x^2 (75, 11250)

The graph shows $A = 300x - 2x^2$ (set up in Example 1). The maximum is at the top of the hump, where $\dfrac{dA}{dx} = 0$ .

So an optimisation problem has two halves:

Modelling: express the quantity to be optimised as a function of a single variable. This usually needs the constraint (fixed perimeter, fixed volume, fixed material) to eliminate a second variable.
Calculus: differentiate, set the derivative equal to zero, solve, and show that you have a maximum or minimum as required.

Solving an optimisation problem

Draw a diagram and label the variables, say $x$ and $y$ .
Write an expression for the quantity $Q$ to be optimised, in terms of $x$ and $y$ .
Write an equation for the constraint, rearrange it for $y$ , and substitute to get $Q$ in terms of $x$ only.
Find $\dfrac{dQ}{dx}$ , set it equal to $0$ and solve. Reject values that are impossible in context (negative lengths, zero volume).
Show the stationary value is the required type, usually with $\dfrac{d^2Q}{dx^2}$ .
Answer the question asked: the optimal dimensions, the maximum or minimum value, or both, with units.

Writing the function

The modelling step is where most students get stuck, and it is also where "show that" questions give you help. A few patterns cover nearly everything.

Situation	Quantity	Constraint	Eliminate
Fencing a rectangle	area $xy$	perimeter fixed	$y$ from the perimeter
Open box from a sheet	volume $x(a - 2x)^2$	sheet size fixed	already one variable
Closed cylinder	surface area $2\pi r^2 + 2\pi rh$	volume $\pi r^2 h$ fixed	$h = \dfrac{V}{\pi r^2}$
Sector of a circle	area $\tfrac{1}{2}r^2\theta$	perimeter $2r + r\theta$ fixed	$\theta$ from the perimeter
Shortest distance to a curve	$d^2 = (x - a)^2 + (y - b)^2$	the point lies on the curve	$y$ from the curve

Two tricks recur.

Minimise the square of a distance. A distance $d = \sqrt{\ldots}$ is awkward to differentiate. Since $d$ is positive, $d$ is smallest exactly when $d^2$ is smallest, so minimise $d^2$ instead.

Ignore positive constant factors. Maximising $\pi(75r - r^3)$ gives the same $r$ as maximising $75r - r^3$ . Keep the constant for the final value, though.

Justifying a maximum or minimum

A stationary value is not automatically the one you want. You must show its nature. The second derivative test is standard: a negative $\dfrac{d^2Q}{dx^2}$ confirms a maximum, a positive one a minimum.

Often the second derivative has an obvious sign without substituting. For $S = 2\pi r^2 + \dfrac{1000}{r}$ , $\dfrac{d^2S}{dr^2} = 4\pi + \dfrac{2000}{r^3}$ is positive for every $r > 0$ , so any stationary point is a minimum. Saying so is enough.

Key result

Express the quantity in one variable using the constraint.
Optimal value where $\dfrac{dQ}{dx} = 0$ .
$\dfrac{d^2Q}{dx^2} < 0$ : maximum. $\dfrac{d^2Q}{dx^2} > 0$ : minimum.
Minimise $d^2$ rather than $d$ .
Reject stationary values that make no physical sense.

Worked examples

Fencing against a river

A farmer has $300$ m of fencing to make a rectangular enclosure. One side is a straight river and needs no fence. Find the dimensions that give the maximum area, and the maximum area.

Solution

Let the side parallel to the river be $y$ m and the two sides perpendicular to it be $x$ m each. The fencing gives $2x + y = 300$ , so $y = 300 - 2x$ .

A = xy = x(300 - 2x) = 300x - 2x^2

\frac{dA}{dx} = 300 - 4x = 0 \quad\Rightarrow\quad x = 75

$\dfrac{d^2A}{dx^2} = -4 < 0$ , so this is a maximum.

Then $y = 300 - 150 = 150$ . The enclosure is $75$ m by $150$ m (the side along the river is $150$ m), and the maximum area is $75 \times 150 = 11\,250\ \text{m}^2$ .

