Build and solve an optimisation model

Formulate an optimisation problem before differentiating

The source’s complete five-step procedure is:

1. Identify variables; if useful, draw a diagram with all given information labelled.
2. Form equations relating those variables.
3. Use all constraints to express the quantity to optimise in a single variable.
4. Differentiate to find stationary values.
5. Test their nature with the first or second derivative test, then answer the original question.

The source describes this as both formulating and solving the mathematical model. Its cone example recommends a sketch to visualise the changing dimensions; its capped-container “Thinking Process” works backwards from the goal, then presents the volume relation, reduced area, stationary calculation and minimum test in a logical order. These are part of the worked steps, rather than a separate teacher-guide section.

Check the feasible domain and boundaries

Independently, restrict the variable to its physically feasible domain and check boundary behaviour before claiming an overall optimum.

A stationary point alone does not establish a global maximum or minimum.

A positive second derivative is sufficient at a stationary point but is not necessary for every minimum; the source also permits a first derivative test.

Count exposed surfaces carefully: a hemisphere-capped cylinder has one flat base, while the source’s ordinary closed can has two ends.

Connect the model to the examples

Example 13 distinguishes the negative height derivative from the positive rate of dropping. Examples 14–16 retain all printed optimisation working, source planning prompts, angle restrictions, sign-table directions and source diagrams. The can calculation also answers the opening question: for the closed-can model, optimal height equals diameter.

Use the mastery checklist to review each skill after lecture and again after tutorial. The practice map below links the printed question references to their worked solutions.

Compare a stationary value with the feasible boundaries

A rectangular region has area $A=x(18-x)$, with a design restriction $2\le x\le6$ metres. Find the greatest feasible area. Explain why solving $A^{\prime}=0$ is not enough.

After your attempt: compare the reasoning

The stationary input is 9, outside the feasible interval. On $2\le x\le6$, $A^{\prime}=18-2x>0$, so the area increases throughout. The largest feasible area is $A(6)=72\ \mathrm{m}^2$ at the boundary. In this model a global optimum can occur at an allowed endpoint even though the derivative there is not zero.