MAT2615 Assignment 3 (COMPLETE ANSWERS) 2025 - DUE 2025; 100% TRUSTED Complete, trusted solutions and explanations.

,MAT2615 Assignment 3 (COMPLETE ANSWERS) 2025 -
DUE 2025 WE WISH YOU ALL THE BEST

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]

2. (Section 10.3)
A delivery company needs the measurements of a rectangular
box such that the length plus twice the
width plus the height should not exceed 300cm. Use the method
of Lagrange to find the maximum
volume of such a box?
[10]


𝑵𝑼𝑴𝑩𝑬𝑹 𝟏: 𝑪𝒓𝒊𝒕𝒊𝒄𝒂𝒍 𝑷𝒐𝒊𝒏𝒕𝒔 𝒐𝒇 𝒕𝒉𝒆 𝑭𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝒇(𝒙, 𝒚)
= 𝒚𝒙 − 𝒍𝒏(𝒙) + 𝟏𝟐𝒚𝟐𝒇(𝒙, 𝒚)
= 𝒚𝒙 − \𝒍𝒏(𝒙) + \𝒇𝒓𝒂𝒄{𝟏}{𝟐} 𝒚^𝟐𝒇(𝒙, 𝒚)
= 𝒚𝒙 − 𝒍𝒏(𝒙) + 𝟐𝟏𝒚𝟐
𝑇𝑜 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑝𝑜𝑖𝑛𝑡𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛, 𝑤𝑒 𝑛𝑒𝑒𝑑 𝑡𝑜 𝑐𝑜𝑚𝑝𝑢𝑡𝑒 𝑡ℎ
𝑺𝒕𝒆𝒑 𝟏: 𝑪𝒐𝒎𝒑𝒖𝒕𝒆 𝒕𝒉𝒆 𝒑𝒂𝒓𝒕𝒊𝒂𝒍 𝒅𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒗𝒆𝒔
1. 𝑷𝒂𝒓𝒕𝒊𝒂𝒍 𝒅𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒗𝒆 𝒘𝒊𝒕𝒉 𝒓𝒆𝒔𝒑𝒆𝒄𝒕 𝒕𝒐 𝒙𝒙𝒙:

, 𝜕𝑓𝜕𝑥 = 𝜕𝜕𝑥(𝑦𝑥 − 𝑙𝑛(𝑥) + 12𝑦2)
= 𝑦 − 1𝑥\𝑓𝑟𝑎𝑐{\𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑓}{\𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑥}
= \𝑓𝑟𝑎𝑐{\𝑝𝑎𝑟𝑡𝑖𝑎𝑙}{\𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑥} \𝑙𝑒𝑓𝑡( 𝑦𝑥
− \𝑙𝑛(𝑥) + \𝑓𝑟𝑎𝑐{1}{2} 𝑦^2 \𝑟𝑖𝑔ℎ𝑡)
= 𝑦 − \𝑓𝑟𝑎𝑐{1}{𝑥}𝜕𝑥𝜕𝑓
= 𝜕𝑥𝜕(𝑦𝑥 − 𝑙𝑛(𝑥) + 21𝑦2) = 𝑦 − 𝑥1
2. 𝑷𝒂𝒓𝒕𝒊𝒂𝒍 𝒅𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒗𝒆 𝒘𝒊𝒕𝒉 𝒓𝒆𝒔𝒑𝒆𝒄𝒕 𝒕𝒐 𝒚𝒚𝒚:
𝜕𝑓𝜕𝑦 = 𝜕𝜕𝑦(𝑦𝑥 − 𝑙𝑛(𝑥) + 12𝑦2)
= 𝑥 + 𝑦\𝑓𝑟𝑎𝑐{\𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑓}{\𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑦}
= \𝑓𝑟𝑎𝑐{\𝑝𝑎𝑟𝑡𝑖𝑎𝑙}{\𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑦} \𝑙𝑒𝑓𝑡( 𝑦𝑥
− \𝑙𝑛(𝑥) + \𝑓𝑟𝑎𝑐{1}{2} 𝑦^2 \𝑟𝑖𝑔ℎ𝑡)
= 𝑥 + 𝑦𝜕𝑦𝜕𝑓 = 𝜕𝑦𝜕(𝑦𝑥 − 𝑙𝑛(𝑥) + 21𝑦2) = 𝑥 + 𝑦
𝑺𝒕𝒆𝒑 𝟐: 𝑺𝒆𝒕 𝒕𝒉𝒆 𝒑𝒂𝒓𝒕𝒊𝒂𝒍 𝒅𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒗𝒆𝒔 𝒆𝒒𝒖𝒂𝒍 𝒕𝒐 𝒛𝒆𝒓𝒐
𝑇𝑜 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑝𝑜𝑖𝑛𝑡𝑠, 𝑤𝑒 𝑛𝑒𝑒𝑑 𝑡𝑜 𝑠𝑜𝑙𝑣𝑒 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑜𝑓 𝑒𝑞𝑢𝑎𝑡𝑖
𝜕𝑓𝜕𝑥 = 0𝑎𝑛𝑑𝜕𝑓𝜕𝑦 = 0\𝑓𝑟𝑎𝑐{\𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑓}{\𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑥}
= 0 \𝑞𝑢𝑎𝑑 \𝑡𝑒𝑥𝑡{𝑎𝑛𝑑} \𝑞𝑢𝑎𝑑 \𝑓𝑟𝑎𝑐{\𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑓}{
\𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑦} = 0𝜕𝑥𝜕𝑓 = 0𝑎𝑛𝑑𝜕𝑦𝜕𝑓 = 0
1. 𝐹𝑟𝑜𝑚 𝜕𝑓𝜕𝑥 = 𝑦 − 1𝑥 = 0\𝑓𝑟𝑎𝑐{\𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑓}{\
𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑥} = 𝑦 − \𝑓𝑟𝑎𝑐{1}{𝑥} = 0𝜕𝑥𝜕𝑓 = 𝑦 − 𝑥1 =
0, 𝑤𝑒 𝑔𝑒𝑡:
𝑦 = 1𝑥𝑦 = \𝑓𝑟𝑎𝑐{1}{𝑥}𝑦 = 𝑥1
2. 𝐹𝑟𝑜𝑚 𝜕𝑓𝜕𝑦 = 𝑥 + 𝑦 = 0\𝑓𝑟𝑎𝑐{\𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑓}{\
𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑦} = 𝑥 + 𝑦 = 0𝜕𝑦𝜕𝑓 = 𝑥 + 𝑦 = 0, 𝑤𝑒 𝑔𝑒𝑡:
𝑦 = −𝑥𝑦 = −𝑥𝑦 = −𝑥