Optimisation problems

What is it about?

The picture changes with the variable, the point on the graph of the objective function follows, and the strip below shows the derivative (green: increasing, red: decreasing). The method: objective function → constraint (one variable left) → domain → $Z' = 0$ → sign change of $Z'$ or sign of $Z''$ → compare the boundary values → answer.

Box: $V(x) = x(30 - 2x)(20 - 2x)$, $0 < x < 10$, is largest for $x \approx 3.92$ cm with $V \approx 1056.3$ cm³; the second zero of $V'$, $x \approx 12.74$, lies outside the domain. Can: $S(r) = 2\pi r^2 + \frac{2V}{r}$ is smallest for $r = \sqrt[3]{V/(2\pi)}$, where $h = 2r$. Cylinder in a cone: $r = \frac{2}{3}R$, $h = \frac{H}{3}$, always $\frac{4}{9}$ of the cone. Lifeguard: $T(x) = \frac{\sqrt{20^2 + x^2}}{v_1} + \frac{\sqrt{30^2 + (60 - x)^2}}{v_2}$; at the optimum $\frac{\sin \alpha_1}{v_1} = \frac{\sin \alpha_2}{v_2}$ – the law of refraction.

Try it

1. Box: drag the handle or the point on the graph, then press “Find the optimum”.
2. Flat sheet: drag a corner of a cut-out, then press “Fold”.
3. 1-litre can and cylinder in a cone: change the data – what stays the same in the “Check” card?
4. Lifeguard: start the “Race”. With equal speeds the straight line is fastest.

Tasks

A sheet measures $16$ cm × $10$ cm. Find the largest box.

Which can holding $500$ cm³ needs the least metal?

Compare $T(24)$, $T(60)$ and the optimum for $v_1 = 7$ m/s, $v_2 = 2$ m/s.

Ideas for the classroom

• Fold paper boxes first and compare their volumes, then let “Find the optimum” show trying and calculating.
• With “Show the solution” off, graph and results are hidden for tasks; “Reveal the solution” uncovers them step by step.
• The lifeguard leads to the law of refraction (Fermat’s principle) – a joint lesson with physics.
• Common stumbling blocks: not using the constraint; keeping solutions outside the domain; not comparing boundary values (for the lifeguard $T(60)$ is only slightly worse than the optimum).