MathsISC Class 12 · Applications of Derivatives

Optimisation Problems: Maxima and Minima in Real-Life Situations

Q: How do I start an optimisation problem?

Draw a diagram, name the variables, and write two equations: one for the quantity to be optimised and one for the condition given. Use the condition to remove all but one variable before differentiating.

Q: Do I always need the second derivative test?

You need some test to show that the critical point is a maximum or a minimum. The second derivative test is usually quickest; if the second derivative is zero there, use the sign of the first derivative on each side instead.

Q: What if the critical point is outside the allowed range?

Reject it and say why, for example "x = 9 gives no box". If the variable lies in a closed interval, also check the end points.

Q: Are profit and cost problems part of this topic for ISC 2027?

Not as the commerce applications of calculus: the 2027 syllabus removed the old Section C unit on cost, revenue and profit functions. General real-life maxima and minima problems, including cost of materials, remain in Applications of Derivatives.

Q: How many marks do optimisation problems carry in the ISC paper?

They are usually long questions. In ISC 2024, two 6-mark alternatives asked for the dimensions of a window admitting maximum light and the least surface area of a closed cylindrical dustbin of given volume.

Updated 15 October 2026 · Fact-checked

To solve an optimisation problem, write the quantity to be maximised or minimised as a function of one variable, using the given condition to eliminate the others. Differentiate, set the derivative to zero to find critical points, confirm with the second derivative test, and answer the question asked, with units.

Follow the five steps in order, remember the maximum has a horizontal tangent where f′(x) = 0, and never skip the f″ test or the units.

Illustration drawn for QuestionBuddy.

Understand Optimisation Problems (Maxima and Minima Word Problems)

An optimisation problem asks for the largest or smallest value of some quantity, such as area, volume, cost or distance, subject to a condition, such as a fixed perimeter, a fixed volume or a fixed length of wire. Calculus finds it because at a smooth maximum or minimum the tangent is horizontal, so the derivative is zero.

The method has five stages.
1. Variables and a diagram. Name the unknown lengths. A labelled sketch prevents most set-up errors.
2. Objective and constraint. Write the quantity to optimise (the objective), and the condition linking the variables (the constraint). For a box made from a square sheet, the objective is volume; the constraint is the size of the sheet.
3. One variable. Use the constraint to express everything in terms of a single variable. This is where the ISC 2024 examiners found most errors: candidates "could not express the area in terms of one variable". Note the allowed range of the variable (lengths positive, cuts less than half the sheet, and so on).
4. Critical points. Differentiate, set the derivative to zero and solve. Reject values outside the allowed range.
5. Test and answer. Use the second derivative test: if $f^{''} (c) < 0$ , $c$ gives a local maximum; if $f^{''} (c) > 0$ , a local minimum. The examiners in 2024 noted candidates who stopped before completing this test, and lost the mark for it. Then answer the actual question: dimensions, the maximum value, or both, with units.

Useful standard results (worth recognising, but derive them in an answer):

For a fixed perimeter, the rectangle of greatest area is a square.
For a closed cylinder of fixed volume, the surface area is least when the height equals the diameter ( $h = 2 r$ ).
For an open box from a square sheet of side $a$ , cutting squares of side $\frac{a}{6}$ gives the greatest volume.
A Norman window (rectangle with a semicircle on top) of fixed perimeter lets in most light when the height of the rectangle equals the radius of the semicircle.

Distance problems. To find the point on a curve nearest to a given point, minimise the square of the distance. The square root is increasing, so it has the same minimum, and the algebra is much simpler.

Closed intervals. If the variable is restricted to a closed interval $[a, b]$ , the absolute maximum or minimum may occur at an end point. Compare the values at the critical points and at $a$ and $b$ .

Why the second derivative test works. If $f^{'} (c) = 0$ and $f^{''} (c) < 0$ , the slope is decreasing through $c$ : positive before and negative after, so the graph rises then falls and $c$ is a peak. The reverse holds for $f^{''} (c) > 0$ . If $f^{''} (c) = 0$ , the test fails and you must look at the sign of $f^{'}$ on each side (the first derivative test).

