Step-by-Step Deconstruction
Topic: Maxima and Minima Optimization (AOD)
(i) Total Boundary Material Equation
The total material includes 2 lengths of x, 2 outer lengths of y, and 1 internal partition of y.
2x + 3y = 300
(ii) Area as a Function of x
From (i): 3y = 300 – 2x → y = 100 – (2/3)x
Area (A) = x × y
A(x) = 100x – (2/3)x²
(iii) Finding Maximum Area
1. Find Critical Points: Differentiate A(x) with respect to x.
A'(x) = 100 – (4/3)x
Set A'(x) = 0 → 100 = (4/3)x → x = 75 metres
2. Second Derivative Test:
A”(x) = -4/3
Since A”(75) < 0, the Area is Maximum at x = 75.
3. Calculate Maximum Area:
A(75) = 100(75) – (2/3)(75)² = 7500 – 3750 = 3750 m²
⚠️ Silly Mistake Audit: The “Partition” Trap
Most students fail at step (i) by using the standard perimeter formula (2x + 2y). Always read the design details! The parallel partition adds an extra length of ‘y’, changing the constraint to 2x + 3y. A mistake here makes the entire calculus process wrong.