MAT2615 Assignment 3 |COMPLETLY SOLVED| 2026

MAT2615
Assignment 3

2026

, MAT2615 Assignment 3 - 2026

Calculus in Higher Dimensions

University of South Africa (UNISA)



Question 1: Critical Points and Extrema

Given: f(x, y) = x² − 6x + 3y² − y³


(a) Find all critical points [5 marks]

Solution:

Critical points occur where ∇f = (0, 0). Calculate the partial derivatives:

fₓ = 2x − 6

f_y = 6y − 3y²

Set both equal to zero:

2x − 6 = 0 ⟹ x = 3

6y − 3y² = 0 ⟹ 3y(2 − y) = 0 ⟹ y = 0 or y = 2

Answer: The two critical points are (3, 0) and (3, 2)


(b) Classify critical points using Theorem 10.2.9 [5 marks]

Solution:

Calculate the second partial derivatives:

fₓₓ = 2

fₓ_y = 0

, f_y_y = 6 − 6y

The discriminant is D = fₓₓ · f_y_y − (fₓ_y)² = 2(6 − 6y) − 0² = 12 − 12y

At (3, 0):

D = 12 − 12(0) = 12 > 0

fₓₓ = 2 > 0

f(3, 0) = 9 − 18 + 0 − 0 = −9

Conclusion: Since D > 0 and fₓₓ > 0, (3, 0) is a local minimum with value f(3, 0) = −9

At (3, 2):

D = 12 − 12(2) = 12 − 24 = −12 < 0

f(3, 2) = 9 − 18 + 3(4) − 8 = 9 − 18 + 12 − 8 = −5

Conclusion: Since D < 0, (3, 2) is a saddle point (minimax) with value f(3, 2) = −5

Global Extrema Analysis:

As x → ±∞, f → +∞ (due to x² term)

As y → −∞, f → +∞ (due to −y³ term)

As y → +∞, f → −∞ (due to −y³ term)

Conclusion: The local minimum at (3, 0) is NOT a global minimum since f → −∞ as y
→ +∞. There are no global extrema.