Mathematics Department
Dr. Mohammad Yasin
Worksheet (4) Solution
Applications of Partial Derivatives
𝟏 Find a first and second-degree polynomial approximation to the following functions near the
given point:
𝑥𝑦 sin 𝑥 𝜋
𝐚. 𝑓(𝑥, 𝑦) = 𝑒 , (1, 2) 𝐛. 𝑓(𝑥, 𝑦) = , ( , 1)
𝑦 2
𝐜. 𝑓(𝑥, 𝑦) = 𝑥 3 + 2𝑦 3 − 𝑥𝑦 , (3,2) 𝐝. 𝑓(𝑥, 𝑦) = ln(𝑥 2 + 𝑦 2 ) , (1,0)
𝜋 𝜋
𝐞. 𝑓(𝑥, 𝑦) = sin 𝑥 sin 𝑦 , ( , )
4 4
Solution:
𝒙𝒚
𝐚. 𝒇(𝒙, 𝒚) = 𝒆 , (1, 2)
First order Taylor series (Linear Approximation) of the function 𝑓(𝑥, 𝑦) is given by
𝑓(𝑥, 𝑦) ≅ 𝑓(𝑎, 𝑏) + [(𝑥 − 𝑎) 𝑓𝑥 (𝑎, 𝑏) + (𝑦 − 𝑏) 𝑓𝑦 (𝑎, 𝑏)]
𝑓(𝑥, 𝑦) = 𝑒 𝑥𝑦 ⇒ 𝑓(1, 2) = 𝑒 2
𝑥𝑦
𝑓𝑥 (𝑥, 𝑦) = 𝑦 𝑒 ⇒ 𝑓𝑥 (1, 2) = 2 𝑒 2
𝑥𝑦
𝑓𝑦 (𝑥, 𝑦) = 𝑥 𝑒 ⇒ 𝑓𝑦 (1, 2) = 𝑒 2
Then the linear approximation is
𝑓(𝑥, 𝑦) ≅ 𝑒 2 + [2 𝑒 2 (𝑥 − 1) + 𝑒 2 (𝑦 − 2)] = 𝑳(𝒙, 𝒚)
Page 1 of 20
Second order Taylor series (Quadratic Approximation) of the function 𝑓(𝑥, 𝑦) is given by
𝑓(𝑥, 𝑦) ≅ 𝑓(𝑎, 𝑏) + [(𝑥 − 𝑎) 𝑓𝑥 (𝑎, 𝑏) + (𝑦 − 𝑏) 𝑓𝑦 (𝑎, 𝑏)]
1
+ [(𝑥 − 𝑎)2 𝑓𝑥𝑥 (𝑎, 𝑏) + 2 (𝑥 − 𝑎) (𝑦 − 𝑏) 𝑓𝑥𝑦 (𝑎, 𝑏) + (𝑦 − 𝑏)2 𝑓𝑦𝑦 (𝑎, 𝑏)]
2!
𝑥𝑦
𝑓𝑥𝑥 (𝑥, 𝑦) = 𝑦 2 𝑒 ⇒ 𝑓𝑥𝑥 (1, 2) = 4 𝑒 2
𝑥𝑦 𝑥𝑦
𝑓𝑥𝑦 (𝑥, 𝑦) = 𝑥𝑦 𝑒 +𝑒 ⇒ 𝑓𝑥𝑦 (1, 2) = 3 𝑒 2
𝑥𝑦
𝑓𝑦𝑦 (𝑥, 𝑦) = 𝑥 2 𝑒 ⇒ 𝑓𝑦𝑦 (1, 2) = 𝑒 2
Then the quadratic approximation is
1
𝑓(𝑥, 𝑦) ≅ 𝑳(𝒙, 𝒚) + [4 𝑒 2 (𝑥 − 1)2 + 6 𝑒 2 (𝑥 − 1) (𝑦 − 2) + 𝑒 2 (𝑦 − 2)2 ]
2!
≅ 𝑒 2 + [2𝑒 2 (𝑥 − 1) + 𝑒 2 (𝑦 − 2)]
1
+ [4 𝑒 2 (𝑥 − 1)2 + 6 𝑒 2 (𝑥 − 1) (𝑦 − 2) + 𝑒 2 (𝑦 − 2)2 ] = 𝑸(𝒙, 𝒚)
2!
