Grade 12 Maths · Calculus · Find a maximum or minimum in context

Find a maximum or minimum in context Grade 12

Write the quantity to maximise or minimise as a function of one variable, set its derivative equal to 0, keep the value that fits the situation and state the largest or smallest value (for example the biggest volume of an open box).

questions use it, all grades · 22 past paper, 37 practice: 59 questions use it, all grades · 22 past paper, 37 practice
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Lesson 7: How To Solve Optimisation Questions | Area, Volume & Cost | Grade 12 CalculusMlungisi Nkosi Maths & Science

Calculus grade 12 Optimisation: ExamKevinmathscience

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Question 10.2 · November 2025 · Paper 1 · National (DBE) · 3 marks

A rectangular metal sheet has dimensions

x

and

h

units, with

x > h

, and a perimeter of

50

units. The metal sheet is rolled into a cylinder with two open ends (top and bottom) and height

h

units. $Diagram for this question$

10.2 Calculate the value of

x

that will maximise the volume of the cylinder.(3)

Work it out on paper first. Then mark yourself with the official memo, one tick at a time.

More past paper questions: find a maximum or minimum in context

Work each one out first, then open the memo and check every tick.

3. Question 4.4.2 · November 2022 · Paper 2 · National (DBE) · 4 marks
In the diagram, $M (3; - 5)$ is the centre of the circle having PN as its diameter. KL is a tangent to the circle at $N (7; - 2)$ . $Diagram for this question$
4.4 Points $A (t; t)$ and B are not shown on the diagram.

From point A, another tangent is drawn to touch the circle with centre M at B.

Answer the questions that follow.
4.4.2 Determine the minimum length of AB.(4)
Show the memo (4 ticks)Memo
1. ✓
  METHOD 1: Substitution into correct formula $t = \frac{- 4}{2 ( 2 )}$
  1 mark · Find turning point of a parabola
2. ✓
  METHOD 1: $t = - 1$ $= - 1$
  Minimum at $t = - 1$
  1 mark · Find turning point of a parabola
3. ✓
  METHOD 1: Substitution $A B = 2 (- 1)^{2} + 4 (- 1) + 9$
  1 mark · Find a maximum or minimum in context
4. ✓
  METHOD 1: Answer $A B = 7$
  1 mark · Find a maximum or minimum in context
4. Question 10.2.2 · September 2026 · Paper 1 · Provincial · 2 marks
10.2 PQRS is a trapezium with PQ parallel to RS and measurements $PQ = 2 x - 3$ , $QR = 5 - x$ and $RS = 4 x + 3$ in meters, as indicated on the diagram.

Answer the questions that follow. $Diagram for question 10.2$
10.2.2 Calculate the value of $x$ if PQRS has a maximum area.(2)
Show the memo (2 ticks)Memo
1. ✓
  Derivative = 0 $A^{'} (x) = 0$
  $15 - 6 x = 0$
  1 mark · Find a maximum or minimum in context
2. ✓
  Answer $6 x = 15$
  $x = 2, 5$ m
  1 mark · Find a maximum or minimum in context
5. Question 9.5 · May/June 2025 · Paper 1 · National (DBE) · 6 marks
Given: $f (x) = x^{3} - 6 x^{2} + 9 x - 4 = (x - 4) (x - k)^{2}$
9.5 Calculate the maximum vertical distance between $f$ and $h$ for $1 < x < 3$ , if $h (x) = - 2 f^{'} (x)$ .(6)
Show the memo (6 ticks)Memo
1. ✓
  Establish the expression for the distance $h (x) = - 2 (3 x^{2} - 12 x + 9) = - 6 x^{2} + 24 x - 18$
  $Distance = h (x) - f (x)$
  $= (- 6 x^{2} + 24 x - 18) - (x^{3} - 6 x^{2} + 9 x - 4)$
  1 mark · Find a horizontal or vertical length
2. ✓
  Simplify the distance expression $= - 6 x^{2} + 24 x - 18 - x^{3} + 6 x^{2} - 9 x + 4$
  $= - x^{3} + 15 x - 14$
  1 mark · Collect like terms
3. ✓
  Differentiate the distance expression and equate to 0To find max distance, let $D^{'} (x) = 0$ .
  $D^{'} (x) = - 3 x^{2} + 15 = 0$
  1 mark · Find a maximum or minimum in context
4. ✓
  Solve the derivative equation for x squared $3 x^{2} = 15$
  $x^{2} = 5$
  1 mark · Solve a linear equation
5. ✓
  Find the correct x-value within the given domain $x = \pm 5$
  But the domain is restricted to $1 < x < 3$ .
  $∴ x = 5$ (since $- 5$ is not in the interval)
  1 mark · Solve a quadratic by factorising
6. ✓
  Calculate the maximum vertical distanceSubstitute $x = 5$ into the distance equation:
  $Max distance = - (5)^{3} + 15 (5) - 14$
  $= - 55 + 155 - 14$
  $= 105 - 14 \approx 8, 36$
  1 mark · Find a maximum or minimum in context
6. Question 5.5 · May/June 2023 · Paper 1 · National (DBE) · 5 marks
The graphs of the functions $f (x) = - (x + 3)^{2} + 4$ and $g (x) = x + 5$ are drawn below. The graphs intersect at A and B. $Diagram for this question$
5.5 The maximum distance between $f$ and $g$ in the interval $x_{A} < x < x_{B}$ is $k$ . If $h (x) = g (x) + k$ , determine the equation of $h$ in the form $h (x) = ...$ (5)
Show the memo (5 ticks)Memo
1. ✓
  Distance $D (x) = f (x) - g (x) = - x^{2} - 7 x - 10$
  1 mark · Find a horizontal or vertical length
2. ✓
  $- 2 x - 7 = 0$ Max: $D^{'} (x) = - 2 x - 7 = 0$
  1 mark · Find a maximum or minimum in context
3. ✓
  $x = - \frac{7}{2}$ $x = - \frac{7}{2}$
  1 mark · Find stationary points
4. ✓
  $k = 2, 25$ $D (- \frac{7}{2}) = - (- \frac{7}{2})^{2} - 7 (- \frac{7}{2}) - 10 = 2, 25$
  $∴ k = 2, 25$
  1 mark · Find a maximum or minimum in context
5. ✓
  $h (x) = x + 7, 25$ $h (x) = x + 5 + 2, 25$
  $∴ h (x) = x + 7, 25$
  1 mark · Shift a graph

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