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64
A square 8×8 maximises this model.
Class 12 · English Version · Mathematics
Solve optimisation and area problems with boundaries checked.
Original Kidemy Class 12 progression. This level has its own text, examples and questions. Curriculum versions share the progression; complete board-specific coverage still needs specification matching and additional material.
Stationary points are candidates, not automatic maxima. A constrained problem needs its domain, stationary values and relevant boundaries checked. For area between curves, integrate upper minus lower on intervals where their order is known.
For a rectangle with perimeter 24, A=x(12−x)=36−(x−6)²,0<x<12. Maximum area 36 occurs at x=6.
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64
A square 8×8 maximises this model.
26
[x³]1 to 3=27−1.
0.5
d/dx ln x=1/x for x>0.
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∫(x−x²)dx=[x²/2−x³/3]0 to 1=1/6.
Identify upper and lower curves.
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The largest or smallest value can occur at an endpoint even without zero derivative there.
Explain candidate comparison.
Write a constrained optimisation solution including domain and boundary reasoning.
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