3. Lagrange Multipliers
Let z = f(x,y) be a function in x and y and g(x, y) = 0 be an
equation in x and y. The graph of z = f(x, y) is a surface and
g(x, y) = 0 is a curve in the xy-plane,
z = f(x, y)
y g(x, y) = 0
x
4. Lagrange Multipliers
Let z = f(x,y) be a function in x and y and g(x, y) = 0 be an
equation in x and y. The graph of z = f(x, y) is a surface and
g(x, y) = 0 is a curve in the xy-plane, and we assume this
curve is in the domain of f(x, y).
z = f(x, y)
Domain
g(x, y) = 0
y x
5. Lagrange Multipliers
Let z = f(x,y) be a function in x and y and g(x, y) = 0 be an
equation in x and y. The graph of z = f(x, y) is a surface and
g(x, y) = 0 is a curve in the xy-plane, and we assume this
curve is in the domain of f(x, y).
The set of points (x, y, z = f(x, y))
z = f(x, y)
with (x, y) on the curve form a
"trail" on the surface defined by
f(x, y).
Domain
g(x, y) = 0
x y
6. Lagrange Multipliers
Let z = f(x,y) be a function in x and y and g(x, y) = 0 be an
equation in x and y. The graph of z = f(x, y) is a surface and
g(x, y) = 0 is a curve in the xy-plane, and we assume this
curve is in the domain of f(x, y).
The set of points (x, y, z = f(x, y))
z = f(x, y)
with (x, y) on the curve form a
"trail" on the surface defined by
f(x, y). Where are the extrema on
this trail? That is, where are the
highest point and the lowest point
on this trail?
Domain
g(x, y) = 0
x y
7. Lagrange Multipliers
Let z = f(x,y) be a function in x and y and g(x, y) = 0 be an
equation in x and y. The graph of z = f(x, y) is a surface and
g(x, y) = 0 is a curve in the xy-plane, and we assume this
curve is in the domain of f(x, y).
The set of points (x, y, z = f(x, y))
z = f(x, y)
with (x, y) on the curve form a
"trail" on the surface defined by
f(x, y). Where are the extrema on
this trail? That is, where are the
highest point and the lowest point
on this trail? The equation
g(x, y) = 0 is called the
Domain constraint equation in the set up.
g(x, y) = 0
x These types of problems are
y
known as the "extrema with
constraints" problems.
9. Lagrange Multipliers
Plot g(x, y) = 0 over the contour lines
z = f(x, y)
of z = f(x, y) in the xy plane,
Level curves z = c
10. Lagrange Multipliers
Plot g(x, y) = 0 over the contour lines
z = f(x, y)
of z = f(x, y) in the xy plane,
at a point P where the tangent to
g(x, y) = 0 is different from the tangent
to the contour line means the trail
cuts across the contour line, thus
P
the trail is still changing altitude at P.
Level curves z = c
11. Lagrange Multipliers
Plot g(x, y) = 0 over the contour lines
z = f(x, y)
of z = f(x, y) in the xy plane,
at a point P where the tangent to
g(x, y) = 0 is different from the tangent
to the contour line means the trail
cuts across the contour line, thus
P
the trail is still changing altitude at P.
Level curves z = c
y
x
Level curves z = c
12. Lagrange Multipliers
Plot g(x, y) = 0 over the contour lines
z = f(x, y)
of z = f(x, y) in the xy plane,
at a point P where the tangent to
g(x, y) = 0 is different from the tangent
to the contour line means the trail
cuts across the contour line, thus
P
the trail is still changing altitude at P.
So P could not be an extremum.
Level curves z = c
y
x
Level curves z = c
13. Lagrange Multipliers
Plot g(x, y) = 0 over the contour lines
z = f(x, y)
of z = f(x, y) in the xy plane,
at a point P where the tangent to
g(x, y) = 0 is different from the tangent
to the contour line means the trail
cuts across the contour line, thus
P
the trail is still changing altitude at P.
So P could not be an extremum.
Thus at an extremum on the trail
over g(x, y) = 0, its tangent must be Level curves z = c
g(x, y) = 0
parallel to the tangent to the contour
y
lines. x
Level curves z = c
14. Lagrange Multipliers
Plot g(x, y) = 0 over the contour lines
z = f(x, y)
of z = f(x, y) in the xy plane,
at a point P where the tangent to
g(x, y) = 0 is different from the tangent
to the contour line means the trail
cuts across the contour line, thus
P
the trail is still changing altitude at P.
So P could not be an extremum.
