What Optimisation Means in IGCSE Maths
Optimisation problems ask you to find the maximum or minimum value of a quantity. In everyday terms, this might mean finding the dimensions of a box that maximise its volume, or the speed that minimises fuel consumption. At IGCSE level, you solve these problems using differentiation — finding the derivative, setting it equal to zero, and determining whether you have a maximum or minimum.
These questions appear on Paper 4 of the Extended syllabus and are typically worth seven to nine marks. They combine algebra, differentiation, and sometimes geometry into a single multi-part question.
The Problem
An open-topped rectangular box is made from a square piece of cardboard with side length 24 cm. Squares of side x cm are cut from each corner, and the sides are folded up. Find the value of x that maximises the volume of the box, and find this maximum volume.
Step 1: Visualise and Set Up
Imagine a 24 cm × 24 cm square of cardboard. When you cut out a square of side x from each corner and fold up the sides:
- The base of the box is a square with side length (24 − 2x) cm
- The height of the box is x cm
Therefore the volume is:
- V = x(24 − 2x)²
Step 2: Expand the Expression
First, expand (24 − 2x)²:
- (24 − 2x)² = 576 − 96x + 4x²
Now multiply by x:
- V = x(576 − 96x + 4x²)
- V = 576x − 96x² + 4x³
It is often neater to write this as V = 4x³ − 96x² + 576x.
Step 3: Differentiate
Find dV/dx:
- dV/dx = 12x² − 192x + 576
We can simplify by dividing through by 12:
- dV/dx = 12(x² − 16x + 48)
Step 4: Find the Turning Points
Set the derivative equal to zero:
- x² − 16x + 48 = 0
Factor this quadratic. We need two numbers that multiply to 48 and add to −16. Those are −12 and −4:
- (x − 12)(x − 4) = 0
- x = 12 or x = 4
Step 5: Determine Which Solution is Valid
Think about the physical constraints. The cardboard is 24 cm wide, and we cut x from each side, leaving (24 − 2x) for the base. For the base to have a positive length:
- 24 − 2x > 0
- x < 12
So x = 12 gives a base length of zero — no box at all. Therefore x = 4 is the only practical solution.
Step 6: Confirm It Is a Maximum
We need to verify that x = 4 gives a maximum, not a minimum. Find the second derivative:
- d²V/dx² = 24x − 192
At x = 4:
- d²V/dx² = 24(4) − 192 = 96 − 192 = −96
Since the second derivative is negative, the turning point at x = 4 is a maximum. This is exactly what we want.
Step 7: Calculate the Maximum Volume
Substitute x = 4 into the volume formula:
- V = 4(4)³ − 96(4)² + 576(4)
- V = 4(64) − 96(16) + 576(4)
- V = 256 − 1536 + 2304
- V = 1024 cm³
The maximum volume is 1024 cm³, achieved when the corner squares have side length 4 cm.
Understanding the Second Derivative Test
The second derivative tells you about the concavity of the curve:
- If d²V/dx² < 0 at a turning point, the curve is concave down (like a hill), so you have a maximum
- If d²V/dx² > 0 at a turning point, the curve is concave up (like a valley), so you have a minimum
- If d²V/dx² = 0, the test is inconclusive and you need to check points either side
At IGCSE level, the second derivative test almost always gives a clear positive or negative value. If you forget this test, you can alternatively substitute values of x slightly less than and slightly greater than the turning point into the original derivative to check whether it changes from positive to negative (maximum) or negative to positive (minimum).
General Strategy for Optimisation Problems
Every optimisation problem at IGCSE follows these steps:
- Read the question and identify what quantity you are maximising or minimising
- Express that quantity as a function of a single variable (you may need to use a constraint to eliminate a second variable)
- Differentiate the function
- Set the derivative to zero and solve for the variable
- Check validity — does the solution make physical sense?
- Use the second derivative to confirm maximum or minimum
- Substitute back to find the maximum or minimum value
Another Example: Minimising Surface Area
A closed cylindrical tin has a volume of 500 cm³. Find the radius that minimises the total surface area.
Setting up: The volume constraint is πr²h = 500, so h = 500/(πr²).
The total surface area is S = 2πr² + 2πrh. Substituting for h:
- S = 2πr² + 2πr × 500/(πr²)
- S = 2πr² + 1000/r
Differentiating: dS/dr = 4πr − 1000/r²
Setting to zero: 4πr = 1000/r², so 4πr³ = 1000, giving r³ = 250/π, and r = (250/π)^(1/3) ≈ 4.30 cm.
Second derivative: d²S/dr² = 4π + 2000/r³, which is always positive. So this is a minimum, as required.
Common Mistakes
- Not expanding before differentiating. You must write V as a polynomial in x before you can differentiate term by term. Alternatively, you can use the product rule, but at IGCSE level, expanding is expected.
- Forgetting the physical constraints. Always check that your answer makes sense. A negative length or a radius larger than the available material is not valid.
- Dropping the second derivative check. The mark scheme awards a mark for determining the nature of the turning point. Do not skip this step.
- Arithmetic errors in expansion. Expanding (24 − 2x)² requires careful application of the identity (a − b)² = a² − 2ab + b². Double-check each term.
- Giving the value of x instead of the quantity asked for. If the question asks for the maximum volume, you must substitute x back in and calculate V.
Practice Question
A farmer has 200 metres of fencing to enclose a rectangular field against a straight river (no fencing needed along the river). Find the dimensions that maximise the area of the field.
Hint: Let x be the side perpendicular to the river. The fencing is used for two widths and one length, so the length is 200 − 2x. The area is A = x(200 − 2x).
Get Expert Help with IGCSE Maths
If differentiation — or any other Extended syllabus topic — is giving you trouble, a specialist IGCSE Maths tutor can help you build confidence and technique in just a few sessions.
Need Help With IGCSE Maths?
Book a free 60-minute trial class with Teacher Rig and get personalised guidance for your IGCSE Maths preparation.