Optimisation Problems (Maxima and Minima Word Problems): formulas to remember

Critical point: $f^{'} (c) = 0$; Candidates for a local maximum or minimum.
Second derivative test: $f^{''} (c) < 0 \Rightarrow max, f^{''} (c) > 0 \Rightarrow min$; If f″(c) = 0, use the first derivative test instead.
Cylinder: $V = π r^{2} h, S_{closed} = 2 π r^{2} + 2 π r h$; Open at one end: πr² + 2πrh.
Cone and sphere: $V_{cone} = \frac{1}{3} π r^{2} h, V_{sphere} = \frac{4}{3} π R^{3}$; Curved surface of a cone: πrl, with l² = r² + h².
Equilateral triangle: $Area = \frac{3}{4} a^{2}$; Needed in window problems; a wrong formula was a common error in ISC 2024.
Distance trick: $minimise D^{2} = (x - x_{0})^{2} + (y - y_{0})^{2}$; Same minimum point as D, easier to differentiate.
Closed interval: $max_{[a, b]} f = max {f (a), f (b), f (c_{i})}$; Compare critical values with end-point values.

How to solve Optimisation Problems (Maxima and Minima Word Problems) questions

Follow these steps for every 5-mark optimisation question; each one usually carries a mark.

1Draw and label a diagram. Name the variables (for example radius $r$ and height $h$ ).
2Write the constraint as an equation (for example $π r^{2} h = 128 π$ ) and the objective as a formula (for example $S = 2 π r^{2} + 2 π r h$ ).
3Use the constraint to eliminate one variable, so the objective depends on one variable only. State its allowed range.
4Differentiate the objective and set the derivative equal to zero. Solve, rejecting values outside the range.
5Find the second derivative and evaluate it at the critical point; state "maximum" or "minimum" with the sign as your reason.
6Find the other dimensions and the optimum value using the constraint.
7Write a final sentence that answers the question in context, with units.

Optimisation Problems (Maxima and Minima Word Problems) in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Minimum surface area for a fixed volume

Suppose a drinks company wants a closed cylindrical can that holds 330 cm³ and uses as little metal as possible (ignore the thickness of the metal and any rims).

Find the radius and height that minimise the surface area, and the ratio of height to radius.

Try it first, then show the solutionSolution

$π r^{2} h = 330 \Rightarrow h = \frac{330}{π r ^{2}}$ .
$S = 2 π r^{2} + 2 π r h = 2 π r^{2} + \frac{660}{r}$ .
$\frac{d S}{d r} = 4 π r - \frac{660}{r ^{2}} = 0 \Rightarrow r^{3} = \frac{330}{2 π} \approx 52.5 \Rightarrow r \approx 3.75$ cm.
$\frac{d ^{2} S}{d r ^{2}} = 4 π + \frac{1320}{r ^{3}} > 0$ , so this is a minimum.
From $4 π r = \frac{660}{r ^{2}}$ we get $660 = 4 π r^{3}$ , so $h = \frac{330}{π r ^{2}} = \frac{2 π r ^{3}}{π r ^{2}} = 2 r \approx 7.49$ cm.

Answer: Radius about 3.75 cm and height about 7.49 cm: the height equals the diameter.

Memory hook: The cheapest closed can is as tall as it is wide: h = 2r.

Minimising area with margins

Imagine a poster that must have 600 cm² of printed matter, with margins of 3 cm at the top and bottom and 2 cm on each side.

Find the dimensions of the printed area that make the total paper area as small as possible.

Try it first, then show the solutionSolution

Let the printed width be $x$ cm; then its height is $\frac{600}{x}$ cm.
Paper area $P = (x + 4) (\frac{600}{x} + 6) = 624 + 6 x + \frac{2400}{x}$ .
$\frac{d P}{d x} = 6 - \frac{2400}{x ^{2}} = 0 \Rightarrow x^{2} = 400 \Rightarrow x = 20$ .
$\frac{d ^{2} P}{d x ^{2}} = \frac{4800}{x ^{3}} > 0$ : minimum.
Printed area $20 \times 30$ cm; paper $24 \times 36 = 864$ cm².

Answer: Print 20 cm wide by 30 cm high; the poster is then 24 cm by 36 cm, using 864 cm² of paper.

Memory hook: Margins make shape matter: balance width against height until the derivative is zero.

Tricks to remember Optimisation Problems (Maxima and Minima Word Problems)

Short memory aids. Say each one aloud once, then use it on a question the same day.

DOCTA
Diagram, Objective, Constraint, Test, Answer: five steps, five marks.
Use it when: Structuring every long optimisation answer.
Sad face, happy face
f″ < 0 is a frown (∩): maximum. f″ > 0 is a smile (∪): minimum.
Use it when: Remembering the second derivative test.
Square the distance
Nearest-point questions: minimise D², never D.
Use it when: Avoiding square roots in the derivative.
Can = cube-ish
Least metal for a closed can: height equals diameter.
Use it when: Checking your cylinder answer.

Optimisation Problems (Maxima and Minima Word Problems): solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Least sum of squares · 3 marks · Short answerBoard level

The sum of two positive numbers is 16. Find the numbers if the sum of their squares is least.