𝐬𝐢𝐧 𝒙 𝜋
𝐛. 𝒇(𝒙, 𝒚) = , ( , 1)
𝒚 2
sin 𝑥 𝜋
𝑓(𝑥, 𝑦) = ⇒ 𝑓 ( , 1) = 1
𝑦 2
cos 𝑥 𝜋
𝑓𝑥 (𝑥, 𝑦) = ⇒ 𝑓𝑥 ( , 1) = 0
𝑦 2
− sin 𝑥 𝜋
𝑓𝑦 (𝑥, 𝑦) = ⇒ 𝑓𝑦 ( , 1) = −1
𝑦2 2
Then the linear approximation is
𝜋
𝑓(𝑥, 𝑦) ≅ 1 + [0. (𝑥 − ) − (𝑦 − 1)] = 2 − 𝑦 = 𝑳(𝒙, 𝒚)
2
Page 2 of 20
− sin 𝑥 𝜋
𝑓𝑥𝑥 (𝑥, 𝑦) = ⇒ 𝑓𝑥𝑥 ( , 1) = −1
𝑦 2
− cos 𝑥 𝜋
𝑓𝑥𝑦 (𝑥, 𝑦) = ⇒ 𝑓𝑥𝑦 ( , 1) = 0
𝑦2 2
2 sin 𝑥 𝜋
𝑓𝑦𝑦 (𝑥, 𝑦) = ⇒ 𝑓𝑦𝑦 ( , 1) = 2
𝑦3 2
Then the quadratic approximation is
1 𝜋 2 𝜋
𝑓(𝑥, 𝑦) ≅ 𝑳(𝒙, 𝒚) + [− (𝑥 − ) + 0. (𝑥 − ) (𝑦 − 1) + 2(𝑦 − 1)2 ]
2! 2 2
1 𝜋 2
≅ 2 − 𝑦 − (𝑥 − ) + (𝑦 − 1)2 = 𝑸(𝒙, 𝒚)
2 2
𝐜. 𝒇(𝒙, 𝒚) = 𝒙𝟑 + 𝟐𝒚𝟑 − 𝒙𝒚 , (3,2)
𝑓(𝑥, 𝑦) = 𝑥 3 + 2𝑦 3 − 𝑥𝑦 ⇒ 𝑓(3, 2) = 37
𝑓𝑥 (𝑥, 𝑦) = 3𝑥 2 − 𝑦 ⇒ 𝑓𝑥 (3, 2) = 25
𝑓𝑦 (𝑥, 𝑦) = 6𝑦 2 − 𝑥 ⇒ 𝑓𝑦 (3, 2) = 21
Then the linear approximation is
𝑓(𝑥, 𝑦) ≅ 37 + [25 (𝑥 − 3) + 21 (𝑦 − 2)] = 25𝑥 + 21𝑦 − 80 = 𝑳(𝒙, 𝒚)
𝑓𝑥𝑥 (𝑥, 𝑦) = 6𝑥 ⇒ 𝑓𝑥𝑥 (3, 2) = 18
𝑓𝑥𝑦 (𝑥, 𝑦) = −1 ⇒ 𝑓𝑥𝑦 (3, 2) = −1
𝑓𝑦𝑦 (𝑥, 𝑦) = 12𝑦 ⇒ 𝑓𝑦𝑦 (3, 2) = 24
Then the quadratic approximation is
1
𝑓(𝑥, 𝑦) ≅ 𝑳(𝒙, 𝒚) + [18 (𝑥 − 3)2 − 2(𝑥 − 3) (𝑦 − 2) + 24 (𝑦 − 2)2 ]
2!
≅ 25𝑥 + 21𝑦 − 80 + 9 (𝑥 − 3)2 − (𝑥 − 3) (𝑦 − 2) + 12 (𝑦 − 2)2 = 𝑸(𝒙, 𝒚)
Page 3 of 20
𝐝. 𝒇(𝒙, 𝒚) = 𝐥𝐧(𝒙𝟐 + 𝒚𝟐 ) , (1,0)
𝑓(𝑥, 𝑦) = ln(𝑥 2 + 𝑦 2 ) ⇒ 𝑓(1,0) = 0
2𝑥
𝑓𝑥 (𝑥, 𝑦) = ⇒ 𝑓𝑥 (1,0) = 2
𝑥2 + 𝑦2
2𝑦
𝑓𝑦 (𝑥, 𝑦) = ⇒ 𝑓𝑦 (1,0) = 0
𝑥2 + 𝑦2
Then the linear approximation is
𝑓(𝑥, 𝑦) ≅ 0 + [2 (𝑥 − 1) + 0. (𝑦 − 0)] = 2𝑥 − 2 = 𝑳(𝒙, 𝒚)
2𝑦 2 − 2𝑥 2
𝑓𝑥𝑥 (𝑥, 𝑦) = 2 ⇒ 𝑓𝑥𝑥 (1, 0) = −2
(𝑥 + 𝑦 2 )2
−4𝑥𝑦
𝑓𝑥𝑦 (𝑥, 𝑦) = ⇒ 𝑓𝑥𝑦 (1, 0) = 0
(𝑥 2 + 𝑦 2 )2
2𝑥 2 − 2𝑦 2
𝑓𝑦𝑦 (𝑥, 𝑦) = 2 ⇒ 𝑓𝑦𝑦 (1, 0) = 2
(𝑥 + 𝑦 2 )2
Then the quadratic approximation is
1
𝑓(𝑥, 𝑦) ≅ 𝑳(𝒙, 𝒚) + [−2 (𝑥 − 1)2 + 0. (𝑥 − 1) (𝑦 − 0) + 2 (𝑦 − 0)2 ]
2!
