Optimization
Practice Flashcards
What is optimisation in calculus?
Track your progress — Sign up free to save your progress and get smart review reminders based on spaced repetition.
All Flashcards in Topic 5.7
Below are all 8 flashcards for this topic. Sign up free to track your progress and get personalized review schedules.
5.7.18 cards
What is optimisation in calculus?
Finding the maximum or minimum value of a quantity. You use derivatives to locate stationary points, then determine if it is a max or min.
Optimise = find the best value (highest or lowest).
What are the steps to solve an optimisation problem?
1) Write an expression for the quantity to optimise. 2) Express it in terms of ONE variable (use a constraint). 3) Differentiate and set f'(x) = 0. 4) Solve and classify (max or min). 5) State the answer with units.
Key step: get to one variable before differentiating.
How do you check if a stationary point is a maximum or minimum in a context problem?
Use a sign diagram of f'(x), OR check the endpoints. In closed-interval problems, also evaluate f at the endpoints.
Sign diagram: + then − = max; − then + = min.
A farmer has 80m of fencing. Maximise the area of a rectangular enclosure against a wall (3 sides fenced).
Let width = x. Then length = 80 − 2x. Area A = x(80−2x) = 80x − 2x². A' = 80 − 4x = 0 → x = 20. A = 20 × 40 = 800 m².
Write Area in terms of x using the fencing constraint.
What is a constraint in an optimisation problem?
A rule that links two or more variables. You use it to eliminate one variable so you can write everything in terms of one unknown.
Constraint lets you go from 2 unknowns to 1.
Revenue R(x) = 40x − x². What value of x maximises revenue?
R'(x) = 40 − 2x = 0 → x = 20. R'(20) = −2 < 0 → local max. Max revenue = R(20) = 40(20)−400 = 400.
Second derivative negative confirms maximum.
In an IB optimisation question, what must you always include in the answer?
1) The optimal VALUE of x. 2) The optimal value of the quantity (max area, min cost, etc.). 3) Confirmation it is a max or min (sign diagram or second derivative). 4) Units if the problem has them.
IB mark schemes reward classification + full answer.
Cost C = 2x² − 12x + 20. Find the minimum cost and the value of x.
C' = 4x − 12 = 0 → x = 3. C'(3) = 4 > 0 → local min. Min cost = 2(9) − 12(3) + 20 = 18 − 36 + 20 = 2.
Positive second derivative = minimum.
Topic 5.7 study notes
Full notes & explanations for Optimization
Math AI SL exam skills
Paper structures, command terms & tips
Want smart review reminders?
Sign up free to track your progress. Our spaced repetition algorithm will tell you exactly which cards to review and when.
Start Free