The largest open box

A square sheet of card has side $30$ cm. A square of side $x$ cm is cut from each corner, and the sides are folded up to make an open box.

(a) Show that the volume of the box is $V = x(30 - 2x)^2$ .

(b) Find the value of $x$ that gives the maximum volume, and the maximum volume.

Solution

(a) After folding, the base is a square of side $30 - 2x$ and the height is $x$ , so $V = x(30 - 2x)^2$ .

(b) Expand: $V = x(900 - 120x + 4x^2) = 4x^3 - 120x^2 + 900x$ .

\frac{dV}{dx} = 12x^2 - 240x + 900 = 12(x^2 - 20x + 75) = 12(x - 5)(x - 15)

So $x = 5$ or $x = 15$ . But $x = 15$ uses the whole sheet and gives $V = 0$ , so reject it. (In context $0 < x < 15$ .)

$\dfrac{d^2V}{dx^2} = 24x - 240 = -120 < 0$ at $x = 5$ , so $x = 5$ gives a maximum.

$V = 5(30 - 10)^2 = 5 \times 400 = 2000\ \text{cm}^3$ .

y = x (30 - 2x)^2 (5, 2000)

A sector with fixed perimeter

A sector of a circle has radius $r$ cm and angle $\theta$ radians. Its perimeter is $20$ cm.

(a) Show that the area of the sector is $A = 10r - r^2$ .

(b) Find the maximum area and the corresponding value of $\theta$ .

Solution

(a) Using circular measure, the arc length is $r\theta$ , so the perimeter is $2r + r\theta = 20$ , giving $r\theta = 20 - 2r$ . The area is

A = \tfrac{1}{2}r^2\theta = \tfrac{1}{2}r(r\theta) = \tfrac{1}{2}r(20 - 2r) = 10r - r^2

(b) $\dfrac{dA}{dr} = 10 - 2r = 0$ gives $r = 5$ . $\dfrac{d^2A}{dr^2} = -2 < 0$ , so this is a maximum.

Maximum area $= 50 - 25 = 25\ \text{cm}^2$ . From $r\theta = 20 - 2r = 10$ , $\theta = 2$ radians.

A can using the least metal

A closed cylindrical can has radius $r$ cm and height $h$ cm, and its volume is $500\ \text{cm}^3$ .

(a) Show that the total surface area is $S = 2\pi r^2 + \dfrac{1000}{r}$ .

(b) Find the value of $r$ for which $S$ is a minimum, and the minimum surface area, giving answers to $3$ significant figures.

Solution

(a) The volume gives $\pi r^2 h = 500$ , so $h = \dfrac{500}{\pi r^2}$ . The surface area is two circles plus the curved surface:

S = 2\pi r^2 + 2\pi r h = 2\pi r^2 + 2\pi r \cdot \frac{500}{\pi r^2} = 2\pi r^2 + \frac{1000}{r}

(b) $S = 2\pi r^2 + 1000r^{-1}$ , so

\frac{dS}{dr} = 4\pi r - \frac{1000}{r^2} = 0 \quad\Rightarrow\quad r^3 = \frac{1000}{4\pi} = \frac{250}{\pi}

r = \sqrt[3]{\frac{250}{\pi}} \approx 4.30

$\dfrac{d^2S}{dr^2} = 4\pi + \dfrac{2000}{r^3} > 0$ for all $r > 0$ , so this is a minimum.

$S = 2\pi(4.301)^2 + \dfrac{1000}{4.301} \approx 116.2 + 232.5 \approx 349\ \text{cm}^2$ .

(The corresponding height is $h \approx 8.60$ cm, twice the radius: the most economical can is as tall as it is wide.)

Shortest distance to a curve

$P$ is a point on the curve $y = \sqrt{x}$ and $A$ is the point $(4, 0)$ . Find the shortest distance from $A$ to the curve.