≅ 2𝑥 − 2 − (𝑥 − 1)2 + 𝑦 2 = 𝑸(𝒙, 𝒚)
𝜋 𝜋
𝐞. 𝒇(𝒙, 𝒚) = 𝐬𝐢𝐧 𝒙 𝐬𝐢𝐧 𝒚 , ( , )
4 4
𝜋 𝜋 1
𝑓(𝑥, 𝑦) = sin 𝑥 sin 𝑦 ⇒ 𝑓( , ) =
4 4 2
𝜋 𝜋 1
𝑓𝑥 (𝑥, 𝑦) = cos 𝑥 sin 𝑦 ⇒ 𝑓𝑥 ( , ) =
4 4 2
𝜋 𝜋 1
𝑓𝑦 (𝑥, 𝑦) = sin 𝑥 cos 𝑦 ⇒ 𝑓𝑦 ( , ) =
4 4 2
Page 4 of 20
Then the linear approximation is
1 1 𝜋 1 𝜋 1 𝜋
𝑓(𝑥, 𝑦) ≅ + [ (𝑥 − ) + (𝑦 − )] = (𝑥 + 𝑦 − + 1) = 𝑳(𝒙, 𝒚)
2 2 4 2 4 2 2
𝜋 𝜋 −1
𝑓𝑥𝑥 (𝑥, 𝑦) = − sin 𝑥 sin 𝑦 ⇒ 𝑓𝑥𝑥 ( , ) =
4 4 2
𝜋 𝜋 1
𝑓𝑥𝑦 (𝑥, 𝑦) = cos 𝑥 cos 𝑦 ⇒ 𝑓𝑥𝑦 ( , ) =
4 4 2
𝜋 𝜋 −1
𝑓𝑦𝑦 (𝑥, 𝑦) = − sin 𝑥 sin 𝑦 ⇒ 𝑓𝑦𝑦 ( , ) =
4 4 2
Then the quadratic approximation is
1 −1 𝜋 2 𝜋 𝜋 1 𝜋 2
𝑓(𝑥, 𝑦) ≅ 𝑳(𝒙, 𝒚) + [ (𝑥 − ) + (𝑥 − ) (𝑦 − ) − (𝑦 − ) ]
2! 2 4 4 4 2 4
1 𝜋 1 𝜋 2 1 𝜋 𝜋 1 𝜋 2
≅ (𝑥 + 𝑦 − + 1) − (𝑥 − ) + (𝑥 − ) (𝑦 − ) − (𝑦 − ) = 𝑸(𝒙, 𝒚)
2 2 4 4 2 4 4 4 4
Page 5 of 20
𝟐 Find the equation of the tangent plane to the following surfaces at the specified point:
𝐚. 𝑧 = 4𝑥 2 − 𝑦 2 + 2𝑦 , (−1, 2,4) 𝐛. 𝑧 = √4 − 𝑥 2 − 2𝑦 2 , (1, −1,1)
𝐜. 𝑧 = 𝑦 sin(𝑥 − 𝑦) , (2,2,0) 𝐝. 𝑥 3 + 𝑦 3 + 𝑧 3 = 6 − 𝑥𝑦𝑧 , (1,2, −1)
𝑥𝑦𝑧
𝐞. 𝑥 + 𝑦 + 𝑧 = 𝑒 , (0,0,1) 𝐟. 𝑥𝑦 + 𝑦𝑧 + 𝑧𝑥 = 5 , (1,2,1)
Solution:
𝐚. 𝑧 = 4𝑥 2 − 𝑦 2 + 2𝑦 , (−1, 2,4)
Equation of the tangent plane to the surface 𝒛 = 𝒇(𝒙, 𝒚) at point 𝑃(𝑎, 𝑏, 𝑐) is
𝑧 − 𝑐 = 𝑓𝑥 |(𝑎,𝑏) (𝑥 − 𝑎) + 𝑓𝑦 |(𝑎,𝑏) (𝑦 − 𝑏)
𝑓(𝑥, 𝑦) = 4𝑥 2 − 𝑦 2 + 2𝑦
𝑓𝑥 (𝑥, 𝑦) = 8𝑥 ⇒ 𝑓𝑥 (−1, 2) = −8
𝑓𝑦 (𝑥, 𝑦) = −2𝑦 + 2 ⇒ 𝑓𝑦 (−1, 2) = −2
Hence, the equation of the tangent plane is
𝑧 − 4 = −8 (𝑥 + 1) − 2 (𝑦 − 2) or 8𝑥 + 2𝑦 + 𝑧 = 0
𝐛. 𝑧 = √4 − 𝑥 2 − 2𝑦 2 , (1, −1,1)
𝑓(𝑥, 𝑦) = √4 − 𝑥 2 − 2𝑦 2
−𝑥
𝑓𝑥 (𝑥, 𝑦) = ⇒ 𝑓𝑥 (1, −1) = −1
√4 − 𝑥 2 − 2𝑦 2
−2𝑦
𝑓𝑦 (𝑥, 𝑦) = ⇒ 𝑓𝑦 (1, −1) = 2
√4 − 𝑥 2 − 2𝑦 2
Hence, the equation of the tangent plane is
𝑧 − 1 = − (𝑥 − 1) + 2 (𝑦 + 1) or 𝑥 − 2𝑦 + 𝑧 − 4 = 0
