Derivatives in real-world problems

A workshop has to make sheet-metal tanks with no lid, with a square base, holding exactly $32$ dm³ (that is, $32$ litres). The less metal a tank uses, the cheaper it is. If the base is $2$ dm wide, the tank must be $8$ dm tall; if the base is $6$ dm wide, it needs to be less than $1$ dm tall. What base side should we choose to use the least metal?

Optimisation problems

Explore: try a few sizes

Let the base side be $x$ (dm). The height is $h=\dfrac{32}{x^2}$ , and the area of metal (the base and four sides) is $S(x)=x^2+4xh=x^2+\frac{128}{x}.$

$x$ (dm)	$2$	$3$	$4$	$5$	$6$
$S(x)$ (dm²)	$68$	$\approx51.67$	$48$	$50.6$	$\approx57.33$

A small base means tall sides, so a lot of metal goes on the sides; a large base uses a lot of metal on the bottom. The best value seems to be near $x=4$ , but the table does not prove it. The derivative will give the exact answer.

Example The tank that uses the least material

Find the base side and the height of the open tank with a square base and volume $32$ dm³ that uses the least area of metal.

Solution. With base side $x>0$ , the area of metal is $S(x)=x^2+\dfrac{128}{x}$ (from the exploration). We find its minimum on $(0, +\infty)$ :

$S'(x)=2x-\frac{128}{x^2}=\frac{2(x^3-64)}{x^2},\qquad S'(x)=0\iff x^3=64\iff x=4.$

For $0<x<4$ , $S'<0$ ; for $x>4$ , $S'>0$ ; so $S$ is smallest at $x=4$ : $S(4)=16+32=48$ . Then $h=\dfrac{32}{16}=2$ .

So the tank should have a $4\times4$ dm base and be $2$ dm tall; the least amount of metal is $48$ dm².

Check. The volume is $4\cdot4\cdot2=32$ dm³, as required.

Area of metal

S(x)=x^2+\dfrac{128}{x}

Business problems have the same shape, but the domain is often a closed interval, so both endpoints must be compared too.

Example The largest profit

A company estimates that if it makes $x$ hundred products a month ( $0\le x\le 15$ ), its profit is $P(x)=-x^3+9x^2+120x-200$ (million đồng). How many products should it make to have the largest profit?

Solution. We find the maximum of $P$ on $[0, 15]$ . The derivative is a quadratic, which we factorise:

$P'(x)=-3x^2+18x+120=-3(x-10)(x+4)=0\iff x=10\ \ (\text{reject } x=-4).$

We compare the values at the critical point and at both endpoints:

$P(0)=-200,\qquad P(10)=-1000+900+1200-200=900,\qquad P(15)=250.$

So the company should make $1000$ products a month ( $x=10$ ); the largest profit is $900$ million đồng.

Velocity, acceleration and rates of change

The derivative $f'(t_0)$ is the instantaneous rate of change of the quantity $f$ at time $t_0$ . For an object moving along a straight line with distance $s(t)$ :

the velocity is $v(t)=s'(t)$ ;
the acceleration is $a(t)=v'(t)=s''(t)$ (see the second derivative in grade 11).

Example The greatest velocity

A test car runs along a straight track and has travelled $s(t)=-t^3+9t^2$ (metres) after $t$ seconds, $0\le t\le 6$ . Find the acceleration at $t=1$ and the greatest velocity of the car.

Solution. We differentiate twice to get the velocity and the acceleration:

$v(t)=s'(t)=-3t^2+18t,\qquad a(t)=v'(t)=-6t+18.$

At $t=1$ : $a(1)=12$ m/s². To find the greatest velocity on $[0, 6]$ , we solve $v'(t)=a(t)=0$ :

$a(t)=0\iff t=3;\qquad v(0)=0,\quad v(3)=-27+54=27,\quad v(6)=-108+108=0.$

So at $t=1$ the acceleration is $12$ m/s²; the greatest velocity is $27$ m/s, reached at $t=3$ seconds.