Solution

Let $P$ be $(x, \sqrt{x})$ . Then

AP^2 = (x - 4)^2 + (\sqrt{x})^2 = x^2 - 8x + 16 + x = x^2 - 7x + 16

Minimise $D = AP^2$ instead of $AP$ :

\frac{dD}{dx} = 2x - 7 = 0 \quad\Rightarrow\quad x = \frac{7}{2}

$\dfrac{d^2D}{dx^2} = 2 > 0$ , a minimum.

$D = \dfrac{49}{4} - \dfrac{49}{2} + 16 = \dfrac{15}{4}$ , so the shortest distance is $\sqrt{\dfrac{15}{4}} = \dfrac{\sqrt{15}}{2} \approx 1.94$ .

A window with a triangular top

A window consists of a rectangle of width $x$ m and height $y$ m, with an equilateral triangle of side $x$ m on top. The total perimeter of the window is $10$ m.

(a) Show that the area of the window is $A = 5x - \dfrac{6 - \sqrt{3}}{4}x^2$ .

(b) Find the exact value of $x$ for which $A$ has a stationary value, and determine its nature.

Solution

(a) The perimeter is the two vertical sides, the bottom and the two slanting sides of the triangle: $2y + 3x = 10$ , so $y = \dfrac{10 - 3x}{2}$ .

The triangle has height $\dfrac{\sqrt{3}}{2}x$ , so its area is $\tfrac{1}{2} \cdot x \cdot \dfrac{\sqrt{3}}{2}x = \dfrac{\sqrt{3}}{4}x^2$ .

A = x \cdot \frac{10 - 3x}{2} + \frac{\sqrt{3}}{4}x^2 = 5x - \frac{3}{2}x^2 + \frac{\sqrt{3}}{4}x^2 = 5x - \frac{6 - \sqrt{3}}{4}x^2

(b)

\frac{dA}{dx} = 5 - \frac{6 - \sqrt{3}}{2}x = 0 \quad\Rightarrow\quad x = \frac{10}{6 - \sqrt{3}}

Rationalising, $x = \dfrac{10(6 + \sqrt{3})}{33} \approx 2.34$ .

$\dfrac{d^2A}{dx^2} = -\dfrac{6 - \sqrt{3}}{2} < 0$ (since $\sqrt{3} < 6$ ), so this is a maximum. The maximum area is about $5.86\ \text{m}^2$ .

Watch out

Differentiating with two variables still present. $A = xy$ cannot be differentiated with respect to $x$ until $y$ has been replaced using the constraint.

Giving only $x$ . If the question asks for the maximum area, you must substitute back to find it. Read the final line of the question again before you stop.

Keeping impossible solutions. In Example 2, $x = 15$ gives zero volume. Reject it with a reason.

No justification. "This is a maximum" without the sign of the second derivative (or another valid argument) loses the mark.

Forgetting a face or edge. Open boxes have no lid; fencing against a wall has one side missing; closed cylinders have two circular ends. Check your expression against the diagram.

Exam tip

"Show that" parts give you the function. Even if you cannot derive it, use the given result for the rest of the question: the calculus marks are still available.
The derivative must be correct before the method mark for solving is awarded in some schemes, so differentiate carefully, rewriting $\dfrac{1000}{r}$ as $1000r^{-1}$ first.
Nature: write " $\dfrac{d^2S}{dr^2} = 4\pi + \dfrac{2000}{r^3} > 0$ , so minimum". If the sign is obvious for all positive values, say why.
Accuracy: when the stationary value is not exact, keep full calculator accuracy for $r$ when computing $S$ . Rounding $r$ to $4.3$ first changes the third significant figure in some problems.
Context: state the answer in words, with units, for example "maximum volume $2000\ \text{cm}^3$ when $x = 5$ ".

