Revision: optimisation and real-life problems

Real-life problems are a speciality of the new exam format. They usually appear in Part III (short answer, with a single number in some unit), sometimes as a whole true/false question in Part II, and are usually worth about $1$ to $2$ marks altogether. The calculus is usually not hard; the hard part is setting up the function and finding the right domain for the variable.

The background is in the lessons on maximum and minimum values and derivatives in real-life problems.

A method for optimisation problems

Example Short answer: an open box

Four equal squares of side $x$ cm are cut from the corners of a square sheet of card of side $30$ cm, and the sides are folded up to make an open box. What is the greatest possible volume of the box, in cm³?

Solution. The base is a square of side $30-2x$ and the height is $x$ , with $0<x<15$ . The volume is

$V(x)=x(30-2x)^2,\qquad V'(x)=(30-2x)^2-4x(30-2x)=(30-2x)(30-6x).$

On $(0, 15)$ , $30-2x>0$ , so $V'$ has the sign of $30-6x$ : positive for $x<5$ , negative for $x>5$ . So $V$ is greatest at $x=5$ :

$V(5)=5\cdot20^2=2000.$

The greatest volume is $2000$ cm³ (cutting squares of side $5$ cm).

Rates of change and motion

Two questions that are often confused: "when is $Q$ greatest" (solve $Q'=0$ ) and "when is $Q$ increasing fastest" (find the maximum of $Q'$ , that is, look at $Q''$ ).

Example Multiple choice: increasing fastest

The number of bacteria in a culture after $t$ hours is $N(t)=1000+30t^2-t^3$ , $0\le t\le20$ . At what time is the number of bacteria increasing fastest? A. $t=10$ . B. $t=20$ . C. $t=0$ . D. $t=15$ .

Solution. The growth rate is $N'(t)=60t-3t^2$ . We find the maximum of $N'$ on $[0, 20]$ :

$N''(t)=60-6t=0\iff t=10;\qquad N'(0)=0,\quad N'(10)=300,\quad N'(20)=0.$

So the bacteria increase fastest at $t=10$ hours: the answer is A. Option B is when $N'(t)=0$ , that is when the number of bacteria is greatest, not when it grows fastest.

Example True/false: motion in a straight line

An object moves in a straight line with displacement $s(t)=-t^3+9t^2$ (m), $t$ in seconds, $0\le t\le6$ . a) The velocity of the object at $t=2$ is $24$ m/s. b) The greatest velocity of the object is $27$ m/s. c) At $t=4$ , the object is speeding up. d) The distance travelled in the $6$ seconds is $108$ m.

Solution. We first find the velocity and acceleration:

$v(t)=-3t^2+18t,\qquad a(t)=-6t+18.$

a) True: $v(2)=-12+36=24$ .

b) True: $v$ is a quadratic with coefficient $-3<0$ , greatest at $t=3$ (where $a=0$ ): $v(3)=-27+54=27$ .

c) False: $v(4)=24>0$ but $a(4)=-6<0$ ; opposite signs, so the object is slowing down.

d) True: $v(t)=3t(6-t)\ge0$ on $[0, 6]$ , so the object never turns back, and the distance is $s(6)-s(0)=-216+324=108$ .

Key: T T F T.

Cost, revenue and profit

Example Multiple choice: output for the greatest profit

A workshop makes $x$ items a day ( $0\le x\le60$ ) and earns a profit of $P(x)=-x^3+45x^2+1200x$ (thousand đồng). How many items a day should it make to maximise profit? A. $40$ . B. $15$ . C. $60$ . D. $10$ .

Solution. We solve $P'(x)=0$ on $(0, 60)$ and compare with the two endpoints:

$P'(x)=-3x^2+90x+1200=-3(x-40)(x+10)=0\iff x=40\ (\text{reject } x=-10).$

$P(0)=0,\qquad P(40)=56\,000,\qquad P(60)=18\,000.$

So it should make $40$ items a day (a profit of $56$ million đồng): the answer is A. Option B is the root of $P''(x)=0$ ; D is the root with the sign changed; C is just an endpoint.

