Download Answers to Homework Questions for Sect 27

Section 27 Answers
Number 1
f (x, y, λ) = 2xy + λ(x + y − 12)
f (x, y, λ) = 2xy + λx + λy − λ12
Partial Derivatives
System of Equations to Solve
fx (x, y, λ) = 2y + λ
fy (x, y, λ) = 2x + λ
fλ (x, y, λ) = x + y − 12
2y + λ = 0
2x + λ = 0
x + y − 12 = 0
• Solving the first 2 equations for λ
2y + λ = 0 =⇒ λ = −2y
2x + λ = 0 =⇒ λ = −2x
• Combining the first 2 equations
−2y = −2x =⇒ x = y
• Substituting into the third equation
x + y − 12 = 0 =⇒ x + x − 12 = 0 =⇒ 2x = 12 =⇒ x = 6
• Using x = 6, solve for y
x = yandx = 6 =⇒ y = 6
• f (x, y) = 2xy, so f (6, 6) = 2(6)(6) = 72
• Therefore the Max is 72
1
NUMBER 2
f (x, y, λ) = x2 + 3y 2 + λ(x − y + 1)
f (x, y, λ) = x2 + 3y 2 + λx − λy + λ
Partial Derivatives
System of Equations to Solve
fx (x, y, λ) = 2x + λ
fy (x, y, λ) = 6y − λ
fλ (x, y, λ) = x − y + 1
2x + λ = 0
6y − λ = 0
x−y+1= 0
• Solve first 2 equations for λ
2x + λ = 0 =⇒ λ = −2x
6y − λ = 0 =⇒ λ = 6y
• Combine previous 2
−2x = 6y =⇒ x = −3y
• Sub into third eqn.
x − y + 1 = 0 =⇒ (−3y) − y + 1 = 0 =⇒ 1 = 4y =⇒ y =
1
4
• Use y to find value for x
x = −3y =⇒ x = −3( 41 ) =⇒ x =
• f (x, y) is minimized when x =
, 1 ) = ( −3
)2 + 3( 14 )2 =
f ( −3
4 4
4
9
16
+
3
16
−3
4
−3
4
=
and y = 41 . The minimum value is
12
16
2
12
16
=
3
4
NUMBER 3
f (x, y, λ) = x2 − y 2 + λ(x2 + y 2 − 4)
f (x, y, λ) = x2 − y 2 + λx2 + λy 2 − λ4
Partial Derivatives
System of Equations to Solve
fx (x, y, λ) = 2x − 2λx
fy (x, y, λ) = −2y + 2λy
fλ (x, y, λ) = x2 + y 2 − 4
2x − 2λx = 0
−2y + 2λy = 0
x2 + y 2 − 4 = 0
• Simplify first 2 equations
2x − 2λx = 0 =⇒ 2x(1 − λ) = 0 =⇒ x = 0 or λ = 1
−2y + 2λy = 0 =⇒ 2y(1 − λ) = 0 =⇒ y = 0 or λ = 1
• IF x = 0
x2 + y 2 − 4 = 0 =⇒ y 2 = 4 =⇒ y = ±2
• IF y = 0
x2 + y 2 − 4 = 0 =⇒ x2 = 4 =⇒ x = ±2
• λ = 1 gives no useful info
• In either Case, the original equation = −4
• f (x, y) is minimized when x = 0 and y = ±2 OR when y = 0 and x = ±2 The
minimum value is −4
3
NUMBER 4
f (x, y, λ) = x + y − x2 − y 2 + λ(x + 2y − 6)
f (x, y, λ) = x + y − x2 − y 2 + λx + λ2y − λ6)
Partial Derivatives
System of Equations to Solve
fx (x, y, λ) = 1 − 2x + λ
fy (x, y, λ) = 1 − 2y + 2λ
fλ (x, y, λ) = x + 2y − 6
1 − 2x + λ = 0
1 − 2y + 2λ = 0
x + 2y − 6 = 0
• Solve first 2 equations for λ
1 − 2x + λ = 0 =⇒ λ = 2x − 1
1 − 2y + 2λ = 0 =⇒ λy −
1
2
• Combine previous 2
=⇒ 2x − 1 = y −
1
2
=⇒ y = 2x −
1
2
• Sub into third eqn.
x+2y−6 = 0 =⇒ x+2(2x− 21 )−6 = 0 =⇒ x+4x−1−6 = 0 =⇒ 5x = 7 =⇒ x =
• Use x to find value for y
x + 2y − 6 = 0 =⇒
7
5
+ 2y − 6 = 0 =⇒ 2y = 6 −
• f (x, y) is minimized when x =
7
5
and y =
23
.
10
7
5
=⇒ y =
The minimum value is −3.55
23
) = x + y − x2 − y 2 = ( 57 ) + ( 23
) − ( 57 )2 − ( 23
)2 = −3.55
f ( 75 , 10
10
10
4
23
10
7
5
NUMBER 5
f (x, y, λ) = x2 − 10y 2 + λ(x + y − 9)
f (x, y, λ) = x2 − 10y 2 + λx + λy − λ9
Partial Derivatives
System of Equations to Solve
fx (x, y, λ) = 2x + λ
fy (x, y, λ) = −20y + λ
fλ (x, y, λ) = x + y − 9
2x + λ = 0
−20y + λ = 0
x+y−9=0
• Solve first 2 equations for λ
2x + λ = 0 =⇒ −2x = λ
−20y + λ = 0 =⇒ 20y = λ
• Combine previous 2
−2x = 20y =⇒ x = −10y
• Sub into third eqn.
x + y − 9 = 0 =⇒ −10y + y − 9 = 0 =⇒ −9y = 9 =⇒ y = −1
• Use y to find value for x
x = −10y =⇒ x = −10(−1) =⇒ x = 10
• f (x, y) is minimized when x = 10 and y = −1. The minimum value is 100−10 =
90
5