Optimisation AI HL

IB Mathematics: Applications and Interpretation HL · Unit 5: Calculus

Lesson 5.7 · Optimisation

Use constraints and derivatives to find the best possible value in a practical situation.

Ambassadors for Christ

Paul wrote, “we are ambassadors for Christ” (2 Corinthians 5:20, KJV). An ambassador carries a message on behalf of another. Jesus has entrusted his followers with the good news that he died for our sins and rose again, inviting people to be reconciled to God through him.

Andrew gives us a personal example. After meeting Jesus, he found his brother Simon and brought him to Jesus (John 1:41–42). Sharing the gospel may begin with a conversation with someone close to us, and it must also reach people beyond our familiar circles. As we make careful choices in an optimisation problem, may we use the opportunities before us to speak faithfully about Christ to every person.

Learning objectives

By the end of this lesson, you should be able to:

identify the quantity to maximise or minimise and the constraint in a problem;
express the quantity as a function of one variable;
state a physically meaningful domain;
find stationary points and decide whether they give a maximum or minimum;
check endpoints or limiting behaviour when the context requires it;
interpret and round the final answer appropriately, including units.

The optimisation method

Define a variable. Say what it measures and give its units.
Write the constraint. Use the given perimeter, volume, surface area or other fixed quantity.
Build one function. Express the area, profit, cost or volume to optimise in terms of a single variable.
State the domain. Exclude negative lengths and impossible designs.
Find candidates. Solve where the derivative is zero, using algebra or technology.
Verify and interpret. Check the candidates, endpoints or limiting behaviour; give the requested dimensions or quantity with units.

A stationary point outside the physical domain cannot be used, even if it appears on the calculator’s graph.

1. Maximising area with fixed fencing

Worked example · A rectangular garden.

A rectangular garden must be enclosed by $100$ m of fencing. Let its width be $x$ m and its length be $L$ m. Find the dimensions that give the greatest area.

The perimeter condition gives:

$2x+2L=100 \quad\Longrightarrow\quad L=50-x$

Both dimensions must be positive, so $0<x<50$ . The area is:

$A(x)=x(50-x)=50x-x^2$

Differentiate and solve:

$A'(x)=50-2x=0 \quad\Longrightarrow\quad x=25$

The area increases before $x=25$ and decreases after it. The length is also $50-25=25$ m, so:

$\boxed{A_{\max}=25(25)=625\text{ m}^2}$

The optimal garden is a $25$ m by $25$ m square. At either limiting value $x=0$ or $x=50$ , the area would be zero, supporting the conclusion that the interior stationary point gives the greatest area.

2. Maximising profit

Worked example · Weekly sales.

A company’s weekly profit from selling $q$ items is modelled by $P(q)=-2q^2+120q-500$ euros, where $0\leq q\leq60$ .

$P'(q)=-4q+120$

Setting $P'(q)=0$ gives $q=30$ . Profit rises before $30$ and falls after $30$ . Check the endpoints and the stationary point:

$P(0)=-500,\qquad P(30)=1300,\qquad P(60)=-500$

The greatest modelled weekly profit is €1300 when 30 items are sold. The answer is a whole number of items, so no further rounding is needed.

3. Minimising material for a container

A closed cylinder has a circular top and bottom. If its radius is $r$ , height is $h$ and fixed volume is $V$ , then:

$V=\pi r^2h,\qquad S=2\pi r^2+2\pi rh$

Use the volume constraint to eliminate $h$ :

$h=\frac{V}{\pi r^2} \quad\Longrightarrow\quad S(r)=2\pi r^2+\frac{2V}{r},\qquad r>0$

Worked example · A closed can of volume $1000$ cm $^3$ .

Here $V=1000$ , so:

$S(r)=2\pi r^2+\frac{2000}{r}$

Differentiate and solve $S'(r)=0$ :

$S'(r)=4\pi r-\frac{2000}{r^2}=0 \quad\Longrightarrow\quad r^3=\frac{500}{\pi}$

Thus $r\approx5.42$ cm. The volume condition gives $h\approx10.8$ cm, and the minimum surface area is approximately $554$ cm $^2$ , all to three significant figures.

As $r$ approaches zero, $2000/r$ becomes very large. As $r$ increases without bound, $2\pi r^2$ becomes very large. The stationary point therefore gives the minimum for $r>0$ .

Closed cylinder or open container?

Container	Surface area	What is included?
Closed cylinder	$2\pi r^2+2\pi rh$	Top, base and curved side
Open-top cylinder	$\pi r^2+2\pi rh$	Base and curved side only

Read the description carefully. Including a nonexistent top changes the model and the optimal dimensions.

4. Making a tray from a rectangular sheet

Squares of side $x$ are cut from each corner of a sheet measuring $a$ by $b$ . After folding, the tray’s dimensions are:

$\text{length}=a-2x,\qquad \text{width}=b-2x,\qquad \text{height}=x$

Therefore:

$\boxed{V(x)=x(a-2x)(b-2x)}$

All three dimensions must be positive, so $0<x<\frac12\min(a,b)$ . When using a calculator, restrict the graph to this domain before selecting a maximum.