Route and installation problems

Here a point $M$ moves along a straight line; the distance or cost has a slanting part (use Pythagoras' theorem) and a straight part.

Laying a cable from the power station

A

to the island

C

Example Short answer: the least installation cost

A power station $A$ stands on a straight coastline. An island $C$ is $3$ km from the shore, and the nearest point of the shore to the island is $B$ , with $AB=8$ km. A cable is laid along the shore from $A$ to a point $M$ on $AB$ , then underwater from $M$ to $C$ . The cable costs $30$ million đồng per km on land and $50$ million đồng per km underwater. How far from $B$ should $M$ be to make the cost least, in km?

Solution. Let $BM=x$ (km), $0\le x\le8$ . Then $MC=\sqrt{x^2+9}$ , $AM=8-x$ , and the cost (million đồng) is

$T(x)=30(8-x)+50\sqrt{x^2+9},\qquad T'(x)=-30+\frac{50x}{\sqrt{x^2+9}}.$

We solve $T'(x)=0$ (this needs $x>0$ , so we may square):

$5x=3\sqrt{x^2+9}\iff25x^2=9x^2+81\iff x^2=\frac{81}{16}\iff x=\frac94.$

Comparing: $T(0)=390$ , $T\left(\frac94\right)=172.5+187.5=360$ , $T(8)=50\sqrt{73}\approx427$ . So $M$ should be $2.25$ km from $B$ , and the least cost is $360$ million đồng.

Key terms

English	Tiếng Việt
optimisation problem	bài toán tối ưu
instantaneous rate of change	tốc độ thay đổi tức thời
velocity / acceleration	vận tốc / gia tốc
cost / revenue / profit	chi phí / doanh thu / lợi nhuận
average cost	chi phí trung bình
constraint on the variable	điều kiện của biến
volume / total surface area	thể tích / diện tích toàn phần
Pythagoras' theorem	định lí Pythagore

Exercises

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

Exercise 1 Multiple choice

A rectangular garden has perimeter $40$ m. Its greatest possible area is A. $100$ m². B. $400$ m². C. $96$ m². D. $200$ m².

What you need and how to start

What you need: if one side is $x$ , the other is $20-x$ .

Steps:

Write $S(x)=x(20-x)$ , $0<x<20$ .
Solve $S'(x)=0$ .

Show solution

$S'(x)=20-2x=0\iff x=10$ , and $S'$ changes from $+$ to $-$ . $S(10)=100$ . Option B is $20^2$ .

Answer. A.

Exercise 2 Short answer

An object moves with $s(t)=t^3-3t^2+5t$ (m), $t$ in seconds. Find its velocity (m/s) at the moment when its acceleration is $0$ .

What you need and how to start

What you need: $v=s'$ , $a=v'$ .

Steps:

Find $v(t)$ and $a(t)$ .
Solve $a(t)=0$ and substitute into $v$ .

Show solution

$v(t)=3t^2-6t+5$ , $a(t)=6t-6=0\iff t=1$ . $v(1)=3-6+5=2$ .

Answer. $2$ .

Exercise 3 Multiple choice

The total cost of making $x$ items is $C(x)=x^2+20x+900$ (thousand đồng), $x>0$ . The least average cost per item is A. $80$ thousand đồng. B. $30$ thousand đồng. C. $60$ thousand đồng. D. $50$ thousand đồng.

What you need and how to start

What you need: the average cost is $\overline C(x)=x+20+\dfrac{900}{x}$ .

Steps:

Solve $\overline C'(x)=1-\dfrac{900}{x^2}=0$ .
Evaluate $\overline C$ at the positive root.

Show solution

$\overline C'(x)=0\iff x=30$ , and $\overline C'$ changes from $-$ to $+$ . $\overline C(30)=30+20+30=80$ . Option B is the number of items $x$ ; C leaves out the term $20$ .

