Example — reviewing the rectangle problem

Example — Reviewing the rectangle problem

Check: sides $4$ and $8$. Perimeter $= 2\cdot(4+8) = 24$ ✓. Area $= 4\cdot 8 = 32$ ✓.

Reflection: the crucial point was turning the two conditions (perimeter, area) into a system of two equations. This transformation is general: any problem with “given perimeter and area” reduces to a system of this kind.

Extension: what happens if the area is $40$? Let us try: $x^2-12x+40=0$, discriminant $144-160 < 0$. No real solution! Geometrically: there is no rectangle with perimeter $24$ and area $40$. For which values of $A$ does one exist? The equation $x^2-12x+A=0$ has real solutions when $\Delta = 144-4A \ge 0$, that is $A\le 36$. Interpretation: with perimeter $24$, the greatest possible area is $36$ (the case of the square with side $6$). A fine result “handed to us” by reflection!

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Topics: Working method
Concepts: Checking the result
Skills: Limiting-case analysis · Modelling