Look back. The greatest velocity happens exactly when the acceleration is $0$ : before that the car speeds up ( $a>0$ ), after it the car slows down ( $a<0$ ).

Rates of change are not only for motion. The next example is about temperature.

Example A cup of tea cools down

The temperature of a cup of tea after $t$ minutes is $T(t)=20+\dfrac{70}{1+0.1t}$ (°C). Find the starting temperature and the rate of change of the temperature at $t=10$ minutes.

Solution. At the start $T(0)=20+70=90$ °C. We write $T(t)=20+70(1+0.1t)^{-1}$ and use the chain rule:

$T'(t)=-70\cdot0.1\cdot(1+0.1t)^{-2}=\frac{-7}{(1+0.1t)^2},\qquad T'(10)=\frac{-7}{2^2}=-1.75.$

So the tea starts at $90$ °C; at $t=10$ minutes its temperature is falling at $1.75$ °C per minute (the minus sign means it is decreasing).

Shortest paths and least cost

This problem has a lot of data, so we solve it in four steps.

A pipeline from the pumping station

A

to the factory

F

Example The cheapest pipeline

A factory $F$ is $3$ km from a straight river bank, and $H$ is the point on the bank closest to $F$ . A pumping station $A$ is on the bank, $8$ km from $H$ . A pipe is laid along the bank from $A$ to a point $M$ on the segment $AH$ , then straight from $M$ to $F$ . Each km of pipe along the bank costs $3$ billion đồng, and each km of pipe over land from $M$ to $F$ costs $5$ billion đồng. Where should $M$ be to make the cost as small as possible?

Solution.

Understand. We must choose the position of $M$ on the segment $AH$ , which is $8$ km long; the cost is made of two pieces of pipe with different prices.

Plan. Let $HM=x$ (km), $0\le x\le 8$ . Then $AM=8-x$ and, by Pythagoras' theorem, $MF=\sqrt{x^2+9}$ . The cost is $C(x)=3(8-x)+5\sqrt{x^2+9}\ \ (\text{billion đồng}).$

Solve. We solve $C'(x)=0$ ; both sides are non-negative, so we may square:

$C'(x)=-3+\frac{5x}{\sqrt{x^2+9}}=0\iff 5x=3\sqrt{x^2+9}\iff 25x^2=9x^2+81\iff x=\frac94.$

For $x=\dfrac94=2.25$ : $\sqrt{x^2+9}=\sqrt{14.0625}=3.75$ , so $C(2.25)=3\cdot5.75+5\cdot3.75=36$ . The endpoints: $C(0)=24+15=39$ , $C(8)=5\sqrt{73}\approx42.72$ .

Look back. Laying the pipe straight from $A$ to $F$ ( $x=8$ ) gives the shortest path but the highest cost, because all of the pipe is over land.

So $M$ should be $2.25$ km from $H$ (that is, $5.75$ km from $A$ ); the least cost is $36$ billion đồng.

Watch out Common mistakes

Forgetting the constraint on the variable. Wrong: for a box made from a $60$ cm sheet by cutting squares of side $x$ , solving $V'(x)=0$ gives $x=10$ and $x=30$ , and both are kept. Why it is wrong: the box exists only if $0<x<30$ . Right: keep only $x=10$ .
Finding a critical point without checking whether it is a maximum or a minimum. Wrong: " $S'(4)=0$ , so $S(4)$ is the largest". Why it is wrong: $S'$ changes sign from $-$ to $+$ at $4$ . Right: $S(4)$ is the smallest.
Ignoring the endpoints on a closed interval. Wrong: concluding the largest profit without computing $P(0)$ and $P(15)$ . Why it is wrong: the maximum can be at an endpoint. Right: compare all three values.
Answering without units or with the wrong quantity. Wrong: "the answer is $x=10$ " when the question asks for the number of products and $x$ counts hundreds. Why it is wrong: $x=10$ means $10$ hundred. Right: $1000$ products, with a profit of $900$ million đồng.