Thus at an extremum on the trail
over g(x, y) = 0, its tangent must be Level curves z = c
g(x, y) = 0
parallel to the tangent to the contour
y
lines. Equivalently, at the extremum, x
the normal vector to the constraint
N = ∇g(x, y), and the gradient
to the surface ∇ f(x, y), must be
parallel,
Level curves z = c
15. Lagrange Multipliers
Plot g(x, y) = 0 over the contour lines
z = f(x, y)
of z = f(x, y) in the xy plane,
at a point P where the tangent to
g(x, y) = 0 is different from the tangent
to the contour line means the trail
cuts across the contour line, thus
P
the trail is still changing altitude at P.
So P could not be an extremum.
Thus at an extremum on the trail
over g(x, y) = 0, its tangent must be Level curves z = c
g(x, y) = 0
parallel to the tangent to the contour
y
lines. Equivalently, at the extremum, x
the normal vector to the constraint
N = ∇g(x, y), and the gradient
to the surface ∇ f(x, y), must be
parallel, i.e. there is a λ such that:
Level curves z = c
∇g(x, y) = λ* ∇ f(x, y).
16. Lagrange Multipliers
Plot g(x, y) = 0 over the contour lines of z = f(x, y) in the xy
plane. At a point P where the tangent to g(x, y) = 0 is
different from the tangent to the contour line means the trail
cuts across the contour line at P, thus it is still changing
altitude. Therefore P could not be an extremum. Thus the
extrema points on g(x, y) = 0 must be tangential to the
contour lines. Equivalently, at the extremum, the gradient to
g(x, y) = 0 and the gradient to the contour line must be the
multiple of each other.
P
g(x, y) = 0
17. Lagrange Multipliers
Plot g(x, y) = 0 over the contour lines of z = f(x, y) in the xy
plane. At a point P where the tangent to g(x, y) = 0 is
different from the tangent to the contour line means the trail
cuts across the contour line at P, thus it is still changing
altitude. Therefore P could not be an extremum. Thus the
extrema points on g(x, y) = 0 must be tangential to the
contour lines. Equivalently, at the extremum, the gradient to
g(x, y) = 0 and the gradient to the contour line must be the
multiple of each other, that is, there is a λ such that:
∇g(x, y) = λ* ∇ f(x, y) or that
P
dg i + dg j = λ ( df i + df j )
dx dy dx dy
N= ∇g(x, y)
so we have gx = λfx
∇f(x, y) the system: gy = λfy
g(x, y) = 0
g(x, y) = 0 We solve this syetem for the extrema.
20. Lagrange Multipliers
The variable λ is called a multiplier. In more complex problems,
we might use many differernt multipliers.
Example: Find the extrema of f(x, y) = xy given the constraint
g(x, y) = x2 + y2 – 2 = 0.
21. Lagrange Multipliers
The variable λ is called a multiplier. In more complex problems,
we might use many differernt multipliers.
Example: Find the extrema of f(x, y) = xy given the constraint
g(x, y) = x2 + y2 – 2 = 0.
We are to solve: gx = λfx
gy = λfy
g(x, y) = 0
22. Lagrange Multipliers
The variable λ is called a multiplier. In more complex problems,
we might use many differernt multipliers.
Example: Find the extrema of f(x, y) = xy given the constraint
g(x, y) = x2 + y2 – 2 = 0.
We are to solve: gx = λfx that is: 2x = λy eq1
gy = λfy 2y = λx eq2
g(x, y) = 0 x2 + y2 = 2 eq3
23. Lagrange Multipliers
The variable λ is called a multiplier. In more complex problems,
we might use many differernt multipliers.
Example: Find the extrema of f(x, y) = xy given the constraint
g(x, y) = x2 + y2 – 2 = 0.
We are to solve: gx = λfx that is: 2x = λy eq1
gy = λfy 2y = λx eq2
g(x, y) = 0 x2 + y2 = 2 eq3
2x
From eq1, we get λ = y ,
24. Lagrange Multipliers
The variable λ is called a multiplier. In more complex problems,
we might use many differernt multipliers.
Example: Find the extrema of f(x, y) = xy given the constraint
g(x, y) = x2 + y2 – 2 = 0.
We are to solve: gx = λfx that is: 2x = λy eq1
gy = λfy 2y = λx eq2
g(x, y) = 0 x2 + y2 = 2 eq3
2x 2y
From eq1, we get λ = y , from eq2, we get λ = x .
25. Lagrange Multipliers
The variable λ is called a multiplier. In more complex problems,
we might use many differernt multipliers.
Example: Find the extrema of f(x, y) = xy given the constraint
g(x, y) = x2 + y2 – 2 = 0.