Try it first, then show the solutionSolution

Let the numbers be $x$ and $16 - x$ . $S = x^{2} + (16 - x)^{2}$ . (1 mark)
$\frac{d S}{d x} = 2 x - 2 (16 - x) = 4 x - 32 = 0 \Rightarrow x = 8$ . (1 mark)
$\frac{d ^{2} S}{d x ^{2}} = 4 > 0$ , so minimum; least value $64 + 64 = 128$ . (1 mark)

Answer: Both numbers are 8; the least sum of squares is 128.

Example 2: Maximum of a quadratic · 1 mark · MCQBoard level

The maximum value of $f (x) = 4 x - x^{2}$ is:
(a) 2
(b) 4
(c) 8
(d) 16

Try it first, then show the solutionSolution

$f^{'} (x) = 4 - 2 x = 0 \Rightarrow x = 2$ ; $f^{''} (x) = - 2 < 0$ ; $f (2) = 4$ .

Answer: (b) 4.

Example 3: Open box from a square sheet · 5 marks · Long answerHarder board level

An open box is made from a square sheet of side 18 cm by cutting equal squares of side $x$ cm from each corner and folding up the sides. Find $x$ for maximum volume and the maximum volume.

Try it first, then show the solutionSolution

$V = x (18 - 2 x)^{2}$ , with $0 < x < 9$ . (1 mark)
$V = 4 x^{3} - 72 x^{2} + 324 x$ , so $\frac{d V}{d x} = 12 x^{2} - 144 x + 324 = 12 (x - 3) (x - 9)$ . (1 mark)
$\frac{d V}{d x} = 0 \Rightarrow x = 3$ or $x = 9$ ; $x = 9$ is rejected (no box). (1 mark)
$\frac{d ^{2} V}{d x ^{2}} = 24 x - 144 = - 72 < 0$ at $x = 3$ : maximum. (1 mark)
$V = 3 \times 1 2^{2} = 432$ cm³. (1 mark)

Answer: Cut squares of side 3 cm; the maximum volume is 432 cm³.

Example 4: Closed cylinder of least surface area · 5 marks · Long answerHarder board level

A closed cylindrical container must hold $128 π$ cm³. Find its radius and height for the least total surface area.

Try it first, then show the solutionSolution

$π r^{2} h = 128 π \Rightarrow h = \frac{128}{r ^{2}}$ . (1 mark)
$S = 2 π r^{2} + 2 π r \cdot \frac{128}{r ^{2}} = 2 π r^{2} + \frac{256 π}{r}$ . (1 mark)
$\frac{d S}{d r} = 4 π r - \frac{256 π}{r ^{2}} = 0 \Rightarrow r^{3} = 64 \Rightarrow r = 4$ . (1 mark)
$\frac{d ^{2} S}{d r ^{2}} = 4 π + \frac{512 π}{r ^{3}} > 0$ : minimum. (1 mark)
$h = \frac{128}{16} = 8$ cm, and $S = 32 π + 64 π = 96 π$ cm². (1 mark)

Answer: Radius 4 cm and height 8 cm (height = diameter); least surface area $96 π$ cm².

Example 5: Norman window for maximum light · 5 marks · Long answerHarder board level

A window is a rectangle surmounted by a semicircle, the diameter of the semicircle being the top side of the rectangle. The total perimeter is 10 m. Find the dimensions that admit the most light.

Try it first, then show the solutionSolution

Let the radius be $r$ and the height of the rectangle $h$ . Perimeter: $2 r + 2 h + π r = 10 \Rightarrow h = \frac{10 - 2 r - π r}{2}$ . (1 mark)
Area $A = 2 r h + \frac{1}{2} π r^{2} = r (10 - 2 r - π r) + \frac{1}{2} π r^{2} = 10 r - 2 r^{2} - \frac{π}{2} r^{2}$ . (1 mark)
$\frac{d A}{d r} = 10 - 4 r - π r = 0 \Rightarrow r = \frac{10}{4 + π}$ . (1 mark)
$\frac{d ^{2} A}{d r ^{2}} = - (4 + π) < 0$ : maximum. (1 mark)
$h = \frac{10 - ( 2 + π ) \frac{10}{4 + π}}{2} = \frac{10}{4 + π} = r$ . Width $= 2 r = \frac{20}{4 + π} \approx 2.80$ m, height of rectangle $\approx 1.40$ m. (1 mark)

Answer: Width $\frac{20}{4 + π} \approx 2.80$ m and rectangle height $\frac{10}{4 + π} \approx 1.40$ m; the maximum area is $\frac{50}{4 + π} \approx 7.00$ m². (ISC 2024 asked a similar question with an equilateral triangle on top.)