Using a TI-84 Plus CE

Enter your one-variable model in Y=.
Set a graph window that displays its physical domain.
Use 2nd → TRACE (CALC) → maximum or minimum. Place the left and right bounds on either side of the candidate point.
Substitute the calculator’s input back into the constraint to find every requested dimension.
Check endpoints, other stationary points and any whole-number restrictions before stating your conclusion.

Do not round an intermediate radius or cut size too early. Keep full calculator precision, then round the final requested values.

IB-style Paper 1 · Skill practice

Show the function, domain and reasoning. Mark allocations are shown in square brackets.

Question 1 · Fenced garden [8 marks]

A rectangular garden is enclosed on all four sides by $100$ m of fencing. Its width is $x$ m.

Show that its length is $50-x$ metres. [2]
Write an expression for its area and state the domain of $x$ . [2]
Find the dimensions that maximise its area. [3]
State the maximum area. [1]

Show solution

(a) $2x+2L=100$ , so $L=50-x$ . (b) $A(x)=x(50-x)$ , with $0<x<50$ . (c) $A'(x)=50-2x=0$ gives $x=25$ m and $L=25$ m. The derivative changes from positive to negative there. (d) The maximum area is $625$ m $^2$ .

Question 2 · Profit [7 marks]

Weekly profit is modelled by $P(q)=-2q^2+120q-500$ euros for $0\leq q\leq60$ , where $q$ is the number of items sold.

Find $P'(q)$ and its stationary value of $q$ . [3]
Show that this gives a maximum on the stated domain. [2]
Find the maximum profit. [2]

Show solution

(a) $P'(q)=-4q+120$ , so $q=30$ . (b) $P'$ is positive for $q<30$ and negative for $q>30$ ; the endpoint profits are both $-€500$ . (c) $P(30)=-2(900)+120(30)-500=\boxed{€1300}$ .

Question 3 · Open-top cylinder [9 marks]

An open-top cylindrical container must have volume $216\pi$ cm $^3$ . Its radius is $r$ cm and its height is $h$ cm.

Show that $h=216/r^2$ . [2]
Show that its surface area is $A(r)=\pi r^2+432\pi/r$ . [2]
Find the radius and height that minimise the surface area. [3]
Find the exact minimum surface area. [2]

Show solution

(a) $\pi r^2h=216\pi$ , so $h=216/r^2$ . (b) The base and side have area $\pi r^2+2\pi rh=\pi r^2+432\pi/r$ . (c) $A'(r)=2\pi r-432\pi/r^2=0$ , giving $r^3=216$ and $r=6$ cm. Hence $h=216/36=6$ cm. This is a minimum since the area becomes large as $r$ tends to zero or infinity. (d) $A_{\min}=36\pi+72\pi=\boxed{108\pi\text{ cm}^2}$ .

IB-style Paper 2 · Applications

Question 4 · Closed cylindrical package [10 marks]

A closed cylindrical package must have volume $1000$ cm $^3$ . Its radius is $r$ cm and its height is $h$ cm. The manufacturer wishes to use as little material as possible.

Write $h$ in terms of $r$ . [2]
Show that the total surface area is $S(r)=2\pi r^2+2000/r$ . [2]
Find the radius and height that minimise $S$ , to three significant figures. [4]
Find the minimum surface area to three significant figures. [2]

Show solution

(a) From $\pi r^2h=1000$ , $h=1000/(\pi r^2)$ . (b) $S=2\pi r^2+2\pi rh=2\pi r^2+2000/r$ . (c) Setting $S'(r)=4\pi r-2000/r^2=0$ gives $r=(500/\pi)^{1/3}\approx\boxed{5.42\text{ cm}}$ . Using the unrounded radius gives $h\approx\boxed{10.8\text{ cm}}$ . (d) $S_{\min}\approx\boxed{554\text{ cm}^2}$ .

Question 5 · An open tray made from A4 card [12 marks]

A sheet measuring $29.7$ cm by $21$ cm is used to make an open tray. Equal squares of side $x$ cm are removed from its corners before the sides are folded up.

State the tray’s length, width and height in terms of $x$ . [3]
Explain why $0<x<10.5$ . [2]
Show that its volume is $V(x)=4x^3-101.4x^2+623.7x$ . [2]
Use your GDC or calculus to find the cut size that gives the greatest volume. [2]
Find that volume and the tray’s three dimensions to three significant figures. [3]

Show solution

(a) The dimensions are $(29.7-2x)$ cm by $(21-2x)$ cm by $x$ cm. (b) $x$ must be positive and $21-2x>0$ , so $0<x<10.5$ . (c) Expanding $x(29.7-2x)(21-2x)$ gives the stated cubic. (d) $V'(x)=12x^2-202.8x+623.7$ . Its stationary values are approximately $4.04234$ and $12.8577$ ; only the first is in the physical domain. The volume rises and then falls around this value. (e) The cut size is $\boxed{4.04\text{ cm}}$ , the greatest volume is approximately $\boxed{1.13\times10^3\text{ cm}^3}$ , and the dimensions are approximately $\boxed{21.6\text{ cm}\times12.9\text{ cm}\times4.04\text{ cm}}$ .

