, 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]
𝟏. 𝑭𝒊𝒏𝒅𝒊𝒏𝒈 𝑪𝒓𝒊𝒕𝒊𝒄𝒂𝒍 𝑷𝒐𝒊𝒏𝒕𝒔 𝒐𝒇 𝒇(𝒙, 𝒚)
= 𝒚𝒙 − 𝒍𝒏 𝒙 + 𝟏𝟐𝒚𝟐𝒇(𝒙, 𝒚)
= \𝒇𝒓𝒂𝒄{𝒚}{𝒙} − \𝒍𝒏 𝒙 + \𝒇𝒓𝒂𝒄{𝟏}{𝟐} 𝒚^𝟐𝒇(𝒙, 𝒚)
= 𝒙𝒚 − 𝒍𝒏𝒙 + 𝟐𝟏𝒚𝟐
𝑇𝑜 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑝𝑜𝑖𝑛𝑡𝑠, 𝑤𝑒 𝑐𝑜𝑚𝑝𝑢𝑡𝑒 𝑡ℎ𝑒 𝒇𝒊𝒓𝒔𝒕
− 𝒐𝒓𝒅𝒆𝒓 𝒑𝒂𝒓𝒕𝒊𝒂𝒍 𝒅𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒗𝒆𝒔 𝑎𝑛𝑑 𝑠𝑒𝑡 𝑡ℎ𝑒𝑚 𝑡𝑜 𝑧𝑒𝑟𝑜:
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) = −𝑦𝑥2 − 1𝑥
= −\𝑓𝑟𝑎𝑐{𝑦}{𝑥^2} − \𝑓𝑟𝑎𝑐{1}{𝑥} = −𝑥2𝑦 − 𝑥1
𝑆𝑒𝑡𝑡𝑖𝑛𝑔 𝑓𝑥 = 0𝑓_𝑥 = 0𝑓𝑥 = 0:
−𝑦𝑥2 − 1𝑥 = 0 −\𝑓𝑟𝑎𝑐{𝑦}{𝑥^2} − \𝑓𝑟𝑎𝑐{1}{𝑥} = 0 − 𝑥2𝑦 − 𝑥1
=0
𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑦𝑖𝑛𝑔 𝑏𝑦 𝑥2𝑥^2𝑥2:
−𝑦 − 𝑥 = 0 ⇒ 𝑦 = −𝑥 − 𝑦 − 𝑥 = 0 \𝑞𝑢𝑎𝑑 \𝑅𝑖𝑔ℎ𝑡𝑎𝑟𝑟𝑜𝑤 \𝑞𝑢𝑎𝑑 𝑦
= −𝑥 − 𝑦 − 𝑥 = 0 ⇒ 𝑦 = −𝑥
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]
𝟏. 𝑭𝒊𝒏𝒅𝒊𝒏𝒈 𝑪𝒓𝒊𝒕𝒊𝒄𝒂𝒍 𝑷𝒐𝒊𝒏𝒕𝒔 𝒐𝒇 𝒇(𝒙, 𝒚)
= 𝒚𝒙 − 𝒍𝒏 𝒙 + 𝟏𝟐𝒚𝟐𝒇(𝒙, 𝒚)
= \𝒇𝒓𝒂𝒄{𝒚}{𝒙} − \𝒍𝒏 𝒙 + \𝒇𝒓𝒂𝒄{𝟏}{𝟐} 𝒚^𝟐𝒇(𝒙, 𝒚)
= 𝒙𝒚 − 𝒍𝒏𝒙 + 𝟐𝟏𝒚𝟐
𝑇𝑜 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑝𝑜𝑖𝑛𝑡𝑠, 𝑤𝑒 𝑐𝑜𝑚𝑝𝑢𝑡𝑒 𝑡ℎ𝑒 𝒇𝒊𝒓𝒔𝒕
− 𝒐𝒓𝒅𝒆𝒓 𝒑𝒂𝒓𝒕𝒊𝒂𝒍 𝒅𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒗𝒆𝒔 𝑎𝑛𝑑 𝑠𝑒𝑡 𝑡ℎ𝑒𝑚 𝑡𝑜 𝑧𝑒𝑟𝑜:
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) = −𝑦𝑥2 − 1𝑥
= −\𝑓𝑟𝑎𝑐{𝑦}{𝑥^2} − \𝑓𝑟𝑎𝑐{1}{𝑥} = −𝑥2𝑦 − 𝑥1
𝑆𝑒𝑡𝑡𝑖𝑛𝑔 𝑓𝑥 = 0𝑓_𝑥 = 0𝑓𝑥 = 0:
−𝑦𝑥2 − 1𝑥 = 0 −\𝑓𝑟𝑎𝑐{𝑦}{𝑥^2} − \𝑓𝑟𝑎𝑐{1}{𝑥} = 0 − 𝑥2𝑦 − 𝑥1
=0
𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑦𝑖𝑛𝑔 𝑏𝑦 𝑥2𝑥^2𝑥2:
−𝑦 − 𝑥 = 0 ⇒ 𝑦 = −𝑥 − 𝑦 − 𝑥 = 0 \𝑞𝑢𝑎𝑑 \𝑅𝑖𝑔ℎ𝑡𝑎𝑟𝑟𝑜𝑤 \𝑞𝑢𝑎𝑑 𝑦
= −𝑥 − 𝑦 − 𝑥 = 0 ⇒ 𝑦 = −𝑥