Example 6: Nearest point on a parabola · 5 marks · Long answerHOTS / competency

Find the point on the curve $y^{2} = 4 x$ nearest to the point $(2, 1)$ , and the least distance.

Try it first, then show the solutionSolution

A point on the curve is $(\frac{y ^{2}}{4}, y)$ . Minimise $D = (\frac{y ^{2}}{4} - 2)^{2} + (y - 1)^{2}$ , the square of the distance. (1 mark)
$\frac{d D}{d y} = 2 (\frac{y ^{2}}{4} - 2) \frac{y}{2} + 2 (y - 1) = \frac{y ^{3}}{4} - 2 y + 2 y - 2 = \frac{y ^{3}}{4} - 2$ . (2 marks)
$\frac{d D}{d y} = 0 \Rightarrow y^{3} = 8 \Rightarrow y = 2$ , so $x = 1$ . (1 mark)
$\frac{d ^{2} D}{d y ^{2}} = \frac{3 y ^{2}}{4} = 3 > 0$ : minimum. Least distance $= (1 - 2)^{2} + (2 - 1)^{2} = 2$ . (1 mark)

Answer: The nearest point is $(1, 2)$ , at distance $2$ .

Where students lose marks in Optimisation Problems (Maxima and Minima Word Problems)

The slips examiners see most often, and what to write instead.

Mistake: Not reducing to one variable
The ISC 2024 examiners reported that candidates could not express the area in terms of one variable in the window question.
Do this: Write the constraint first, solve it for one variable, and substitute before you differentiate.
Usually costs: Most of the question: every later step depends on it.
Mistake: Skipping the second derivative test
A critical point may be a maximum or a minimum; examiners in 2024 penalised answers that did not complete the test.
Do this: Always compute f″ at the critical point and state its sign and the conclusion.
Usually costs: 1 mark.
Mistake: Wrong area or volume formula
In 2024 candidates used a wrong formula for the area of an equilateral triangle (it is √3a²/4).
Do this: Write the formula you use in words before substituting.
Usually costs: 2 to 4 marks.
Mistake: Stopping at the critical value
The question asks for dimensions or the maximum value, not just x.
Do this: Finish with every required quantity and a sentence with units.
Usually costs: 1 mark.

Exam tips

Draw a labelled diagram even if one is given; it helps you write the constraint correctly.
Give the range of the variable (for example $0 < x < 9$ ); it justifies rejecting impossible critical values.
Keep $π$ and surds exact until the final line, then give a decimal approximation if useful.
If $f^{''} (c) = 0$ , switch to the first derivative test and say so.
For distance problems, state "minimising $D^{2}$ is equivalent to minimising $D$ " before you begin.

Practice questions on Optimisation Problems (Maxima and Minima Word Problems)

