MAT2615 Assignment 3 solutions 2026

MAT2615 Assignment 3 Solutions 2026
Full Solutions By TA tutor iQ level
UNISA
DUE: 31-07-2026

,Question 1: Critical Points and Extrema

f(x, y) = x² − 6x + 3y² − y³



Part (a): Finding Critical Points

Critical points occur where both partial derivatives equal zero simultaneously.

Compute partial derivatives:
∂𝑓
𝑓𝑥 = = 2𝑥 − 6
∂𝑥
∂𝑓
𝑓𝑦 = = 6𝑦 − 3𝑦 2
∂𝑦


Set equal to zero:

𝑓𝑥 = 0 ⟹ 2𝑥 − 6 = 0 ⟹ 𝑥 = 3
𝑓𝑦 = 0 ⟹ 6𝑦 − 3𝑦 2 = 0 ⟹ 3𝑦(2 − 𝑦) = 0 ⟹ 𝑦 = 0 or 𝑦 = 2


Critical points: (3, 0) and (3, 2)



Part (b): Classify using Theorem 10.2.9 (Second Derivative Test)

Compute second-order partial derivatives:

𝑓𝑥𝑥 = 2, 𝑓𝑦𝑦 = 6 − 6𝑦, 𝑓𝑥𝑦 = 0


The discriminant (Hessian determinant) is:

𝐻(𝑥, 𝑦) = 𝑓𝑥𝑥 ⋅ 𝑓𝑦𝑦 − (𝑓𝑥𝑦 )2 = (2)(6 − 6𝑦) − 0 = 12 − 12𝑦




At (3, 0):

𝐻(3,0) = 12 − 12(0) = 12 > 0


Since 𝐻 > 0and 𝑓𝑥𝑥 = 2 > 0: (3, 0) is a local minimum.

𝑓(3,0) = 9 − 18 + 0 − 0 = −9

, At (3, 2):

𝐻(3,2) = 12 − 12(2) = 12 − 24 = −12 < 0


Since 𝐻 < 0: (3, 2) is a saddle point (minimax value).

𝑓(3,2) = 9 − 18 + 12 − 8 = −5




Global extrema (by inspection):

As 𝑥 → +∞with 𝑦 = 0: 𝑓(𝑥, 0) = 𝑥 2 − 6𝑥 → +∞

As 𝑦 → +∞with 𝑥 = 3: 𝑓(3, 𝑦) = −9 + 3𝑦 2 − 𝑦 3 → −∞

Therefore:

• The local minimum at (3, 0) with value −9 is NOT a global minimum (f is
unbounded below)

• There is no global maximum (f is unbounded above)