Page 6 of 20
𝐜. 𝑧 = 𝑦 sin(𝑥 − 𝑦) , (2,2,0)
𝑓(𝑥, 𝑦) = 𝑦 sin(𝑥 − 𝑦)
𝑓𝑥 (𝑥, 𝑦) = 𝑦 cos(𝑥 − 𝑦) ⇒ 𝑓𝑥 (2, 2) = 2
𝑓𝑦 (𝑥, 𝑦) = −𝑦 cos(𝑥 − 𝑦) + sin(𝑥 − 𝑦) ⇒ 𝑓𝑦 (2, 2) = −2
Hence, the equation of the tangent plane is
𝑧 − 0 = 2 (𝑥 − 2) − 2 (𝑦 − 2) or 2𝑥 − 2𝑦 − 𝑧 = 0
𝐝. 𝑥 3 + 𝑦 3 + 𝑧 3 = 6 − 𝑥𝑦𝑧 , (1,2, −1)
Equation of the tangent plane to the surface 𝑭(𝒙, 𝒚, 𝒛) = 𝟎 at point 𝑃(𝑎, 𝑏, 𝑐) is
𝐹𝑥 |(𝑎,𝑏,𝑐) (𝑥 − 𝑎) + 𝐹𝑦 |(𝑎,𝑏,𝑐) (𝑦 − 𝑏) + 𝐹𝑧 |(𝑎,𝑏,𝑐) (𝑧 − 𝑐) = 0
𝐹(𝑥, 𝑦, 𝑧) = 𝑥 3 + 𝑦 3 + 𝑧 3 + 𝑥𝑦𝑧 − 6 = 0
𝐹𝑥 = 3𝑥 2 + 𝑦𝑧, 𝐹𝑦 = 3𝑦 2 + 𝑥𝑧 and 𝐹𝑧 = 3𝑧 2 + 𝑥𝑦
∴ 𝐹𝑥 |(1,2,−1) = 1, 𝐹𝑦 |(1,2,−1) = 11 and 𝐹𝑧 |(1,2,3) = 5
Hence, the equation of the tangent plane is
(𝑥 − 1) + 11 (𝑦 − 2) + 5 (𝑧 + 1) = 0
𝑥𝑦𝑧
𝐞. 𝑥 + 𝑦 + 𝑧 = 𝑒 , (0,0,1)
𝑥𝑦𝑧
𝐹(𝑥, 𝑦, 𝑧) = 𝑥 + 𝑦 + 𝑧 − 𝑒 =0
𝑥𝑦𝑧 𝑥𝑦𝑧 𝑥𝑦𝑧
𝐹𝑥 = 1 + 𝑦𝑧 𝑒 , 𝐹𝑦 = 1 + 𝑥𝑧 𝑒 , and 𝐹𝑧 = 1 + 𝑥𝑦 𝑒 ,
∴ 𝐹𝑥 |(0,0,1) = 1, 𝐹𝑦 |(0,0,1) = 1 and 𝐹𝑧 |(0,0,1) = 1
Hence, the equation of the tangent plane is
𝑥+𝑦+𝑧−1=0
Page 7 of 20
𝐟. 𝑥𝑦 + 𝑦𝑧 + 𝑧𝑥 = 5 , (1,2,1)
𝐹(𝑥, 𝑦, 𝑧) = 𝑥𝑦 + 𝑦𝑧 + 𝑧𝑥 − 5 = 0
𝐹𝑥 = 𝑦 + 𝑧, 𝐹𝑦 = 𝑥 + 𝑧, and 𝐹𝑧 = 𝑦 + 𝑥,
∴ 𝐹𝑥 |(1,2,1) = 3, 𝐹𝑦 |(1,2,1) = 2 and 𝐹𝑧 |(1,2,1) = 3
Hence, the equation of the tangent plane is
3 (𝑥 − 1) + 2 (𝑦 − 2) + 3 (𝑧 − 1) = 0
𝟑 Find the total differentials of the following functions:
𝑟 𝑥 2 +𝑦 2
𝐚. 𝑢 = 𝐛. 𝑧 = 𝑒
𝑠 + 2𝑡
𝐜. 𝑢 = tan(3𝑥 − 𝑦) + 6
𝑦+𝑧 𝑢𝑣
𝐝. 𝑧 = ln ( )
𝑠𝑡
Solution:
𝑟
𝐚. 𝑢 =
𝑠 + 2𝑡
𝜕𝑢 𝜕𝑢 𝜕𝑢
𝑑𝑢 = 𝑑𝑟 + 𝑑𝑠 + 𝑑𝑡
𝜕𝑟 𝜕𝑠 𝜕𝑡
1 𝑟 2𝑟
𝑑𝑢 = 𝑑𝑟 − 𝑑𝑠 − 𝑑𝑡
𝑠 + 2𝑡 (𝑠 + 2𝑡)2 (𝑠 + 2𝑡)2
𝑥 2 +𝑦 2
𝐛. 𝑧 = 𝑒
𝜕𝑧 𝜕𝑧
𝑑𝑧 = 𝑑𝑥 + 𝑑𝑦
𝜕𝑥 𝜕𝑦
𝑥 2 +𝑦 2 𝑥 2 +𝑦 2
𝑑𝑧 = 2𝑥 𝑒 𝑑𝑥 + 2𝑦 𝑒 𝑑𝑦
Page 8 of 20
𝑦+𝑧
𝐜. 𝑢 = tan(3𝑥 − 𝑦) + 6
𝜕𝑢 𝜕𝑢 𝜕𝑢
𝑑𝑢 = 𝑑𝑥 + 𝑑𝑦 + 𝑑𝑧
𝜕𝑥 𝜕𝑦 𝜕𝑧
𝑦+𝑧 𝑦+𝑧
