Maxima, minima
Strategy and geometry formulas
General approach:
1. Form an equation linking the variables in the question
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:
$ \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) and to compare sides in similar triangles (see Q1).
Questions
Geometry problem
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 $y = 60 - {3 \over 4}x$.
(ii) Express the area, $Z$ cm2, of rectangle $CPQR$ in terms of $x$.
Answer: $ Z = 60x - {3 \over 4}x^2 $
(iii) Given that $x$ can vary, find the maximum value of $Z$.
Answer: $ Z = 1200 $
Maximum/minimum gradient
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) $
Past year O level questions
| 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 |