Answer. A.

Exercise 4 Written answer

An open tank with a square base has volume $32$ dm³. Find its dimensions so that it uses the least material (the area of the base plus the four sides is least).

What you need and how to start

What you need: base side $x$ , height $h=\dfrac{32}{x^2}$ .

Steps:

Write $S(x)=x^2+4xh=x^2+\dfrac{128}{x}$ , $x>0$ .
Draw the table of variations on $(0, +\infty)$ .

Show solution

$S'(x)=2x-\dfrac{128}{x^2}=\dfrac{2(x^3-64)}{x^2}=0\iff x=4$ ; $S'<0$ for $0<x<4$ and $S'>0$ for $x>4$ . Then $h=\dfrac{32}{16}=2$ and $S(4)=16+32=48$ dm².

Answer. Base side $4$ dm, height $2$ dm (material area $48$ dm²).

Exercise 5 True/False

A shop sells shirts at $200$ thousand đồng each and sells $50$ a month; for every $1$ thousand đồng reduction in price it sells $1$ more shirt. Each shirt costs the shop $120$ thousand đồng. Let $x$ (thousand đồng) be the price reduction, $0\le x\le80$ .

a) At a price of $190$ thousand đồng, the shop sells $60$ shirts a month.

b) The monthly profit is $P(x)=(80-x)(50+x)$ (thousand đồng).

c) The profit is greatest when the price is $170$ thousand đồng.

d) The greatest profit is $4225$ thousand đồng.

What you need and how to start

What you need: profit = (selling price − cost price) × quantity.

Steps:

Price $200-x$ , profit per shirt $80-x$ , quantity $50+x$ .
Find the maximum of $P$ on $[0, 80]$ .

Show solution

a) True: a reduction of $10$ thousand gives $50+10=60$ shirts.

b) True: profit per shirt $200-x-120=80-x$ , times $50+x$ .

c) False: $P'(x)=30-2x=0\iff x=15$ , so the best price is $185$ thousand đồng.

d) True: $P(15)=65\cdot65=4225$ , more than $P(0)=4000$ and $P(80)=0$ .

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

Exercise 6 Short answer

A ball is thrown straight up; its height after $t$ seconds is $h(t)=1.5+19.6t-4.9t^2$ (m). Find the greatest height of the ball, in metres.

What you need and how to start

What you need: the height is greatest when the velocity $h'(t)=0$ .

Steps:

Solve $h'(t)=19.6-9.8t=0$ .
Substitute into $h(t)$ .

Show solution

$h'(t)=0\iff t=2$ , and $h'$ changes from $+$ to $-$ . $h(2)=1.5+39.2-19.6=21.1$ .

Answer. $21.1$ .

Exercise 7 Multiple choice

A cylindrical can has volume $16\pi$ cm³. To use the least material (least total surface area), the radius of its base must be A. $2$ cm. B. $4$ cm. C. $8$ cm. D. $1$ cm.

What you need and how to start

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

Steps:

Substitute $h=\dfrac{16}{r^2}$ to get $S(r)=2\pi r^2+\dfrac{32\pi}{r}$ .
Solve $S'(r)=0$ .

Show solution

$S'(r)=4\pi r-\dfrac{32\pi}{r^2}=0\iff r^3=8\iff r=2$ ; $S'$ changes from $-$ to $+$ . Then $h=4$ . Option B is the height, C is $r^3$ .

Answer. A.

Exercise 8 Short answer

A person at point $A$ on a lake is $4$ km from a straight shore ( $B$ is the point of the shore nearest to $A$ ). They want to reach point $C$ on the shore with $BC=10$ km: they row at $3$ km/h to a point $M$ on $BC$ , then walk at $5$ km/h to $C$ . How far from $B$ should $M$ be to make the time least, in km?

What you need and how to start

What you need: time = distance ÷ speed; $AM=\sqrt{x^2+16}$ with $x=BM$ .