We are to solve: gx = λfx that is: 2x = λy eq1
gy = λfy 2y = λx eq2
g(x, y) = 0 x2 + y2 = 2 eq3
2x 2y
From eq1, we get λ = y , from eq2, we get λ = x .
2y 2x
Set them equal: x = y
26. Lagrange Multipliers
The variable λ is called a multiplier. In more complex problems,
we might use many differernt multipliers.
Example: Find the extrema of f(x, y) = xy given the constraint
g(x, y) = x2 + y2 – 2 = 0.
We are to solve: gx = λfx that is: 2x = λy eq1
gy = λfy 2y = λx eq2
g(x, y) = 0 x2 + y2 = 2 eq3
2x 2y
From eq1, we get λ = y , from eq2, we get λ = x .
2y 2x
Set them equal: x = y
we get 2y2 = 2x2
27. Lagrange Multipliers
The variable λ is called a multiplier. In more complex problems,
we might use many differernt multipliers.
Example: Find the extrema of f(x, y) = xy given the constraint
g(x, y) = x2 + y2 – 2 = 0.
We are to solve: gx = λfx that is: 2x = λy eq1
gy = λfy 2y = λx eq2
g(x, y) = 0 x2 + y2 = 2 eq3
2x 2y
From eq1, we get λ = y , from eq2, we get λ = x .
2y 2x
Set them equal: x = y
we get 2y2 = 2x2 y = ±x
28. Lagrange Multipliers
The variable λ is called a multiplier. In more complex problems,
we might use many differernt multipliers.
Example: Find the extrema of f(x, y) = xy given the constraint
g(x, y) = x2 + y2 – 2 = 0.
We are to solve: gx = λfx that is: 2x = λy eq1
gy = λfy 2y = λx eq2
g(x, y) = 0 x2 + y2 = 2 eq3
2x 2y
From eq1, we get λ = y , from eq2, we get λ = x .
2y 2x
Set them equal: x = y
we get 2y2 = 2x2 y = ±x
Substitute y = x into eq3, we get 2x2 = 2
29. Lagrange Multipliers
The variable λ is called a multiplier. In more complex problems,
we might use many differernt multipliers.
Example: Find the extrema of f(x, y) = xy given the constraint
g(x, y) = x2 + y2 – 2 = 0.
We are to solve: gx = λfx that is: 2x = λy eq1
gy = λfy 2y = λx eq2
g(x, y) = 0 x2 + y2 = 2 eq3
2x 2y
From eq1, we get λ = y , from eq2, we get λ = x .
2y 2x
Set them equal: x = y
we get 2y2 = 2x2 y = ±x
Substitute y = x into eq3, we get 2x2 = 2 x = ±1
30. Lagrange Multipliers
The variable λ is called a multiplier. In more complex problems,
we might use many differernt multipliers.
Example: Find the extrema of f(x, y) = xy given the constraint
g(x, y) = x2 + y2 – 2 = 0.
We are to solve: gx = λfx that is: 2x = λy eq1
gy = λfy 2y = λx eq2
g(x, y) = 0 x2 + y2 = 2 eq3
2x 2y
From eq1, we get λ = y , from eq2, we get λ = x .
2y 2x
Set them equal: x = y
we get 2y2 = 2x2 y = ±x
Substitute y = x into eq3, we get 2x2 = 2 x = ±1
If y = x, there are two pairs of solutions (1, 1), (-1, -1).
31. Lagrange Multipliers
The variable λ is called a multiplier. In more complex problems,
we might use many differernt multipliers.
Example: Find the extrema of f(x, y) = xy given the constraint
g(x, y) = x2 + y2 – 2 = 0.
We are to solve: gx = λfx that is: 2x = λy eq1
gy = λfy 2y = λx eq2
g(x, y) = 0 x2 + y2 = 2 eq3
2x 2y
From eq1, we get λ = y , from eq2, we get λ = x .
2y 2x
Set them equal: x = y
we get 2y2 = 2x2 y = ±x
Substitute y = x into eq3, we get 2x2 = 2 x = ±1
If y = x, there are two pairs of solutions (1, 1), (-1, -1).
Substitute y = -x into eq3, we get 2x2 = 2
32. Lagrange Multipliers
The variable λ is called a multiplier. In more complex problems,
we might use many differernt multipliers.
Example: Find the extrema of f(x, y) = xy given the constraint
g(x, y) = x2 + y2 – 2 = 0.
We are to solve: gx = λfx that is: 2x = λy eq1
gy = λfy 2y = λx eq2
g(x, y) = 0 x2 + y2 = 2 eq3
2x 2y
From eq1, we get λ = y , from eq2, we get λ = x .
