Calculus III
Contents
3 Dimensional space
Partial derivatives
Multiple integrals
Vector Functions
Line integrals
Surface integrals
Vector operators
Applications
© The scientific sentence. 2010
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Calculus III:
Functions of two variables
Lagrange Multipliers
Optimizing a function means to find its absolute extrema,
that is its absolute maximums and minimums.
So far, to optimize a function f, we find the absolute maximum value
and the absolute minimum value of this function f on a region D that contains
its boundary.
Now we are going to use another process to optimize a function, but
subject to given constraint(s).
The constraint(s) may be the equation(s)
that describe the boundary of a region. They may be also equation(s), or
inequation(s) that set conditions on the variables themeselves.
This process is called method of Lagrange multipliers.
1. Equations of Lagrange Multipliers
It cosists of to compare the variation of a function
with respect to the varaiation of the associated constaints.
That is solving the following system of four equations:
∇f(x,y,z) = λ ∇g(x,y,z)
g(x,y,z) = The constraintes
Three equations for the gradient vector ∇f along
with one equation for the constraintes with four unknowns
x, y, z, and λ.
The constante ∇ is called the Lagrange multiplier.
2. Method of Lagrange Multipliers
Once the above system is solved, we substitute. if they
exist, all the solutions (x,y,z) in the function f(x,y,z),
and identify the minimum and maximum values.
We need always to go back and verify that the found answers
make sense.
Example 1
Find three real numbers strictly positive with largest product if
their product in pairs is equal to 27.
So, in other words,
We want to find three real nombers x, y, y, with largest product xyz if
xy + yz + xz = 27.
Then
f(x,y,z) = xyz
g(x,y,z) = the constraint = xy + yz + xz = 27.
Restriction: x, y , and > 0
fx = yz
fy = xz
fz = xy
gx = y + z
gy = x + z
gz = x + y
The system of equations is
∇ f = λ ∇g
g(x,y,z)= xy + yz + xz
x, y , and > 0
That is
fx = λ gx
fy = λ gy
fz = λ gz
xy + yz + xz = 27
x, y , and > 0
| yz = λ (y + z) | (1) |
| xz = λ (x + z) | (2) |
| xy = λ (x + y) | (3) |
| xy + yz + xz = 27 | (4) |
| x, y , and > 0 | (5) |
Multiplying (1) by x, (2) by y , and (3) by z gives:
| xyz = λ x (y + z) | (1') |
| xyz = λ y (x + z) | (2') |
| xyz = λ z (x + y) | (3') |
| xy + yz + xz = 27 | (4) |
| x, y , and > 0 | (5) |
Equating (1') and (2') gives
x λ (y + z) = y λ (x + z). That is
λ (xz - yz) = 0 ⇒ λ = 0 or xz = yz
We get then two possibilities:
• λ = 0 leads, for the equation (1) to
y = 0 0r z = 0 . According to the constraint (5) neither
of these are possible. So we discount λ = 0.
• xz = yz leads to x = y, because z≠ 0 , according, again to
the constraint (5).
Likewise, equating (2') and (3') gives y = z
Therefore x = y = z
The constraint (5) becomes:
x2 + y2 + z2 = 27
3x2 = 27 ⇒ x = ± 3
According, again to the constraint (5), we discount - 3.
Conclusion
The three nombres are equal to 3.
Example 2
Find the maximum and minimum values of f(x,y) = 4 x - 3 y, subject to the
constraint x2 + y2 = 25.
From the constraint, the region of possible solutions lies
on a disk of radius √25 = 5. This disk is
a closed and bounded region. So, according to the Extreme
value theorem, the extrema values will exist.
We have
f(x,y) = 4 x - 3 y
g(x,y,z) = the constraint = x2 + y2 = 25
The partial derives are:
fx = 4, and
fy = - 3 .
gx = 2 x , and
gy = 2 y
The system of equations is
∇ f = λ ∇g
g(x,y)= x2 + y2
That is
4 = 2λx
- 3 = 2λy
x2 + y2 = 25
• λ = 0 does not satisfy the first two equations.
Hence λ ≠ 0.
Therefore:
x = 2/λ, and
y = - 3/2λ
The constraint becomes:
(2/λ)2 + (- 3/2λ)2 = 36
Solving for λ gives:
λ2 = 1/4
⇒
λ = ± 1/2
λ = ± 1/2
• λ = - 1/2 :
x = - 4 , and
y = 3
⇒ f(- 4,3) = - 25
• λ = + 1/2 :
x = 4 , and
y = - 3
⇒ f(4,- 3) = 25
Conclusion
The maximum is (4, - 3).
The minimum is (- 4, 3).
Example 3
Find the maximum and minimum values of f(x,y,z) = xyz, subject to the
x + y + z = 1, and to the restriction x, y, z ≥ 0.
f(x,y,z) = xyz
g(x,y,z) = the constraint = x + y + z = 1.