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The minimum value of $x^{2} - 6 x + 10$ is:
(a) 0
(b) 1
(c) 3
(d) 10
Hint
Critical point at $x = 3$ .
Answer
(b) 1.
Q2 · 3 marks · Short answerBoard level
A rectangle has perimeter 36 m. Find its dimensions for maximum area.
Hint
Let the sides be $x$ and $18 - x$ .
Answer
$A = x (18 - x)$ , $A^{'} = 18 - 2 x = 0 \Rightarrow x = 9$ , $A^{''} = - 2 < 0$ . A $9 \times 9$ square, area 81 m².
Q3 · 5 marks · Long answerHarder board level
A wire 28 m long is cut into two pieces. One is bent into a square and the other into a circle. Find the length of each piece so that the total area is least.
Hint
If the square uses $x$ m, its side is $\frac{x}{4}$ ; the circle has circumference $28 - x$ .
Answer
$A = \frac{x ^{2}}{16} + \frac{( 28 - x ) ^{2}}{4 π}$ . $A^{'} = \frac{x}{8} - \frac{28 - x}{2 π} = 0 \Rightarrow x = \frac{112}{π + 4} \approx 15.68$ m. $A^{''} = \frac{1}{8} + \frac{1}{2 π} > 0$ : minimum. Square piece $\frac{112}{π + 4}$ m, circle piece $\frac{28 π}{π + 4} \approx 12.32$ m.
Q4 · 3 marks · Short answerHarder board level
Show that of all rectangles inscribed in a circle of radius 5 cm, the square has the greatest area, and find that area.
Hint
If one side is $x$ , the other is $100 - x^{2}$ ; maximise $A^{2}$ .
Answer
$A^{2} = x^{2} (100 - x^{2})$ . $\frac{d}{d x} (A^{2}) = 200 x - 4 x^{3} = 0 \Rightarrow x^{2} = 50$ , so both sides are $50$ : a square. Second derivative $200 - 12 x^{2} = - 400 < 0$ . Area 50 cm².
Q5 · 3 marks · Short answerHarder board level
An open tank with a rectangular base is 3 m deep and holds 12 m³. The base costs Rs 70 per m² and the sides Rs 45 per m². Find the least cost.
Hint
The base area is fixed at 4 m²; minimise $l + b$ with $l b = 4$ .
Answer
Cost $= 70 (4) + 45 \times 3 \times 2 (l + b) = 280 + 270 (l + \frac{4}{l})$ . Derivative $270 (1 - \frac{4}{l ^{2}}) = 0 \Rightarrow l = 2 = b$ . Least cost $= 280 + 270 \times 4 =$ Rs 1360.
Q6 · 5 marks · Long answerHOTS / competency
Show that the height of the cone of greatest volume inscribed in a sphere of radius $R$ is $\frac{4 R}{3}$ , and find that volume.
Hint
If the cone has height $h$ , its base radius satisfies $r^{2} = 2 R h - h^{2}$ .
Answer
$V = \frac{1}{3} π (2 R h - h^{2}) h = \frac{π}{3} (2 R h^{2} - h^{3})$ . $V^{'} = \frac{π}{3} (4 R h - 3 h^{2}) = 0 \Rightarrow h = \frac{4 R}{3}$ . $V^{''} = \frac{π}{3} (4 R - 6 h) = - \frac{4 π R}{3} < 0$ . $r^{2} = \frac{8 R ^{2}}{9}$ , so $V = \frac{1}{3} π \cdot \frac{8 R ^{2}}{9} \cdot \frac{4 R}{3} = \frac{32 π R ^{3}}{81}$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a farmer has 200 m of fencing to enclose a rectangular field beside a straight river; no fence is needed along the river.
(i) If each side perpendicular to the river is $x$ m, express the area $A$ in terms of $x$ . (1 mark)
(ii) Find the critical point. (1 mark)
(iii) Use the second derivative to show it gives a maximum. (1 mark)
(iv) Find the greatest area and the length along the river. (1 mark)
Hint
The side parallel to the river is $200 - 2 x$ .
Answer
(i) $A = x (200 - 2 x) = 200 x - 2 x^{2}$ . (ii) $A^{'} = 200 - 4 x = 0 \Rightarrow x = 50$ . (iii) $A^{''} = - 4 < 0$ . (iv) 100 m along the river; greatest area $50 \times 100 = 5000$ m².

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the second derivative test for a maximum.Show answer
If $f^{'} (c) = 0$ and $f^{''} (c) < 0$ , then $f$ has a local maximum at $c$ .
2.For a closed cylinder of fixed volume and least surface area, how are $h$ and $r$ related?Show answer
$h = 2 r$ .
3.For a fixed perimeter, which rectangle has the greatest area?Show answer
A square.
4.Why minimise $D^{2}$ instead of $D$ ?Show answer
The square root is increasing, so both have the same minimum point, and $D^{2}$ has no root to differentiate.
5.Write the volume of an open box made by cutting squares of side $x$ from a square sheet of side $a$ .Show answer
$V = x (a - 2 x)^{2}$ .
6.What must you compare to find an absolute maximum on $[a, b]$ ?Show answer
The values at the critical points and at the end points $a$ and $b$ .

Optimisation Problems (Maxima and Minima Word Problems): frequently asked questions

How do I start an optimisation problem?

Draw a diagram, name the variables, and write two equations: one for the quantity to be optimised and one for the condition given. Use the condition to remove all but one variable before differentiating.

Do I always need the second derivative test?

You need some test to show that the critical point is a maximum or a minimum. The second derivative test is usually quickest; if the second derivative is zero there, use the sign of the first derivative on each side instead.

What if the critical point is outside the allowed range?

Reject it and say why, for example "x = 9 gives no box". If the variable lies in a closed interval, also check the end points.

Are profit and cost problems part of this topic for ISC 2027?

Not as the commerce applications of calculus: the 2027 syllabus removed the old Section C unit on cost, revenue and profit functions. General real-life maxima and minima problems, including cost of materials, remain in Applications of Derivatives.

How many marks do optimisation problems carry in the ISC paper?

They are usually long questions. In ISC 2024, two 6-mark alternatives asked for the dimensions of a window admitting maximum light and the least surface area of a closed cylindrical dustbin of given volume.