Steps:

Write $T(x)=\dfrac{\sqrt{x^2+16}}{3}+\dfrac{10-x}{5}$ , $0\le x\le10$ .
Solve $T'(x)=0$ and compare with the two endpoints.

Show solution

$T'(x)=\dfrac{x}{3\sqrt{x^2+16}}-\dfrac15=0\iff5x=3\sqrt{x^2+16}\iff16x^2=144\iff x=3$ . $T(0)\approx3.33$ , $T(3)=\dfrac53+\dfrac75\approx3.07$ , $T(10)\approx3.59$ (hours).

Answer. $3$ .

Exercise 9 True/False

Four squares of side $x$ cm are cut from the corners of a $16$ cm × $10$ cm rectangular sheet of card, which is then folded into an open box.

a) The constraint on $x$ is $0<x<5$ .

b) The volume of the box is $V(x)=x(16-x)(10-x)$ .

c) The volume is greatest when $x=\dfrac{20}{3}$ .

d) The greatest volume of the box is $144$ cm³.

What you need and how to start

What you need: the base of the box measures $(16-2x)$ by $(10-2x)$ .

Steps:

Write $V(x)$ and find $V'(x)$ .
Keep only the roots inside the domain of $x$ .

Show solution

$V(x)=x(16-2x)(10-2x)=4x^3-52x^2+160x$ , $V'(x)=12x^2-104x+160=4(x-2)(3x-20)$ .

a) True: we need $10-2x>0$ and $x>0$ .

b) False: each side of the base loses two squares, so it must be $16-2x$ and $10-2x$ .

c) False: $\dfrac{20}{3}\approx6.67$ is not in $(0, 5)$ ; the root to use is $x=2$ .

d) True: $V'$ changes from $+$ to $-$ at $x=2$ , and $V(2)=2\cdot12\cdot6=144$ .

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

Exercise 10 Challenge

A metal sheet $30$ cm wide is folded into a gutter: a base $10$ cm wide and two sides $10$ cm wide, each making an angle $\theta$ with the horizontal, $0<\theta\le\dfrac{\pi}{2}$ (the cross-section is an isosceles trapezium). Find $\theta$ so that the cross-sectional area is greatest, and find that area to 1 decimal place.

What you need and how to start

What you need: area of a trapezium $=\dfrac{(\text{sum of the parallel sides})\cdot\text{height}}{2}$ ; $\cos2\theta=2\cos^2\theta-1$ .

Steps:

The height is $10\sin\theta$ and the top is $10+20\cos\theta$ , so $A(\theta)=100\sin\theta(1+\cos\theta)$ .
Find $A'(\theta)$ and factorise it in terms of $\cos\theta$ .

Show solution

$A'(\theta)=100\left(\cos\theta+\cos^2\theta-\sin^2\theta\right)=100(2\cos^2\theta+\cos\theta-1)=100(2\cos\theta-1)(\cos\theta+1).$ On $\left(0, \dfrac{\pi}{2}\right]$ , $\cos\theta+1>0$ , so $A'=0\iff\cos\theta=\dfrac12\iff\theta=\dfrac{\pi}{3}$ ; $A'>0$ for $\theta<\dfrac{\pi}{3}$ and $A'<0$ for $\theta>\dfrac{\pi}{3}$ . $A\left(\dfrac{\pi}{3}\right)=100\cdot\dfrac{\sqrt3}{2}\cdot\dfrac32=75\sqrt3\approx129.9$ .

Answer. $\theta=\dfrac{\pi}{3}$ ( $60^\circ$ ), with greatest area $75\sqrt3\approx129.9$ cm².

A method for optimisation problems

Rates of change and motion

Cost, revenue and profit

Route and installation problems

Key terms

Exercises

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer

a) The constraint on $x$ is $0<x<60$ .	True	False
b) The area of the plot is $S(x)=x(120-2x)$ .	True	False
c) The area is greatest when $x=30$ .	True	False
d) The greatest area is $3600$ m².	True	False