Question 6 · Coating an open water drum [11 marks]

An open-top cylindrical water drum must hold $3$ m $^3$ . Its inner base and curved side will be coated with a protective material. Let its radius and height be $r$ cm and $h$ cm.

Convert $3$ m $^3$ into cm $^3$ . [1]
Write the coating area $A$ in terms of $r$ and $h$ . [2]
Use the volume condition to express $A$ in terms of $r$ only. [2]
Find the radius and minimum area to three significant figures. [4]
One tin coats $2000$ cm $^2$ . Find the least number of tins required, assuming no wastage. [2]

Show solution

(a) $3$ m $^3=3\,000\,000$ cm $^3$ . (b) With no top, $A=\pi r^2+2\pi rh$ . (c) $h=3\,000\,000/(\pi r^2)$ , so $A(r)=\pi r^2+6\,000\,000/r$ . (d) $A'(r)=2\pi r-6\,000\,000/r^2=0$ gives $r^3=3\,000\,000/\pi$ . Thus $r\approx\boxed{98.5\text{ cm}}$ and $A_{\min}\approx\boxed{9.14\times10^4\text{ cm}^2}$ . The corresponding height is also approximately $98.5$ cm. (e) The unrounded area divided by $2000$ is approximately $45.7$ , so the manufacturer must buy $\boxed{46\text{ tins}}$ .

Question 7 · Maximum volume from fixed material [10 marks]

An open-top box has a square base of side $x$ cm and height $h$ cm. Exactly $300$ cm $^2$ of material is available for its base and four sides. Ignore seams and material thickness.

Show that $h=(300-x^2)/(4x)$ and state the possible values of $x$ . [3]
Show that $V(x)=75x-x^3/4$ . [2]
Find the dimensions that maximise the volume. [3]
Find the maximum volume. [2]

Show solution

(a) The material area is $x^2+4xh=300$ , giving $h=(300-x^2)/(4x)$ . Positive dimensions require $0<x<\sqrt{300}$ . (b) $V=x^2h=x(300-x^2)/4=75x-x^3/4$ . (c) $V'(x)=75-3x^2/4=0$ gives $x=10$ cm within the domain. Then $h=(300-100)/40=5$ cm. The volume increases before $x=10$ and decreases after it. (d) $V_{\max}=10^2(5)=\boxed{500\text{ cm}^3}$ .

Exam checklist

State the constraint and physical domain before differentiating. Keep all relevant digits during calculation. Reject stationary points that cannot represent the design. Check endpoints or limiting behaviour, then give the requested dimensions, maximum or minimum value, units and practical rounding.

Lesson 5.7 · Optimisation

Ambassadors for Christ

Learning objectives

The optimisation method

1. Maximising area with fixed fencing

2. Maximising profit

3. Minimising material for a container

Closed cylinder or open container?

4. Making a tray from a rectangular sheet

Using a TI-84 Plus CE

IB-style Paper 1 · Skill practice

Question 1 · Fenced garden [8 marks]

Question 2 · Profit [7 marks]

Question 3 · Open-top cylinder [9 marks]

IB-style Paper 2 · Applications

Question 4 · Closed cylindrical package [10 marks]

Question 5 · An open tray made from A4 card [12 marks]

Question 6 · Coating an open water drum [11 marks]

Question 7 · Maximum volume from fixed material [10 marks]

Exam checklist

Related

Company

Links

Support

Recommend

alphy school publishers

IB Mathematics: Applications and Interpretation HL 6 Sections80 LessonsLifetime

Curriculum

Optimisation AI HL

Ambassadors for Christ

Learning objectives

The optimisation method

1. Maximising area with fixed fencing

2. Maximising profit

3. Minimising material for a container

Closed cylinder or open container?

4. Making a tray from a rectangular sheet

Using a TI-84 Plus CE

IB-style Paper 1 · Skill practice

Question 1 · Fenced garden [8 marks]

Question 2 · Profit [7 marks]

Question 3 · Open-top cylinder [9 marks]

IB-style Paper 2 · Applications

Question 4 · Closed cylindrical package [10 marks]

Question 5 · An open tray made from A4 card [12 marks]

Question 6 · Coating an open water drum [11 marks]

Question 7 · Maximum volume from fixed material [10 marks]

Exam checklist

Related

Leave A Reply Cancel reply

Company

Links

Support

Recommend

alphy school publishers

Customise Consent Preferences

Log in with your account

Modal title