𝑑𝑢 = 3 sec 2 (3𝑥 − 𝑦) 𝑑𝑥 + [− sec 2 (3𝑥 − 𝑦) + (6 )(ln 6)] 𝑑𝑦 + (6 )(ln 6) 𝑑𝑧
𝑢𝑣
𝐝. 𝑧 = ln ( ) → 𝑧 = ln 𝑢 + ln 𝑣 − ln 𝑠 − ln 𝑡
𝑠𝑡
𝜕𝑧 𝜕𝑧 𝜕𝑧 𝜕𝑧
𝑑𝑧 = 𝑑𝑢 + 𝑑𝑣 + 𝑑𝑠 + 𝑑𝑡
𝜕𝑢 𝜕𝑣 𝜕𝑠 𝜕𝑡
1 1 1 1
𝑑𝑧 = 𝑑𝑢 + 𝑑𝑣 − 𝑑𝑠 − 𝑑𝑡
𝑢 𝑣 𝑠 𝑡
𝒙𝒚
𝟒 The total resistance 𝑅 of 2 resistors (𝑥, 𝑦) connected in parallel is 𝑹 = . Suppose 𝒙
𝒙+𝒚
and 𝑦 are measured to be 200 Ω (ohms) and 400 Ω (ohms), respectively. If 𝒙 decreases by 1 Ω
and 𝑦 increases by 6 Ω, estimate the maximum error in the calculated 𝑹 using the total differential.
Solution:
𝜕𝑅 𝜕𝑅
𝑑𝑅 = 𝑑𝑥 + 𝑑𝑦
𝜕𝑥 𝜕𝑦
𝑦(𝑥 + 𝑦) − 𝑥𝑦 𝑥(𝑥 + 𝑦) − 𝑥𝑦 𝑦2 𝑥2
𝑑𝑅 = 𝑑𝑥 + 𝑑𝑦 = 𝑑𝑥 + 𝑑𝑦
(𝑥 + 𝑦)2 (𝑥 + 𝑦)2 (𝑥 + 𝑦)2 (𝑥 + 𝑦)2
To find the maximum error in the value of 𝑅 , we therefore use
𝑑𝑥 = −1 and 𝑑𝑦 = 6
and with the values of 𝑥 = 200 and 𝑦 = 400, we get
(400)2 (200)2 2
𝑑𝑅 = . (−1) + . (6) = ≅ 0.22
(200 + 400)2 (200 + 400)2 9
Page 9 of 20
𝟓 If 𝑧 = 𝑥 2 − 𝑥𝑦 + 𝑦 2 and (𝑥, 𝑦) changes from (3, −1) to (2.96, −0.95), compare the values
of ∆𝒛 and 𝒅𝒛.
Solution:
First Approach:
By using the definition of the total differential of the function 𝒛 = 𝒇(𝒙, 𝒚) is given by
𝜕𝑧 𝜕𝑧
𝑑𝑧 = 𝑑𝑥 + 𝑑𝑦
𝜕𝑥 𝜕𝑦
𝑑𝑧 = (2𝑥 − 𝑦) 𝑑𝑥 + (−𝑥 + 2𝑦) 𝑑𝑦
To find the value of 𝑑𝑧 , we therefore use
𝑑𝑥 = −0.04 and 𝑑𝑦 = 0.05
and with the values of 𝑥 = 3 and 𝑦 = −1, we get
𝑑𝑧 = [2(3) − (−1)] [−0.04] + [−3 + 2(−1)] [0.05] = −0.53
Second Approach:
Consider a function of two variables, 𝒛 = 𝒇(𝒙, 𝒚) and suppose 𝑥 changes from 𝑎 to 𝑎 + Δ𝑥
and 𝑦 changes from 𝑏 to 𝑏 + Δ𝑦. Then the corresponding increment of 𝑧 is
Δ𝑧 = 𝑓(𝑎 + Δ𝑥, 𝑏 + Δ𝑦) − 𝑓(𝑎, 𝑏)
𝛥𝑧 = 𝑓(2.96, −0.95) − 𝑓(3, −1)
= [(2.96)2 − (2.96)(−0.95) + (−0.95)2 ] − [(3)2 − (3)(−1) + (−1)2 ] = −0.5239
Notice that: Δ𝑧 ≅ 𝑑𝑧 but 𝑑𝑧 is easier to compute.