Summary

Optimisation = write the quantity in one variable, then find its stationary point.
Use the constraint to eliminate the second variable.
Set $\dfrac{dQ}{dx} = 0$ , solve, and reject values impossible in context.
Justify the nature, usually with $\dfrac{d^2Q}{dx^2}$ (negative for maximum, positive for minimum).
Minimise $d^2$ instead of $d$ for shortest distances.
Finish by answering exactly what was asked: dimensions, the optimal value, or both, with units.

Practice questions

Question

The positive numbers $x$ and $y$ satisfy $x + y = 12$ . Find the maximum value of $P = xy^2$ .
An open box has a square base of side $x$ cm and height $h$ cm, and its volume is $32\ \text{cm}^3$ . Show that the external surface area is $S = x^2 + \dfrac{128}{x}$ , and find the dimensions that minimise $S$ .
A rectangle has two vertices on the $x$ -axis and two on the curve $y = 12 - x^2$ , symmetrical about the $y$ -axis. Find the maximum area of the rectangle.
A closed cuboid has a base measuring $x$ cm by $2x$ cm and a volume of $72\ \text{cm}^3$ . Show that its surface area is $S = 4x^2 + \dfrac{216}{x}$ , and find the minimum surface area.
Find the coordinates of the points on the curve $y = x^2$ that are closest to the point $(0, 2)$ , and the shortest distance.
A solid cylinder has radius $r$ cm and height $h$ cm, and its total surface area is $150\pi\ \text{cm}^2$ . Show that its volume is $V = \pi(75r - r^3)$ , and find the maximum volume.
A sector of a circle has radius $r$ cm, angle $\theta$ radians and area $16\ \text{cm}^2$ . Show that its perimeter is $P = 2r + \dfrac{32}{r}$ , and find the minimum perimeter and the corresponding $\theta$ .
A window is a rectangle of width $2r$ m with a semicircle of radius $r$ m on top. The perimeter of the window is $8$ m. Show that its area is $A = 8r - \left(2 + \dfrac{\pi}{2}\right)r^2$ and find the maximum area, giving an exact answer.
A straight line with negative gradient passes through the point $(2, 4)$ and meets the positive $x$ - and $y$ -axes at $A$ and $B$ . Let the gradient be $-m$ , where $m > 0$ . Show that the area of triangle $OAB$ is $8 + 2m + \dfrac{8}{m}$ , and find the minimum area.

