IB Mathematics AA SL Local maximum and minimum values Study Notes - New Syllabus

IB Mathematics AA SL Local maximum and minimum values Study Notes

LEARNING OBJECTIVE

Local maximum and minimum points.

Key Concepts:

Local Maximum and Minimum Points
Optimization
Points of Inflexion

MAA HL and SL Notes – All topics

Local Maximum and Minimum Points

Concept:

A local maximum occurs when a function changes from increasing to decreasing.
A local minimum occurs when a function changes from decreasing to increasing.
Use the first derivative test or second derivative test to identify them.

Important Rules:

If $ f'(x) = 0 $ and $ f”(x) > 0 $, there is a local minimum.
If $ f'(x) = 0 $ and $ f”(x) < 0 $, there is a local maximum.

Example

Find the local maximum and minimum points of the function:
$ f(x) = x^3 – 3x^2 + 4 $

▶️ Answer/Explanation

$ f'(x) = 3x^2 – 6x $
Set $ f'(x) = 0 $ ⇒ $ x(x – 2) = 0 $ ⇒ $ x = 0 $, $ x = 2 $

Second derivative: $ f”(x) = 6x – 6 $
$ f”(0) = -6 $ ⇒ maximum at $ x = 0 $
$ f”(2) = 6 $ ⇒ minimum at $ x = 2 $

Local maximum: $ f(0) = 4 $
Local minimum: $ f(2) = 0 $

Optimization

Concept:

Used to find maximum or minimum values in real-world problems.
Applications: maximize area, minimize cost, maximize profit, etc.

Steps:

Write the quantity to optimize.
Use constraints to reduce variables.
Differentiate and solve $ f'(x) = 0 $.
Use the second derivative test or endpoints to confirm.

Example

A rectangle is to be drawn with a fixed perimeter of 40 cm. Find the dimensions that give the maximum area.

▶️ Answer/Explanation

Let width = $ x $, then length = $ 20 – x $ (since perimeter = 2(x + length))

Area = $ A = x(20 – x) = 20x – x^2 $

Derivative: $ A'(x) = 20 – 2x $
Set to zero: $ A'(x) = 0 \Rightarrow x = 10 $
So length = 10, width = 10

Maximum area: when rectangle is a square (10 × 10 cm)

Example

A company sells a product for $20 per unit. The cost to produce x units is given by $ C(x) = 5x + 100 $. Find the number of units that must be sold to maximize profit.

▶️ Answer/Explanation

Revenue $ R(x) = 20x $
Cost $ C(x) = 5x + 100 $
Profit $ P(x) = R(x) – C(x) = 20x – (5x + 100) = 15x – 100 $

Derivative: $ P'(x) = 15 $
Since derivative is constant, profit increases with more units. No maximum within a restricted domain, but check context limits.
If the question had a domain (e.g., 0 ≤ x ≤ 50), you’d check endpoints.

Example

A closed cylindrical can is to be made with a fixed surface area of 600 cm². Find the dimensions (radius and height) that give the maximum volume.

▶️ Answer/Explanation

Let radius = $ r $, height = $ h $
Surface area: $ 2\pi r^2 + 2\pi rh = 600 $
Solve for $ h $: $ h = \frac{600 – 2\pi r^2}{2\pi r} $

Volume: $ V = \pi r^2 h $
Substitute $ h $:
$ V = \pi r^2 \cdot \frac{600 – 2\pi r^2}{2\pi r} = \frac{r(600 – 2\pi r^2)}{2} $

Differentiate and solve $ V'(r) = 0 $ to find max volume.
This leads to optimal ratio when height = 2 × radius.

Conclusion: For maximum volume, the height equals the diameter (i.e., $ h = 2r $)

Points of Inflexion

Concept:

A point where the graph changes concavity (from concave up to down or vice versa).
Occurs when $ f”(x) = 0 $ and changes sign.
$ f”(x) = 0 $ alone is not sufficient – sign change is necessary.