Page 10 of 20
𝟔 The base radius and height of a right circular cone are measured as 10 cm and 25 cm,
respectively, with a possible error in measurement of as much as 0.1 cm in each. Use the total
differential to estimate the maximum error in the calculated volume of the cone.
Solution:
Let the radius and the height of the right circular cone be 𝒓 and 𝒉, respectively. Therefore, its
volume is
𝜋 𝑟2 ℎ
𝑉=
3
and so
𝜕𝑉 𝜕𝑉
𝑑𝑉 = 𝑑𝑟 + 𝑑ℎ
𝜕𝑟 𝜕ℎ
2𝜋𝑟ℎ 𝜋 𝑟2
∴ 𝑑𝑉 = 𝑑𝑟 + 𝑑ℎ
3 3
To find the maximum error in the volume, we therefore use
𝑑𝑟 = 0.1 and 𝑑ℎ = 0.1
and with the values of 𝑟 = 10 and ℎ = 25, we get
500𝜋 100𝜋
𝑑𝑉 = (0.1) + (0.1) = 20𝜋
3 3
i.e. ∆𝑉 ≅ 20𝜋
Page 11 of 20
𝟕 Apply the second partial derivative test to find relative extrema and saddle points of the
following functions (if exist):
𝐚. 𝑓(𝑥, 𝑦) = 2𝑥 2 + 2𝑥𝑦 + 𝑦 2 + 2𝑥 − 3
1 1
𝐛. 𝑓(𝑥, 𝑦) = 𝑥𝑦 − 𝑥 3 − 𝑦 3
3 3
𝑦
𝐜. 𝑓(𝑥, 𝑦) = ln 𝑥 + ln 𝑦 − 𝑥 − +5
2
𝐝. 𝑓(𝑥, 𝑦) = 𝑒 𝑥 + 𝑒 𝑦 − 𝑥 − 2𝑦 + 1
Solution:
𝐚. 𝑓𝑥 (𝑥, 𝑦) = 4𝑥 + 2𝑦 + 2 and 𝑓𝑦 (𝑥, 𝑦) = 2𝑥 + 2𝑦
𝑓𝑥 (𝑥, 𝑦) = 0 → 4𝑥 + 2𝑦 + 2 = 0 (1)
𝑓𝑦 (𝑥, 𝑦) = 0 → 2𝑥 + 2𝑦 = 0 (2)
Subtracting (2) from (1)
2𝑥 + 2 = 0 ⇒ 𝑥 = −1
∴ 𝑦=1
𝑓𝑥𝑥 (𝑥, 𝑦) = 4 , 𝑓𝑦𝑦 (𝑥, 𝑦) = 2 and 𝑓𝑥𝑦 (𝑥, 𝑦) = 2
2
𝐷 = 𝑓𝑥𝑥 (𝑥, 𝑦) 𝑓𝑦𝑦 (𝑥, 𝑦) − [ 𝑓𝑥𝑦 (𝑥, 𝑦)]
Critical Point 𝐷 𝑓𝑥𝑥 (𝑥, 𝑦) Conclusion 𝑓(𝑥, 𝑦)
(−1 , 1) 4>0 4>0 Relative Minimum −4
Page 12 of 20
𝐛. 𝑓𝑥 (𝑥, 𝑦) = 𝑦 − 𝑥 2 and 𝑓𝑦 (𝑥, 𝑦) = 𝑥 − 𝑦 2
𝑓𝑥 (𝑥, 𝑦) = 0 → 𝑦 − 𝑥2 = 0 (1)
𝑓𝑦 (𝑥, 𝑦) = 0 → 𝑥 − 𝑦2 = 0 (2)
Substituting (1) into (2)
𝑥 − 𝑥4 = 0 ⇒ 𝑥 = 0, 1 and 𝑦 = 0, 1
𝑓𝑥𝑥 (𝑥, 𝑦) = −2𝑥 , 𝑓𝑦𝑦 (𝑥, 𝑦) = −2𝑦 and 𝑓𝑥𝑦 (𝑥, 𝑦) = 1
𝐷 = 4𝑥𝑦 − 1
Critical Point 𝐷 𝑓𝑥𝑥 (𝑥, 𝑦) Conclusion 𝑓(𝑥, 𝑦)
(0 , 0) −1 < 0 --- Saddle point 0
(1 , 1) 3>0 −2 < 0 Relative Maximum 1/3
1 1 1
𝐜. 𝑓𝑥 (𝑥, 𝑦) = −1 and 𝑓𝑦 (𝑥, 𝑦) = −
𝑥 𝑦 2
1
𝑓𝑥 (𝑥, 𝑦) = 0 → −1=0 ⇒ 𝑥 =1
𝑥
1 1
𝑓𝑦 (𝑥, 𝑦) = 0 → − =0 ⇒ 𝑦=2
𝑦 2
−1 −1
𝑓𝑥𝑥 (𝑥, 𝑦) = , 𝑓𝑦𝑦 (𝑥, 𝑦) = and 𝑓𝑥𝑦 (𝑥, 𝑦) = 0
𝑥2 𝑦2
1
𝐷=
𝑥2 𝑦2
Critical Point 𝐷 𝑓𝑥𝑥 (𝑥, 𝑦) Conclusion 𝑓(𝑥, 𝑦)
1
(1 , 2) >0 −1 < 0 Relative Maximum 3.69
4
Page 13 of 20
𝐝. 𝑓𝑥 (𝑥, 𝑦) = 𝑒 𝑥 − 1 and 𝑓𝑦 (𝑥, 𝑦) = 𝑒 𝑦 − 2
𝑓𝑥 (𝑥, 𝑦) = 0 → 𝑒𝑥 − 1 = 0 ⇒ 𝑥=0 (1)
𝑓𝑦 (𝑥, 𝑦) = 0 → 𝑒𝑦 − 2 = 0 ⇒ 𝑦 = ln 2 (2)