Answers

$x = 12 - y$ , so $P = (12 - y)y^2 = 12y^2 - y^3$ . $\dfrac{dP}{dy} = 24y - 3y^2 = 3y(8 - y) = 0$ gives $y = 8$ ( $y = 0$ is rejected). $\dfrac{d^2P}{dy^2} = 24 - 6y = -24 < 0$ , maximum. $x = 4$ , $P = 4 \times 64 = 256$ .
$x^2 h = 32$ , so $h = \dfrac{32}{x^2}$ . Open box: base plus four sides, $S = x^2 + 4xh = x^2 + \dfrac{128}{x}$ . $\dfrac{dS}{dx} = 2x - \dfrac{128}{x^2} = 0$ gives $x^3 = 64$ , $x = 4$ . $\dfrac{d^2S}{dx^2} = 2 + \dfrac{256}{x^3} > 0$ , minimum. The base is $4$ cm by $4$ cm and the height is $2$ cm ( $S = 48\ \text{cm}^2$ ).
With vertices $(\pm x, 0)$ and $(\pm x, 12 - x^2)$ , $A = 2x(12 - x^2) = 24x - 2x^3$ . $\dfrac{dA}{dx} = 24 - 6x^2 = 0$ gives $x = 2$ . $\dfrac{d^2A}{dx^2} = -12x = -24 < 0$ , maximum. $A = 4 \times 8 = 32$ .
$2x^2 h = 72$ , so $h = \dfrac{36}{x^2}$ . $S = 2(2x^2) + 2(xh) + 2(2xh) = 4x^2 + 6xh = 4x^2 + \dfrac{216}{x}$ . $\dfrac{dS}{dx} = 8x - \dfrac{216}{x^2} = 0$ gives $x^3 = 27$ , $x = 3$ . $\dfrac{d^2S}{dx^2} = 8 + \dfrac{432}{x^3} > 0$ , minimum. $S = 36 + 72 = 108\ \text{cm}^2$ .
For a point $(x, x^2)$ , $D = d^2 = x^2 + (x^2 - 2)^2 = x^4 - 3x^2 + 4$ . $\dfrac{dD}{dx} = 4x^3 - 6x = 2x(2x^2 - 3) = 0$ gives $x = 0$ or $x^2 = \tfrac{3}{2}$ . $\dfrac{d^2D}{dx^2} = 12x^2 - 6$ : at $x = 0$ it is $-6$ (a local maximum of distance, $d = 2$ ); at $x^2 = \tfrac{3}{2}$ it is $12 > 0$ , minimum. $D = \tfrac{9}{4} - \tfrac{9}{2} + 4 = \tfrac{7}{4}$ , so the shortest distance is $\dfrac{\sqrt{7}}{2} \approx 1.32$ , at $\left(\pm\sqrt{\tfrac{3}{2}}, \tfrac{3}{2}\right)$ .
$2\pi r^2 + 2\pi rh = 150\pi$ , so $h = \dfrac{75 - r^2}{r}$ . $V = \pi r^2 h = \pi r(75 - r^2) = \pi(75r - r^3)$ . $\dfrac{dV}{dr} = \pi(75 - 3r^2) = 0$ gives $r = 5$ . $\dfrac{d^2V}{dr^2} = -6\pi r < 0$ , maximum. $V = \pi(375 - 125) = 250\pi \approx 785\ \text{cm}^3$ .
$\tfrac{1}{2}r^2\theta = 16$ , so $r\theta = \dfrac{32}{r}$ and $P = 2r + r\theta = 2r + \dfrac{32}{r}$ . $\dfrac{dP}{dr} = 2 - \dfrac{32}{r^2} = 0$ gives $r = 4$ . $\dfrac{d^2P}{dr^2} = \dfrac{64}{r^3} > 0$ , minimum. $P = 8 + 8 = 16$ cm, and $\theta = \dfrac{32}{16} = 2$ radians.
The perimeter is the base $2r$ , two sides $h$ and the semicircular arc $\pi r$ : $2r + 2h + \pi r = 8$ , so $2h = 8 - 2r - \pi r$ . $A = 2rh + \tfrac{1}{2}\pi r^2 = r(8 - 2r - \pi r) + \tfrac{1}{2}\pi r^2 = 8r - \left(2 + \tfrac{\pi}{2}\right)r^2$ . $\dfrac{dA}{dr} = 8 - (4 + \pi)r = 0$ gives $r = \dfrac{8}{4 + \pi}$ . $\dfrac{d^2A}{dr^2} = -(4 + \pi) < 0$ , maximum. At this $r$ , $\left(2 + \tfrac{\pi}{2}\right)r^2 = \tfrac{1}{2}(4 + \pi)r \cdot r = 4r$ , so $A = 8r - 4r = 4r = \dfrac{32}{4 + \pi} \approx 4.48\ \text{m}^2$ .
The line is $y - 4 = -m(x - 2)$ . At $y = 0$ : $x = 2 + \dfrac{4}{m}$ . At $x = 0$ : $y = 4 + 2m$ . Area $= \tfrac{1}{2}\left(2 + \dfrac{4}{m}\right)(4 + 2m) = \tfrac{1}{2}\left(8 + 4m + \dfrac{16}{m} + 8\right) = 8 + 2m + \dfrac{8}{m}$ . $\dfrac{dA}{dm} = 2 - \dfrac{8}{m^2} = 0$ gives $m = 2$ . $\dfrac{d^2A}{dm^2} = \dfrac{16}{m^3} > 0$ , minimum. Minimum area $= 8 + 4 + 4 = 16$ .

The idea

Writing the function

Justifying a maximum or minimum

Worked examples

Practice questions

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