Revision notes

How to solve maxima and minima problems: strategy and geometry formulas

Problem solving strategy

1. Form an equation linking the variables in the question (if not provided)

2. Find the first derivative and equate it to zero (stationary value)

3. Find information required by question and check if value is maximum/minimum (by first/second derivative test)

Geometry formulas: Area, Volume, Surface area

$ \text{Area of triangle} = {1 \over 2} \times b \times h $

$ \text{Area of trapezium} = {1 \over 2} \times \text{Sum of parallel sides} \times h $

$ \text{Area of circle} = \pi r^2 $

$ \text{Circumference of circle} = 2\pi r = \pi d $


$ \text{Volume of cuboid} = l \times b \times h $

$ \text{Volume of cylinder} = \pi r^2 h $

$ \text{Volume of cone} = {1 \over 3} \pi r^2 h $

$ \text{Curved surface area of cone} = \pi r l $

$ \text{Volume of pyramid} = {1 \over 3} \times \text{Base area} \times h $

$ \text{Volume of sphere} = {4 \over 3} \pi r^3 $

$ \text{Surface area of sphere} = 4 \pi r^2 $


Additionally you may need:

  • Trigonometry formulas (Pythagoras theorem, TOA CAH SOH)
  • Compare corresponding sides in similar triangles (see Question 1 below)

Practice questions

Maxima and minima in geometry problems

Geometry problem

diagram

1. The diagram shows a triangular piece of cardboard in which angle $ACB = 90^\circ$, $AC = 60$ cm and $BC = 80$ cm. A rectangle $CPQR$, where $P$, $Q$ and $R$ due on $AC$, $AB$ and $BC$ respectively, is to be cut out. $PQ = x$ cm and $QR = y$ cm.

(from think! A Maths Workbook Worksheet 12D)

(i) Show that $y = 60 - {3 \over 4}x$.

Solutions

(ii) Express the area, $Z$ cm2, of rectangle $CPQR$ in terms of $x$.

Answer: $ Z = 60x - {3 \over 4}x^2 $

Solutions

(iii) Given that $x$ can vary, find the maximum value of $Z$.

Answer: $ Z = 1200 $

Solutions

Maximum or minimum gradient using third derivative

The first derivative ${dy \over dx}$ is the rate of change of $y$ with respect to $x$.

The second derivative ${d^2 y \over dx}$ is the rate of change of ${dy \over dx}$ with respect to $x$. Thus,

  • For maximum gradient, ${d^2 y \over dx^2} = 0$ and ${d^3 y \over dx^3} < 0 $
  • For minimum gradient, ${d^2 y \over dx^2} = 0$ and ${d^3 y \over dx^3} > 0 $

2. Point $A$ moves along the curve $y = 2x^3 - 12x^2 + x - 5$. Find the coordinates of $A$ such that the gradient to the curve at $A$ is a minimum value.

Answer: $ A(2, -35) $

Solutions

O Level past year questions on maxima and minima

Year & paper Comments
2024 P1 Question 8 Geometry problem
2023 P2 Question 9 Distance, speed & time problem
2023 P1 Question 9b Maximum gradient
2022 P2 Question 4 Geometry problem
2021 P1 Question 8 Geometry problem
2019 P2 Question 11 Geometry problem
2018 P1 Question 10 Geometry problem
2017 P1 Question 6 Geometry problem
2016 P1 Question 7 Distance, speed & time problem
2015 P1 Question 11 Geometry problem (with trigonometry)
2014 P1 Question 10 Geometry problem
2013 P2 Question 7 Geometry problem (Link - Subscription required)
2011 P2 Question 6 Geometry problem
2010 P2 Question 2 Geometry problem
2009 P2 Question 11 Geometry problem
2008 P1 Question 13 Geometry problem
2003 P2 Question 12 Either & Or Geometry problem
2002 P2 Question 12 Either Geometry problem

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