𝑓𝑥𝑥 (𝑥, 𝑦) = 𝑒 𝑥 , 𝑓𝑦𝑦 (𝑥, 𝑦) = 𝑒 𝑦 and 𝑓𝑥𝑦 (𝑥, 𝑦) = 0
𝐷 = 𝑒 𝑥 𝑒𝑦
Critical Point 𝐷 𝑓𝑥𝑥 (𝑥, 𝑦) Conclusion 𝑓(𝑥, 𝑦)
(0 , ln 2) 2>0 1>0 Relative Minimum 2.61
Page 14 of 20
𝟖 Find the maximum and minimum values of 𝑓(𝑥, 𝑦) = 81𝑥 2 + 𝑦 2 subject to the constraint
4𝑥 2 + 𝑦 2 = 9.
Solution:
To determine the constraint function, we must first subtract 9 from both sides of the constraint.
This gives 4𝑥 2 + 𝑦 2 − 9 = 0. The constraint function is equal to the left-hand side, so
g(𝑥, 𝑦) = 4𝑥 2 + 𝑦 2 − 9
The problem asks us to solve for the minimum and maximum values of 𝑓, subject to the constraint.
4𝑥 2 + 𝑦 2 = 9 (1)
Next, set up the second equation using the following template:
𝑓𝑥 𝑓𝑦 162𝑥 2𝑦
= ⇒ =
g𝑥 g𝑦 8𝑥 2𝑦
162𝑥𝑦 = 8𝑥𝑦 ⇒ 154𝑥𝑦 = 0
Note that: Cancellation of 𝒚 in this equation is a fatal mistake. Since,
162𝑥
=1 ⇒ 162𝑥 = 8𝑥 ⇒ 162 = 8
8𝑥
Either 𝑦 = 0 or 𝑥 = 0, Substitute them both in the constrain to get the possible points for the
absolute extrema.
𝒚=𝟎:
3 3 −3
4𝑥 2 = 9 ⇒ 𝑥 = ± ⇒ ( , 0) and ( , 0) (𝟐)
2 2 2
𝒙=𝟎:
𝑦 2 = 9 ⇒ 𝑦 = ±3 ⇒ (0,3) and (0, −3) (𝟑)
Page 15 of 20
Substitute these points into the function 𝑓(𝑥, 𝑦) = 81𝑥 2 + 𝑦 2 , we have
Point (3/2 , 0) (−3/2 , 0) (0, 3) (0, −3)
𝑓(𝑥, 𝑦) 182.5 182.5 9 9
So,
3 −3
( , 0) and ( , 0) are absoulte maximum points with value 182.5
2 2
(0,3) and (0, −3) are absoulte minimum points with value 9
𝟗 Find the maximum and minimum values of 𝑓(𝑥, 𝑦, 𝑧) = 𝑦 2 − 10𝑧 subject to the constraint
𝑥 2 + 𝑦 2 + 𝑧 2 = 36.
Solution:
To determine the constraint function, we must first subtract 36 from both sides of the constraint.
This gives 𝑥 2 + 𝑦 2 + 𝑧 2 − 36 = 0. The constraint function is equal to the left-hand side, so
g(𝑥, 𝑦, 𝑧) = 𝑥 2 + 𝑦 2 + 𝑧 2 − 36
The problem asks us to solve for the minimum and maximum values of 𝑓, subject to the constraint.
𝑥 2 + 𝑦 2 + 𝑧 2 = 36 (1)
Next, set up the second equation using the following template:
𝑓𝑥 𝑓𝑦 𝑓𝑧 0 2𝑦 −10
= = ⇒ = =
g𝑥 g𝑦 g𝑧 2𝑥 2𝑦 2𝑧
0 ∗ 2𝑧 = −20𝑥 ⇒ 𝑥=0 (𝟐)
2𝑧𝑦 = −10𝑦 ⇒ 𝑦(𝑧 + 5) = 0
Page 16 of 20
Either 𝑦 = 0 or 𝑧 = −5, Substitute them both in the constrain to get the possible points for the
absolute extrema.