Restriction: x, y , and ≥ 0
Note that negatives variables require other constraintes.
fx = yz,
fy = xz,
fz = xy
gx = 1 ,
gy = 1 ,
gz = 1
The system of equations to solve is
∇ f = λ ∇g
g(x,y,z) = x + y + z
That is
| yz = λ | (1) |
| xz = λ | (2) |
| xy = λ | (3) |
| x + y + z = 1 | (4) |
Equating (1) and (2) gives
yz = xz. That is
z (x - y) = 0 ⇒ z = 0 or x = y
We get then two possibilities:
1. First possibility: z = 0 leads, for the equation (1) and (2) to
λ = 0 . According to the equation (3),
we have x = 0 or y = 0.
So z = 0 , x = 0 . According to the constraint (4),
we have then y = 1 ⇒ (0,1,0)
z = 0 , y = 0 . According to the constraint (4),
we have then x = 1 ⇒ (1,0,0)
2. Second possibility: x = y leads two possibilities:
x = y = 0 , and x = y ≠ 0
• x = y = 0. According to the constraint (4) z = 1.
⇒ (0,0,1).
• x = y ≠ 0. According to the equations (2) and (3)
xz = xy ⇒ x(y - z) = 0 ⇒ x = 0 or y = z.
But x ≠ 0, then x = y = z ≠ 0.
According to the constraint (4) , 3x = 1 ⇒ x = y = z = 1/3.
⇒ (1/3,1/3,1/3).
Likewise, if we set the equation (1) and (3) , then
(2) and (3), we will get the four same equations as
above.
Conclusion
The three four values are:
f(0,1,0) =
f(1,0,0) =
f(0,0,1) = 0 : Minimums
(1/3,1/3,1/3) = 1/27 : Maximum.
Example 3
Now, we have a constraint that is an inequation. The inequality is
used only to find critical points. For the absolute extrema, this
enequation is replaced by its associated equation.
Find the maximum and minimum values of
f(x,y) = 2x2 + 5y2
on the disk x2 + y2 ≤ 9.
The disk of radius 3 is a closed and bounded region. Then according to the
extrem value theorem, the minimum value and the maximum value exist.
• Critical points in the disk:
f(x,y) = 2x2 + 5y2
g(x,y) = the constraint = x2 + y2 ≤ 9.
fx = 4x = 0 ⇒ x = 0
fy = 10y = 0 ⇒ y = 0
The only crical point is (0,0), that
satisfies the inequality.
• Absolute extrema:
gx = 2x ,
gy = 2y
The system of equations to solve is
∇ f = λ ∇g
g(x,y) = x2 + y2
That is
| 4x = 2λx | (1) |
| 10y = 2λy | (2) |
| x2 + y2 = 9 | (3) |
From the equation (1), we get: 2x(2 - λ) = 0 ⇒ x = 0, λ = 2.
For x = 0, the equation (3) gives y = ± 3.
We have then the two points: (0, - 3) and (0, + 3)
For λ = 2, the equation (2) gives 10y = 4y ⇒ y = 0
the equation (3) gives x = ± 3.
We have then
the two points: (- 3 , 0) and (3, 0)
Conclusion:
f(0,0) = 0 Minimum
f(- 3 , 0) =
f(3, 0) = 18
f(0, - 3) =
f(0, + 3) = 45 Maximum
Example 4
Now, we have more than one constraint. We will use two
constraint, but we can extend the process to more than
two constraints.
Find the maximum and minimum values of
f(x,y,z) = 2y - z, subject to the constraints
x - y - z = 1 and x2 + y2 = 4.
The disk of radius 2 is a closed and bounded region. Then according to the
extrem value theorem, the minimum value and the maximum value exist.
• Critical points in the disk:
f(x,y,z) = 2y - z
g(x,y,z) = the first constraint = x - y - z = 1
h(x,y,z) = the second constraint = x2 + y2 = 4
fx = 0 ,
fy = 2 , and
fz = - 1
gx = 1 ,
gy = - 1 , and
gz = - 1
hx = 2x ,
hy = 2y , and
hz = 0
The system of equations to solve is
∇ f = λ ∇g + μ ∇h
g(x,y,z) = x - y - z = 1
h(x,y,z) = x2 + y2 = 4
That is
| 0 = λ + 2μ x | (1) |
| 2 = - 2λ + 2 μ y | (2) |
| - 1 = - λ + μ 0 | (3) |
| x - y - z = 1 | (4) |
| x2 + y2 = 4 | (5) |
From the equation (3), we get: λ = 1. So the equation
(1) and (2) becomes:
- 1 = 2μ x ⇒ x = - 1/2μ
4 = 2 μ y ⇒ y = 2/μ
The equation (5) gives:
(- 1/2μ)2 + (2/μ)2 = 4
1/4μ2 + 4/μ2 = 4
μ = ± √17/4
• μ = √17/4 ⇒ x = - 2/√17 , y = 8/√17
The equation (4) gives z = x - y - 1 = - 2/√17 - 8/√17 - 1 =
- 10/√17 - 1.
• μ = - √17/4 ⇒ x = 2/√17 , y = - 8/√17
The equation (4) gives z = x - y - 1 = 2/√17 + 8/√17 - 1 =
10/√17 - 1.
Therefore
f(- 2/√17 , 8/√17, - 10/√17 - 1) = 16/√17 + 10/√17 + 1 =
26/√17 + 1 : maximum.
f( 2/√17 , - 8/√17, 10/√17 - 1) = - 16/√17 - 10/√17 + 1 =
- 26/√17 + 1 : minimum.
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