Key terms

English	Tiếng Việt
optimisation problem	bài toán tối ưu
objective function	hàm mục tiêu
constraint on the variable	điều kiện của ẩn
velocity	vận tốc
acceleration	gia tốc
instantaneous rate of change	tốc độ thay đổi tức thời
cost	chi phí
profit	lợi nhuận
revenue	doanh thu

Summary

Four steps: choose the variable and its constraint → set up the function → find the maximum or minimum on the domain → conclude with units.
The domain is an open interval: use a table of variations. A closed interval: compare the values at the critical points and at both endpoints.
Motion: $v(t)=s'(t)$ , $a(t)=v'(t)=s''(t)$ ; the greatest velocity is usually reached when $a(t)=0$ .
$f'(t_0)$ is the instantaneous rate of change of $f$ at $t_0$ ; a minus sign means the quantity is decreasing.
Always check the result: does it satisfy the constraint, and does it make sense in real life?

Practice

Each exercise comes with what you need and some steps to start. Try it yourself first, then open the solution.

Exercise 1

An object moves along a straight line and has travelled $s(t)=t^3-6t^2+9t$ (metres) after $t$ seconds. Find the times when the velocity is $0$ , and the acceleration at those times.

What you need and how to start

What you need: $v(t)=s'(t)$ , $a(t)=s''(t)$ .

Steps:

Compute $v(t)$ and solve $v(t)=0$ .
Compute $a(t)=6t-12$ at each root.

Show solution

$v(t)=3t^2-12t+9=3(t-1)(t-3)=0\iff t=1$ or $t=3$ . $a(t)=6t-12$ : $a(1)=-6$ , $a(3)=6$ .

Answer. $t=1$ s (acceleration $-6$ m/s²) and $t=3$ s (acceleration $6$ m/s²).

Exercise 2

Mr Hùng fences a rectangular garden of area $200$ m², with one side against an existing wall (which needs no fence). Find the dimensions that make the total length of fence as small as possible.

What you need and how to start

What you need: the steps for an optimisation problem; find the minimum on $(0, +\infty)$ with a variation table.

Steps:

Let $x$ be the side perpendicular to the wall; the side parallel to the wall is $\dfrac{200}{x}$ .
The fence has length $L(x)=2x+\dfrac{200}{x}$ ; find its minimum.

Show solution

$L'(x)=2-\dfrac{200}{x^2}=0\iff x=10$ . $L'<0$ for $0<x<10$ and $L'>0$ for $x>10$ , so $L$ is smallest at $x=10$ : $L(10)=20+20=40$ . The side parallel to the wall is $\dfrac{200}{10}=20$ m.

Answer. The garden is $10$ m $\times$ $20$ m (the $20$ m side against the wall); the shortest fence is $40$ m.

Exercise 3

From a square sheet of card with side $60$ cm, four equal squares of side $x$ cm are cut from the corners and the sides are folded up to make an open box. Find $x$ to make the volume of the box as large as possible.

What you need and how to start

What you need: the box has a square base of side $60-2x$ and height $x$ , with $0<x<30$ .

Steps:

Set up $V(x)=x(60-2x)^2$ .
Compute $V'(x)$ , factorise it, and keep only the roots in $(0, 30)$ .

Show solution

$V'(x)=(60-2x)^2-4x(60-2x)=(60-2x)(60-6x)=12(x-30)(x-10)$ . The only root in $(0, 30)$ is $x=10$ ; $V'>0$ for $0<x<10$ and $V'<0$ for $10<x<30$ . $V(10)=10\cdot40^2=16\,000$ .

Answer. $x=10$ cm; the largest volume is $16\,000$ cm³ ( $16$ litres).

Exercise 4

The population of a town $t$ years after 2025 is estimated as $N(t)=\dfrac{26t+10}{t+5}$ (thousand people). Find the rate of population growth at $t=5$ . In the long run, what does the population approach?

What you need and how to start

What you need: the instantaneous rate of change is $N'(t)$ ; the long-run value is $\lim\limits_{t\to+\infty}N(t)$ (the horizontal asymptote).