𝒚=𝟎:
0 + 0 + 𝑧 2 = 36 ⇒ 𝑧 = ± 6 ⇒ (0, 0, 6) and (0, 0, −6) (𝟑)
𝒛 = −𝟓 :
0 + 𝑦 2 + (−5)2 = 36 ⇒ 𝑦 = ±√11 ⇒ (0, √11, −5) and (0, −√11, −5) (𝟒)
Substitute these points into the function 𝑓(𝑥, 𝑦, 𝑧) = 𝑦 2 − 10𝑧 , we have
Point (0, 0, 6) (0, 0, −6) (0, √11, −5) (0, −√11, −5)
𝑓(𝑥, 𝑦) −60 60 61 61
So,
(0, √11, −5) and (0, −√11, −5) are absoulte maximum points with value of 61
(0, 0, 6) is absoulte minimum point with value of − 60
𝟏𝟎 Find three real numbers whose sum is 9 and the sum of whose squares is a small as possible.
Solution:
We want to find the minimum of the squares of the three numbers
𝑓(𝑥, 𝑦, 𝑧) = 𝑥 2 + 𝑦 2 + 𝑧 2
Provided that the sum of the three numbers equals 9.
𝑥+𝑦+𝑧 =9 (1)
Therefore,
g(𝑥, 𝑦, 𝑧) = 𝑥 + 𝑦 + 𝑧 − 9
Page 17 of 20
Next, set up the second equation using the following template:
𝑓𝑥 𝑓𝑦 𝑓𝑧 2𝑥 2𝑦 2𝑧
= = ⇒ = =
g𝑥 g𝑦 g𝑧 1 1 1
𝑥=𝑦=𝑧 (𝟐)
Substitute them both in the constrain to get the possible points for the absolute extrema.
𝒙=𝒚=𝒛:
𝑥+𝑥+𝑥 =9 ⇒ 𝑥 =𝑦 =𝑧 =3
𝟏𝟏 Find the dimensions of the closed rectangular box with maximum volume that can be
inscribed in the unit sphere.
Solution:
We want to find the volume of a rectangular box
𝑓(𝑥, 𝑦, 𝑧) = (2𝑥)(2𝑦)(2𝑧) = 8𝑥𝑦𝑧
Provided that the box can be inscribed in a unit sphere.
𝑥2 + 𝑦2 + 𝑧2 = 1 (1)
Therefore,
g(𝑥, 𝑦, 𝑧) = 𝑥 2 + 𝑦 2 + 𝑧 2 − 1
Next, set up the second equation using the following template:
𝑓𝑥 𝑓𝑦 𝑓𝑧 8𝑦𝑧 8𝑥𝑧 8𝑥𝑦
= = ⇒ = =
g𝑥 g𝑦 g𝑧 2𝑥 2𝑦 2𝑧
𝑦2𝑧 = 𝑥 2𝑧 ⇒ 𝑧(𝑦 2 − 𝑥 2 ) = 0 (𝟐)
𝑥𝑧 2 = 𝑥𝑦 2 ⇒ 𝑥(𝑧 2 − 𝑦 2 ) = 0 (𝟑)
Page 18 of 20
For this application, note that neither one of the three dimensions can equal zero.
𝑥 2 = 𝑦 2 and 𝑧 2 = 𝑦 2
Therefore,
𝑥=𝑦=𝑧 (𝟒)
Substitute them in the constrain to get the maximum volume.
𝒙=𝒚=𝒛:
1
𝑥2 + 𝑥2 + 𝑥2 = 1 ⇒ 𝑥 = 𝑦 = 𝑧 =
√3
𝟏𝟐 A firm manufactures a commodity at two different factories. The total cost of manufacturing
depends on the quantities, 𝒒𝟏 and 𝒒𝟐 , supplied by each factory, and is expressed by thejoint
cost function,
𝐶 = 𝑓(𝑞1 , 𝑞2 ) = 2𝑞1 2 + 𝑞1 𝑞2 + 𝑞2 2 + 500
The company’s objective is to produce 200 units, while minimizing production costs.
How many units should be supplied by each factory?
Solution:
We want to find the minimize
𝐶 = 𝑓(𝑞1 , 𝑞2 ) = 2𝑞1 2 + 𝑞1 𝑞2 + 𝑞2 2 + 500
Provided that the objective is to produce 200 units.
𝑞1 + 𝑞2 = 200 (1)
Therefore,
g(𝑞1 , 𝑞2 ) = 𝑞1 + 𝑞2 − 200
Page 19 of 20
Next, set up the second equation using the following template:
𝑓𝑞1 𝑓𝑞 4𝑞1 + 𝑞2 2𝑞2 + 𝑞1
= 2 ⇒ =
g𝑞 g𝑞 1 1
1 2
4𝑞1 + 𝑞2 = 2𝑞2 + 𝑞1 ⇒ 3𝑞1 = 𝑞2 (𝟐)
Substitute 3𝑞1 = 𝑞2 in the constrain to get the minimum production cost.
𝟑𝒒𝟏 = 𝒒𝟐 :
𝑞1 + 3𝑞1 = 200 ⇒ 𝑞1 = 50 and 𝑞2 = 150
Page 20 of 20
