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Flip to reveal answersWhat is optimisation in calculus?
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All 8 Flashcards — Optimisation in Context
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Question
What is optimisation in calculus?
Answer
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.
💡 Hint
Optimise = find the best value (highest or lowest).
Question
What are the steps to solve an optimisation problem?
Answer
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.
💡 Hint
Key step: get to one variable before differentiating.
Question
How do you check if a stationary point is a maximum or minimum in a context problem?
Answer
Use a sign diagram of f'(x), OR check the endpoints. In closed-interval problems, also evaluate f at the endpoints.
💡 Hint
Sign diagram: + then − = max; − then + = min.
Question
A farmer has 80m of fencing. Maximise the area of a rectangular enclosure against a wall (3 sides fenced).
Answer
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.
💡 Hint
Write Area in terms of x using the fencing constraint.
Question
What is a constraint in an optimisation problem?
Answer
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.
💡 Hint
Constraint lets you go from 2 unknowns to 1.
Question
Revenue R(x) = 40x − x². What value of x maximises revenue?
Answer
R'(x) = 40 − 2x = 0 → x = 20. R'(20) = −2 < 0 → local max. Max revenue = R(20) = 40(20)−400 = 400.
💡 Hint
Second derivative negative confirms maximum.
Question
In an IB optimisation question, what must you always include in the answer?
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.
💡 Hint
IB mark schemes reward classification + full answer.
Question
Cost C = 2x² − 12x + 20. Find the minimum cost and the value of x.
Answer
C' = 4x − 12 = 0 → x = 3. C'(3) = 4 > 0 → local min. Min cost = 2(9) − 12(3) + 20 = 18 − 36 + 20 = 2.
💡 Hint
Positive second derivative = minimum.
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