Steps:

Compute $N'(t)$ with the quotient rule.
Substitute $t=5$ and compute the limit.

Show solution

$N'(t)=\dfrac{26(t+5)-(26t+10)}{(t+5)^2}=\dfrac{120}{(t+5)^2}$ , so $N'(5)=\dfrac{120}{100}=1.2$ thousand people per year. $\lim\limits_{t\to+\infty}N(t)=26$ .

Answer. It grows by $1.2$ thousand people a year (that is, $1200$ people a year); in the long run the population approaches $26$ thousand people.

Exercise 5

A hotel has $50$ rooms. At $400$ thousand đồng a night it is full; each increase of $20$ thousand đồng in the price leaves one more room empty. Let $x$ be the number of price increases ( $0\le x\le 50$ ) and $R(x)$ the revenue per night (thousand đồng). Is each claim true or false?

a) $R(x)=(400+20x)(50-x)$ .

b) $R$ is increasing on the interval $(0, 15)$ .

c) The revenue is largest when the room price is $700$ thousand đồng.

d) The largest revenue per night is $24\,000\,000$ đồng.

What you need and how to start

What you need: revenue = price of one room $\times$ number of rooms let.

Steps:

Write the price and the number of rooms in terms of $x$ .
Compute $R'(x)$ and find where $R'(x)=0$ .
Compute the room price and the revenue at that point.

Show solution

a) True: the price is $400+20x$ and the number of rooms let is $50-x$ .

b) True: $R(x)=-20x^2+600x+20\,000$ , and $R'(x)=600-40x>0$ when $x<15$ .

c) True: $R'$ changes sign from $+$ to $-$ at $x=15$ , so $R$ is largest at $x=15$ ; the price is $400+300=700$ thousand đồng.

d) False: $R(15)=700\cdot35=24\,500$ thousand đồng, that is $24\,500\,000$ đồng.

Answer. a) True; b) True; c) True; d) False.

Exercise 6

A cylindrical can with a lid has volume $1000$ cm³. Find the base radius $r$ (cm) that uses the least material (round to two decimal places).

What you need and how to start

What you need: the volume $V=\pi r^2h$ ; the total surface area $S=2\pi r^2+2\pi rh$ .

Steps:

From $\pi r^2h=1000$ , write $h$ in terms of $r$ and set up $S(r)=2\pi r^2+\dfrac{2000}{r}$ .
Solve $S'(r)=0$ and check the sign.

Show solution

$S'(r)=4\pi r-\dfrac{2000}{r^2}=0\iff r^3=\dfrac{500}{\pi}\iff r=\sqrt[3]{\dfrac{500}{\pi}}\approx5.419$ . $S'<0$ for $r$ below this value and $S'>0$ above it, so this is the minimum.

Answer. $r\approx5.42$ .

Exercise 7

During a flu outbreak, the number of people who have caught the illness by day $t$ is estimated as $f(t)=30t^2-t^3$ ( $0\le t\le 20$ ). The rate of spread is $f'(t)$ (people per day). On which day is the rate of spread greatest, and what is it?

What you need and how to start

What you need: find the maximum of $g(t)=f'(t)$ on $[0, 20]$ .

Steps:

Compute $g(t)=f'(t)=60t-3t^2$ .
Solve $g'(t)=0$ and compare with both endpoints.

Show solution

$g'(t)=60-6t=0\iff t=10$ . $g(0)=0$ , $g(10)=600-300=300$ , $g(20)=1200-1200=0$ .

Answer. Day $10$ ; the greatest rate of spread is $300$ people per day.

Exercise 8

A lorry travels $300$ km at a constant speed $v$ (km/h), $40\le v\le 90$ . Each hour of driving costs $500$ thousand đồng (wages, wear) plus $\dfrac{v^2}{5}$ thousand đồng of fuel. Choose $v$ to make the total cost of the trip as small as possible.

What you need and how to start

What you need: the driving time is $\dfrac{300}{v}$ hours; the total cost is the cost per hour times the time.

