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.
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.
