MAT2615 Assignment 3 (COMPLETE ANSWERS) 2025 - DUE 2025; 100% correct solutions and explanations.

,MAT2615 Assignment 3 (COMPLETE ANSWERS) 2025 - DUE 2025; 100%
TRUSTED Complete, trusted solutions and explanations.
1. (Section 10.2)
Consider the R2 � R function defined by
f(x; y) = y
x
� ln x +
1
2y2:
Find the critical points of f and their nature.
[10]

𝑇𝑜 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑝𝑜𝑖𝑛𝑡𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛:

𝑓(𝑥, 𝑦) = 𝑦𝑥 − 𝑙𝑛 𝑥 + 12𝑦2𝑓(𝑥, 𝑦)
= \𝑓𝑟𝑎𝑐{𝑦}{𝑥} − \𝑙𝑛 𝑥 + \𝑓𝑟𝑎𝑐{1}{2} 𝑦^2𝑓(𝑥, 𝑦)
= 𝑥𝑦 − 𝑙𝑛𝑥 + 21𝑦2

𝑤𝑒 𝑓𝑜𝑙𝑙𝑜𝑤 𝑡ℎ𝑒𝑠𝑒 𝑠𝑡𝑒𝑝𝑠:

𝑺𝒕𝒆𝒑 𝟏: 𝑪𝒐𝒎𝒑𝒖𝒕𝒆 𝒕𝒉𝒆 𝑭𝒊𝒓𝒔𝒕 − 𝑶𝒓𝒅𝒆𝒓 𝑷𝒂𝒓𝒕𝒊𝒂𝒍 𝑫𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒗𝒆𝒔

𝑇ℎ𝑒 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑝𝑜𝑖𝑛𝑡𝑠 𝑜𝑐𝑐𝑢𝑟 𝑤ℎ𝑒𝑟𝑒 𝑏𝑜𝑡ℎ 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒𝑠 𝑓𝑥𝑓_𝑥𝑓𝑥
𝑎𝑛𝑑 𝑓𝑦𝑓_𝑦𝑓𝑦 𝑎𝑟𝑒 𝑧𝑒𝑟𝑜.

𝑷𝒂𝒓𝒕𝒊𝒂𝒍 𝒅𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒗𝒆 𝒘𝒊𝒕𝒉 𝒓𝒆𝒔𝒑𝒆𝒄𝒕 𝒕𝒐 𝒙𝒙𝒙:

𝑓𝑥 = 𝜕𝜕𝑥(𝑦𝑥 − 𝑙𝑛 𝑥 + 12𝑦2)𝑓_𝑥
= \𝑓𝑟𝑎𝑐{\𝑝𝑎𝑟𝑡𝑖𝑎𝑙}{\𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑥} \𝑙𝑒𝑓𝑡( \𝑓𝑟𝑎𝑐{𝑦}{𝑥} − \𝑙𝑛 𝑥
+ \𝑓𝑟𝑎𝑐{1}{2} 𝑦^2 \𝑟𝑖𝑔ℎ𝑡)𝑓𝑥 = 𝜕𝑥𝜕(𝑥𝑦 − 𝑙𝑛𝑥 + 21𝑦2)

𝑈𝑠𝑖𝑛𝑔 𝑡ℎ𝑒 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑟𝑢𝑙𝑒𝑠:

𝑓𝑥 = −𝑦𝑥2 − 1𝑥𝑓_𝑥 = −\𝑓𝑟𝑎𝑐{𝑦}{𝑥^2} − \𝑓𝑟𝑎𝑐{1}{𝑥}𝑓𝑥 = −𝑥2𝑦 − 𝑥1

𝑷𝒂𝒓𝒕𝒊𝒂𝒍 𝒅𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒗𝒆 𝒘𝒊𝒕𝒉 𝒓𝒆𝒔𝒑𝒆𝒄𝒕 𝒕𝒐 𝒚𝒚𝒚:

