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Optimisation problems
This test models geometric and economic optimisation situations. Students construct an objective function, determine its feasible domain, use derivatives to find an extremum and interpret the result in context.
Questions: 10Estimated time: 21 minutes
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Question 1out of 10
What general strategy is suitable for a word optimisation problem?
Define the variable and feasible values from the context. Build the objective function using the constraints. Find candidates, compare relevant values and check the result and units.
Question 2out of 10
A rectangle has perimeter 32 cm. What should one side be for maximum area?
Let one side be x and the other P/2 − x. Area S(x) = x(P/2 − x), for 0 < x < P/2. Setting S′(x) = P/2 − 2x = 0 gives x = P/4; a square maximizes area.
Question 3out of 10
An open-top box has square base side x and volume V = 200. Which function gives surface area?
Volume V = x²h, so h = V/x² for x > 0. An open-top box has one base and four walls: S(x) = x² + 4xh = x² + 4V/x. Do not include a top.
Question 4out of 10
For x > 0, f(x) = x + 25/x has its minimum at which x?
For a > 0 and x > 0, f′(x) = 1 − a/x². Setting it to zero gives x = √a. The derivative is negative before this point and positive after, giving a minimum.
Question 5out of 10
An open-top box has square base side x and volume V = 400. Which function gives surface area?
Volume V = x²h, so h = V/x² for x > 0. An open-top box has one base and four walls: S(x) = x² + 4xh = x² + 4V/x. Do not include a top.
Question 6out of 10
For x > 0, f(x) = x + 100/x has its minimum at which x?
For a > 0 and x > 0, f′(x) = 1 − a/x². Setting it to zero gives x = √a. The derivative is negative before this point and positive after, giving a minimum.
Question 7out of 10
A rectangle against a wall uses 48 m of fence on three sides. What perpendicular side maximises area?
Two sides perpendicular to the wall have length x; the parallel side is f − 2x. Area S(x) = x(f − 2x), for 0 < x < f/2. Setting S′ = f − 4x = 0 gives x = f/4.
Question 8out of 10
Profit is P(x) = −3x² + 168x. Which x maximises profit?
P′(x) = −2ax + b, giving critical point x = b/(2a). For a > 0, the parabola opens downward, so this is a maximum. Check the feasible domain.
Question 9out of 10
What must first be defined when building an optimisation model?
Define the variable and feasible values from the context. Build the objective function using the constraints. Find candidates, compare relevant values and check the result and units.
Question 10out of 10
Why must endpoints also be checked on a closed interval?
On a closed interval, evaluate the function at both endpoints and all critical points inside it. Compare function values, not x values. For positive m, m(x − r)² is greatest at the endpoint farther from r.
Time left: 00:00:00
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Sample questions
- A rectangle has perimeter ___ cm. What should one side be for maximum area?
- Profit is P(x) = −___x² + ___x. Which x maximises profit?
- An open-top box has square base side x and volume V = ___. Which function gives surface area?
- A rectangle against a wall uses ___ m of fence on three sides. What perpendicular side maximises area?
- For x > 0, f(x) = x + ___/x has its minimum at which x?
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