2y 2x
Set them equal: x = y
we get 2y2 = 2x2 y = ±x
Substitute y = x into eq3, we get 2x2 = 2 x = ±1
If y = x, there are two pairs of solutions (1, 1), (-1, -1).
Substitute y = -x into eq3, we get 2x2 = 2 x = ±1
33. Lagrange Multipliers
The variable λ is called a multiplier. In more complex problems,
we might use many differernt multipliers.
Example: Find the extrema of f(x, y) = xy given the constraint
g(x, y) = x2 + y2 – 2 = 0.
We are to solve: gx = λfx that is: 2x = λy eq1
gy = λfy 2y = λx eq2
g(x, y) = 0 x2 + y2 = 2 eq3
2x 2y
From eq1, we get λ = y , from eq2, we get λ = x .
2y 2x
Set them equal: x = y
we get 2y2 = 2x2 y = ±x
Substitute y = x into eq3, we get 2x2 = 2 x = ±1
If y = x, there are two pairs of solutions (1, 1), (-1, -1).
Substitute y = -x into eq3, we get 2x2 = 2 x = ±1
If y = -x, this gives two more pairs of solutions (1, -1), (-1, 1).
35. Lagrange Multipliers
To find out which is a max, a min,
or neither, evaluate each with f(x, y) = xy:
f(1, 1) = 1, f(-1, -1) =1,
f(-1, 1) = -1, f(1, -1) = -1,
36. Lagrange Multipliers
To find out which is a max, a min,
or neither, evaluate each with f(x, y) = xy:
f(1, 1) = 1, f(-1, -1) =1,
f(-1, 1) = -1, f(1, -1) = -1,
So (1, 1) , (-1, -1) are maxima,
and (-1, 1), (1, -1) are minima.
37. Lagrange Multipliers
To find out which is a max, a min,
or neither, evaluate each with f(x, y):
f(1, 1) = 1, f(-1, -1) =1,
f(-1, 1) = -1, f(1, -1) = -1,
So (1, 1) , (-1, -1) are maxima,
and (-1, 1), (1, -1) are minima.
38. Lagrange Multipliers
To find out which is a max, a min,
or neither, evaluate each with f(x, y):
f(1, 1) = 1, f(-1, -1) =1,
f(-1, 1) = -1, f(1, -1) = -1,
So (1, 1) , (-1, -1) are maxima,
and (-1, 1), (1, -1) are minima.
The procedure above is called the method of "Lagrange
Multipliers". It can be used in the case of finding extrema of a
three varaible function w=f(x, y, z) with a constraint
g(x, y, z) = 0. In the three variabe case, we solve the systems:
gx = λfx
gy = λfy
gz = λfz
g(x, y, z) = 0
41. Lagrange Multipliers
Example: Find the largest volume possibe of a box with a
fixed total surface area.
The volume is V = xyz.
The constraint is 2xy + 2xz + 2yz = c or xy + xz + yz = c/2
42. Lagrange Multipliers
Example: Find the largest volume possibe of a box with a
fixed total surface area.
The volume is V = xyz.
The constraint is 2xy + 2xz + 2yz = c or xy + xz + yz = c/2
Hence g(x, y, z) = xy + xz + yz – c/2 = 0.
43. Lagrange Multipliers
Example: Find the largest volume possibe of a box with a
fixed total surface area.
The volume is V = xyz.
The constraint is 2xy + 2xz + 2yz = c or xy + xz + yz = c/2
Hence g(x, y, z) = xy + xz + yz – c/2 = 0. Set up the system:
gx = λfx
gy = λfy
gz = λfz
g(x, y, z) = 0
44. Lagrange Multipliers
Example: Find the largest volume possibe of a box with a
fixed total surface area.
The volume is V = xyz.
The constraint is 2xy + 2xz + 2yz = c or xy + xz + yz = c/2
Hence g(x, y, z) = xy + xz + yz – c/2 = 0. Set up the system:
gx = λfx y + z = λyz eq1
gy = λfy
gz = λfz
g(x, y, z) = 0
45. Lagrange Multipliers
Example: Find the largest volume possibe of a box with a
fixed total surface area.
The volume is V = xyz.
The constraint is 2xy + 2xz + 2yz = c or xy + xz + yz = c/2
Hence g(x, y, z) = xy + xz + yz – c/2 = 0. Set up the system:
gx = λfx y + z = λyz eq1
gy = λfy x + z = λxz eq2
gz = λfz
g(x, y, z) = 0
46. Lagrange Multipliers
Example: Find the largest volume possibe of a box with a
fixed total surface area.
The volume is V = xyz.