𝑓𝑦 = 𝜕𝜕𝑦(𝑦𝑥 − 𝑙𝑛 𝑥 + 12𝑦2)𝑓_𝑦
= \𝑓𝑟𝑎𝑐{\𝑝𝑎𝑟𝑡𝑖𝑎𝑙}{\𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑦} \𝑙𝑒𝑓𝑡( \𝑓𝑟𝑎𝑐{𝑦}{𝑥} − \𝑙𝑛 𝑥
+ \𝑓𝑟𝑎𝑐{1}{2} 𝑦^2 \𝑟𝑖𝑔ℎ𝑡)𝑓𝑦 = 𝜕𝑦𝜕(𝑥𝑦 − 𝑙𝑛𝑥 + 21𝑦2) 𝑓𝑦
= 1𝑥 + 𝑦𝑓_𝑦 = \𝑓𝑟𝑎𝑐{1}{𝑥} + 𝑦𝑓𝑦 = 𝑥1 + 𝑦

, 𝑺𝒕𝒆𝒑 𝟐: 𝑺𝒐𝒍𝒗𝒆 𝒇𝒐𝒓 𝑪𝒓𝒊𝒕𝒊𝒄𝒂𝒍 𝑷𝒐𝒊𝒏𝒕𝒔

𝑆𝑒𝑡𝑡𝑖𝑛𝑔 𝑓𝑥 = 0𝑓_𝑥 = 0𝑓𝑥 = 0:

−𝑦𝑥2 − 1𝑥 = 0 −\𝑓𝑟𝑎𝑐{𝑦}{𝑥^2} − \𝑓𝑟𝑎𝑐{1}{𝑥} = 0 − 𝑥2𝑦 − 𝑥1 = 0 − 𝑦𝑥2
= 1𝑥 −\𝑓𝑟𝑎𝑐{𝑦}{𝑥^2} = \𝑓𝑟𝑎𝑐{1}{𝑥} − 𝑥2𝑦 = 𝑥1 𝑦 = −𝑥𝑦
= −𝑥𝑦 = −𝑥

𝑆𝑒𝑡𝑡𝑖𝑛𝑔 𝑓𝑦 = 0𝑓_𝑦 = 0𝑓𝑦 = 0:

1𝑥 + 𝑦 = 0\𝑓𝑟𝑎𝑐{1}{𝑥} + 𝑦 = 0𝑥1 + 𝑦 = 0 𝑦 = −1𝑥𝑦 = −\𝑓𝑟𝑎𝑐{1}{𝑥}𝑦
= −𝑥1

𝑺𝒕𝒆𝒑 𝟑: 𝑭𝒊𝒏𝒅 𝑰𝒏𝒕𝒆𝒓𝒔𝒆𝒄𝒕𝒊𝒐𝒏

𝐸𝑞𝑢𝑎𝑡𝑖𝑛𝑔 𝑦 = −𝑥𝑦 = −𝑥𝑦 = −𝑥 𝑎𝑛𝑑 𝑦 = −1𝑥𝑦 = −\𝑓𝑟𝑎𝑐{1}{𝑥}𝑦 = −𝑥1:

−𝑥 = −1𝑥 − 𝑥 = −\𝑓𝑟𝑎𝑐{1}{𝑥} − 𝑥 = −𝑥1 𝑥 = 1𝑥𝑥 = \𝑓𝑟𝑎𝑐{1}{𝑥}𝑥
= 𝑥1 𝑥2 = 1𝑥^2 = 1𝑥2 = 1 𝑥 = ±1𝑥 = \𝑝𝑚1𝑥 = ±1

𝐹𝑜𝑟 𝑥 = 1𝑥 = 1𝑥 = 1:

𝑦 = −1𝑦 = −1𝑦 = −1

𝐹𝑜𝑟 𝑥 = −1𝑥 = −1𝑥 = −1:

𝑦 = 1𝑦 = 1𝑦 = 1

𝑇ℎ𝑢𝑠, 𝑡ℎ𝑒 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑝𝑜𝑖𝑛𝑡𝑠 𝑎𝑟𝑒 (1, −1)(1, −1)(1, −1) 𝑎𝑛𝑑 (−1,1)(−1,1)(−1,1).

𝑺𝒕𝒆𝒑 𝟒: 𝑪𝒐𝒎𝒑𝒖𝒕𝒆 𝒕𝒉𝒆 𝑺𝒆𝒄𝒐𝒏𝒅 − 𝑶𝒓𝒅𝒆𝒓 𝑷𝒂𝒓𝒕𝒊𝒂𝒍 𝑫𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒗𝒆𝒔

𝑺𝒆𝒄𝒐𝒏𝒅 𝒅𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒗𝒆𝒔: