Calculus Optimisation
Calculus optimisation uses differentiation to find maximum and minimum values of functions, with applications to real-world problems in economics, engineering, and science.
What You Need to Know
Key Concept Diagram
A maximum or minimum occurs where the derivative f'(x) = 0 (stationary points)
The second derivative test: if f''(x) < 0 the point is a maximum; if f''(x) > 0 it is a minimum
Optimisation problems require setting up a function, differentiating, then solving f'(x) = 0
Always check endpoints and the nature of stationary points
Applications include maximising area, minimising cost, and finding optimal dimensions
Key Vocabulary
Derivative
The rate of change of a function; the gradient of the tangent at any point
Stationary point
A point where the derivative equals zero; could be a maximum, minimum, or inflection
Optimisation
Finding the maximum or minimum value of a function
Second derivative
The derivative of the derivative; used to determine the nature of stationary points
Knowledge Check
Select the correct answer for each question. Click "Check Answer" to see if you are right.
Question 1
Find the stationary point of f(x) = x^2 - 6x + 8.
Question 2
If f''(3) = 4 at a stationary point, the point is:
Question 3
A rectangle has perimeter 20 cm. What width maximises the area?
Key Concepts Summary
- ●A maximum or minimum occurs where the derivative f'(x) = 0 (stationary points)
- ●The second derivative test: if f''(x) < 0 the point is a maximum; if f''(x) > 0 it is a minimum
- ●Optimisation problems require setting up a function, differentiating, then solving f'(x) = 0
- ●Always check endpoints and the nature of stationary points
- ●Applications include maximising area, minimising cost, and finding optimal dimensions