Optimisation uses a fixed condition to express an objective quantity in one variable, then uses diff

Question

Optimisation uses a fixed condition to express an objective quantity in one variable, then uses differentiation to locate and classify a stationary value.

Method

Define the variables clearly.
Identify the fixed condition, such as a fixed volume, perimeter or length.
Form an equation linking the variables using the fixed condition.
Write the objective $V$ in one variable, for example $V=V(x)$.
Solve $\dfrac{\mathrm{d}V}{\mathrm{d}x}=0$, then classify each admissible stationary value. In particular, $\dfrac{\mathrm{d}^2V}{\mathrm{d}x^2}<0$ gives a local maximum and $\dfrac{\mathrm{d}^2V}{\mathrm{d}x^2}>0$ gives a local minimum.
Evaluate the quantity requested, respecting the physical domain and any endpoints.

Solid	Volume	Curved surface area
Cylinder
Cone
Sphere

Compare admissible endpoints and relevant limiting values for a global optimum. Discard infeasible values; an excluded endpoint may give a limit that is never attained. State the requested quantity with units, and add bases when a total surface area is required.

Accepted Answer

Express the quantity to be optimized in one variable, solve $V'(x)=0$, classify the stationary value, and state the required optimum with units.

Optimisation

Optimisation

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