How do I find the dimensions that give the maximum area (or minimum material) — a real optimisation problem?
Turn the requirement into a single-variable function using the constraint.
Turn the requirement into a single-variable function using the constraint. Say you fence a rectangle with a fixed perimeter: express the area A in terms of one side x, then differentiate and set dA/dx = 0 to find the critical width. Check it's a maximum with the second derivative (A'' < 0). For a fixed perimeter this gives a square. The recipe — build the function, set the derivative to zero, test — solves every 'largest/smallest' problem.
Key point
Model the quantity as f(x), solve f'(x)=0, then confirm max/min with f''.
Common mistake
guessing the optimum instead of setting the derivative to zero.
Memory tip
optimisation = 'derivative zero' + a second-derivative check.
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