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MAT2615 Assignment 2 Semester 2 2020
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- University Of South Africa (Unisa)
This document contains MAT2615 Assignment 2 Semester 2 2020 SOLUTIONS.All workings are shown.
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By: PurpleLogic • 2 year ago
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MAT2615ASSIGNMENT 2 SEMESTER 2 2020
QUESTION A
4 3
𝑓(𝑥, 𝑦) = 𝑥 − 4𝑥𝑦 − 2𝑦 2 + 𝑦 + 1
3
𝑓𝑦 (𝑥, 𝑦) = −4𝑥 − 4𝑦 + 1
𝑓𝑥 (𝑥, 𝑦) = 4𝑥 2 − 4𝑦
𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑝𝑜𝑖𝑛𝑡𝑠 𝑜𝑐𝑐𝑢𝑟𝑠 𝑤ℎ𝑒𝑛: 𝑓𝑥 (𝑥, 𝑦) = 0 𝑎𝑛𝑑 𝑓𝑦 (𝑥, 𝑦) = 0
𝑓𝑦 (𝑥, 𝑦) = 0
−4𝑥 − 4𝑦 + 1 = 0
4𝑥 + 4𝑦 = 1 ∙∙∙∙ 1
𝑓𝑥 (𝑥, 𝑦) = 0
4𝑥 2 − 4𝑦 = 0
𝑥2 − 𝑦 = 0 ∙∙∙∙ 2
𝑓𝑟𝑜𝑚 ∙∙∙∙ 2 𝑦 = 𝑥2
𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑦 𝑖𝑛𝑡𝑜 ∙∙∙∙ 1
4𝑥 + 4𝑦 = 1 ∙∙∙∙ 1
4𝑥 + 4𝑥 2 = 1
4𝑥 2 + 4𝑥 − 1 = 0
𝑈𝑠𝑖𝑛𝑔 𝑄𝑢𝑎𝑑𝑟𝑎𝑡𝑖𝑐 𝑓𝑜𝑟𝑚𝑢𝑙𝑎
, −4 ± √(4)2 − 4(4)(−1)
𝑥=
2(4)
−4 ± √32
𝑥=
8
−4 ± 4√2
𝑥=
8
−1 ± √2
𝑥=
2
−1 + √2 −1 − √2
𝑥1 = , 𝑥2 =
2 2
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑥 𝑖𝑛𝑡𝑜
𝑥2 − 𝑦 = 0 ∙∙∙∙ 2
𝑦 = 𝑥2
𝑦1 = (𝑥1 )2
2
−1 + √2 1
𝑦1 = ( ) = (−1 + √2)(−1 + √2)
2 4
1
𝑦1 = (3 − 2√2)
4
𝑦2 = (𝑥2 )2
2
−1 − √2 1
𝑦2 = ( ) = (−1 − √2)(−1 − √2)
2 4
1
𝑦2 = (3 + 2√2)
4
, −1 + √2 3 − 2√2 −1 − √2 3 + 2√2
𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑝𝑜𝑖𝑛𝑡𝑠: ( , ) 𝑎𝑛𝑑 ( , )
2 4 2 4
𝑓𝑥𝑥 (𝑥, 𝑦) = 8𝑥
𝑓𝑦𝑦 (𝑥, 𝑦) = −4
𝑓𝑥𝑦 (𝑥, 𝑦) = −4
−1 + √2 3 − 2√2
∴ 𝐹𝑜𝑟 𝑝𝑜𝑖𝑛𝑡 ( , )
2 4
2
𝐷 = 𝑓𝑥𝑥 (𝑥0 , 𝑦0 )𝑓𝑦𝑦 (𝑥0 , 𝑦0 ) − [𝑓𝑥𝑦 (𝑥0 , 𝑦0 )]
−1 + √2 3 − 2√2
(𝑥0 , 𝑦0 ) = ( , )
2 4
−1 + √2 3 − 2√2
𝑓𝑥𝑥 (𝑥0 , 𝑦0 ) = 𝑓𝑥𝑥 ( , )
2 4
−1 + √2
𝑓𝑥𝑥 (𝑥0 , 𝑦0 ) = 8 ( )
2
𝑓𝑥𝑥 (𝑥0 , 𝑦0 ) = 4(√2 − 1)
𝐷 = 4(√2 − 1)(−4) − (−4)2
𝐷 = 16(√2 − 1) − 16
𝐷 = 16(√2 − 1 − 1)
𝐷 = 16(√2 − 2)
𝑊𝑒 𝑘𝑛𝑜𝑤 √2 < 2 𝑤ℎ𝑖𝑐ℎ 𝑚𝑒𝑎𝑛𝑠 √2 − 2 < 0
𝐷<0
𝐵𝑦 𝑆𝑒𝑐𝑜𝑛𝑑 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑡𝑒𝑠𝑡
, −𝟏 + √𝟐 𝟑 − 𝟐√𝟐
𝒑𝒐𝒊𝒏𝒕 ( , ) 𝒊𝒔 𝒂 𝒔𝒂𝒅𝒅𝒍𝒆 𝒑𝒐𝒊𝒏𝒕
𝟐 𝟒
−1 − √2 3 + 2√2
∴ 𝐹𝑜𝑟 𝑝𝑜𝑖𝑛𝑡 ( , )
2 4
2
𝐷 = 𝑓𝑥𝑥 (𝑥0 , 𝑦0 )𝑓𝑦𝑦 (𝑥0 , 𝑦0 ) − [𝑓𝑥𝑦 (𝑥0 , 𝑦0 )]
−1 − √2 3 + 2√2
(𝑥0 , 𝑦0 ) = ( , )
2 4
−1 − √2 3 + 2√2
𝑓𝑥𝑥 (𝑥0 , 𝑦0 ) = 𝑓𝑥𝑥 ( , )
2 4
−1 − √2
𝑓𝑥𝑥 (𝑥0 , 𝑦0 ) = 8 ( )
2
𝑓𝑥𝑥 (𝑥0 , 𝑦0 ) = −4(1 + √2)
𝑓𝑦𝑦 (𝑥0 , 𝑦0 ) = −4
𝑓𝑥𝑦 (𝑥0 , 𝑦0 ) = −4
𝐷 = −4(1 + √2)(−4) − (−4)2
𝐷 = 16(1 + √2) − 16
𝐷 = 16(1 + √2 − 1)
𝐷 = 16√2
𝐷>0
𝑓𝑥𝑥 (𝑥0 , 𝑦0 ) = −4(1 + √2) < 0
𝐵𝑦 𝑆𝑒𝑐𝑜𝑛𝑑 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑡𝑒𝑠𝑡
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