Steps:

Set up $C(v)=\dfrac{300}{v}\left(500+\dfrac{v^2}{5}\right)=\dfrac{150\,000}{v}+60v$ .
Find its minimum on $[40, 90]$ .

Show solution

$C'(v)=-\dfrac{150\,000}{v^2}+60=0\iff v^2=2500\iff v=50$ . $C(40)=3750+2400=6150$ , $C(50)=3000+3000=6000$ , $C(90)=\dfrac{5000}{3}+5400\approx7066.67$ .

Answer. $v=50$ km/h; the least cost is $6000$ thousand đồng ( $6$ million đồng).

Exercise 9 Review

A metal sheet $30$ cm wide is folded into a gutter: the bottom is $10$ cm wide and each of the two sides is $10$ cm wide, each side tilted outwards at an angle $\theta$ to the vertical ( $0\le\theta<\dfrac{\pi}{2}$ ). Find $\theta$ that makes the cross-section area of the gutter as large as possible.

What you need and how to start

What you need: the cross-section is a trapezium with height $10\cos\theta$ , shorter base $10$ and longer base $10+20\sin\theta$ ; the derivatives of $\sin$ and $\cos$ and the formula $\cos2\theta=1-2\sin^2\theta$ (grade 11).

Steps:

Set up $A(\theta)=100\cos\theta(1+\sin\theta)$ .
Compute $A'(\theta)$ and turn it into an equation in $\sin\theta$ .
Compare with the value at $\theta=0$ .

Show solution

$A=\dfrac{10+10+20\sin\theta}{2}\cdot10\cos\theta=100\cos\theta(1+\sin\theta)$ . $A'(\theta)=100(\cos^2\theta-\sin\theta-\sin^2\theta)=100(1-\sin\theta-2\sin^2\theta)=-100(2\sin\theta-1)(\sin\theta+1)$ . So $A'=0\iff\sin\theta=\dfrac12\iff\theta=\dfrac{\pi}{6}$ ; $A'>0$ before this point and $A'<0$ after it. $A\left(\dfrac{\pi}{6}\right)=100\cdot\dfrac{\sqrt3}{2}\cdot\dfrac32=75\sqrt3\approx129.9$ , which is larger than $A(0)=100$ .

Answer. $\theta=\dfrac{\pi}{6}$ ( $30^\circ$ ); the largest cross-section area is $75\sqrt3\approx129.9$ cm².

Exercise 10 Challenge

A rectangular window is set inside a semicircular frame of radius $10$ dm, with one side of the window on the diameter. Find the dimensions of the window with the largest area.

What you need and how to start

What you need: if half the width of the window is $x$ , its top corners lie on the semicircle, so its height is $\sqrt{100-x^2}$ (Pythagoras' theorem).

Steps:

Set up $S(x)=2x\sqrt{100-x^2}$ for $0<x<10$ .
Compute $S'(x)$ and solve $S'(x)=0$ .

Show solution

$S'(x)=2\sqrt{100-x^2}-\dfrac{2x^2}{\sqrt{100-x^2}}=\dfrac{2(100-2x^2)}{\sqrt{100-x^2}}=0\iff x=5\sqrt2$ . $S'>0$ for $x<5\sqrt2$ and $S'<0$ for $x>5\sqrt2$ . The width is $2x=10\sqrt2$ , the height is $\sqrt{100-50}=5\sqrt2$ , and the area is $10\sqrt2\cdot5\sqrt2=100$ .

Answer. The window is $10\sqrt2\approx14.14$ dm wide and $5\sqrt2\approx7.07$ dm tall; the largest area is $100$ dm².

Optimisation problems

Explore: try a few sizes

Velocity, acceleration and rates of change

Shortest paths and least cost

Key terms

Summary

Practice

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer

a) The velocity of the ball is $v(t)=20-10t$ (m/s).	True	False
b) The ball reaches its greatest height at $t=2$ seconds.	True	False
c) The greatest height of the ball is $20$ m.	True	False
d) The speed of the ball when it hits the ground is more than $20$ m/s.	True	False