The constraint is 2xy + 2xz + 2yz = c or xy + xz + yz = c/2
Hence g(x, y, z) = xy + xz + yz – c/2 = 0. Set up the system:
gx = λfx y + z = λyz eq1
gy = λfy x + z = λxz eq2
gz = λfz x + y = λxy eq3
xy + xz + yz = c/2 eq4
g(x, y, z) = 0
47. Lagrange Multipliers
Example: Find the largest volume possibe of a box with a
fixed total surface area.
The volume is V = xyz.
The constraint is 2xy + 2xz + 2yz = c or xy + xz + yz = c/2
Hence g(x, y, z) = xy + xz + yz – c/2 = 0. Set up the system:
gx = λfx y + z = λyz eq1
gy = λfy x + z = λxz eq2
gz = λfz x + y = λxy eq3
xy + xz + yz = c/2 eq4
g(x, y, z) = 0
From eq1, we solve for z:
y + z = λyz
48. Lagrange Multipliers
Example: Find the largest volume possibe of a box with a
fixed total surface area.
The volume is V = xyz.
The constraint is 2xy + 2xz + 2yz = c or xy + xz + yz = c/2
Hence g(x, y, z) = xy + xz + yz – c/2 = 0. Set up the system:
gx = λfx y + z = λyz eq1
gy = λfy x + z = λxz eq2
gz = λfz x + y = λxy eq3
xy + xz + yz = c/2 eq4
g(x, y, z) = 0
From eq1, we solve for z:
y + z = λyz y = λyz – z
49. Lagrange Multipliers
Example: Find the largest volume possibe of a box with a
fixed total surface area.
The volume is V = xyz.
The constraint is 2xy + 2xz + 2yz = c or xy + xz + yz = c/2
Hence g(x, y, z) = xy + xz + yz – c/2 = 0. Set up the system:
gx = λfx y + z = λyz eq1
gy = λfy x + z = λxz eq2
gz = λfz x + y = λxy eq3
xy + xz + yz = c/2 eq4
g(x, y, z) = 0
From eq1, we solve for z:
y + z = λyz y = λyz – z
y = (λy – 1)z
50. Lagrange Multipliers
Example: Find the largest volume possibe of a box with a
fixed total surface area.
The volume is V = xyz.
The constraint is 2xy + 2xz + 2yz = c or xy + xz + yz = c/2
Hence g(x, y, z) = xy + xz + yz – c/2 = 0. Set up the system:
gx = λfx y + z = λyz eq1
gy = λfy x + z = λxz eq2
gz = λfz x + y = λxy eq3
xy + xz + yz = c/2 eq4
g(x, y, z) = 0
From eq1, we solve for z:
y + z = λyz y = λyz – z
y = (λy – 1)z
y
=z
λy – 1
51. Lagrange Multipliers
Example: Find the largest volume possibe of a box with a
fixed total surface area.
The volume is V = xyz.
The constraint is 2xy + 2xz + 2yz = c or xy + xz + yz = c/2
Hence g(x, y, z) = xy + xz + yz – c/2 = 0. Set up the system:
gx = λfx y + z = λyz eq1
gy = λfy x + z = λxz eq2
gz = λfz x + y = λxy eq3
xy + xz + yz = c/2 eq4
g(x, y, z) = 0
From eq1, we solve for z: From eq3, we solve for x:
y + z = λyz y = λyz – z x + y = λxy
y = (λy – 1)z
y
=z
λy – 1
52. Lagrange Multipliers
Example: Find the largest volume possibe of a box with a
fixed total surface area.
The volume is V = xyz.
The constraint is 2xy + 2xz + 2yz = c or xy + xz + yz = c/2
Hence g(x, y, z) = xy + xz + yz – c/2 = 0. Set up the system:
gx = λfx y + z = λyz eq1
gy = λfy x + z = λxz eq2
gz = λfz x + y = λxy eq3
xy + xz + yz = c/2 eq4
g(x, y, z) = 0
From eq1, we solve for z: From eq3, we solve for x:
y + z = λyz y = λyz – z x + y = λxy y = λxy – x
y = (λy – 1)z
y
=z
λy – 1
53. Lagrange Multipliers
Example: Find the largest volume possibe of a box with a
fixed total surface area.
The volume is V = xyz.
The constraint is 2xy + 2xz + 2yz = c or xy + xz + yz = c/2
Hence g(x, y, z) = xy + xz + yz – c/2 = 0. Set up the system:
gx = λfx y + z = λyz eq1
gy = λfy x + z = λxz eq2
gz = λfz x + y = λxy eq3
xy + xz + yz = c/2 eq4
g(x, y, z) = 0
From eq1, we solve for z: From eq3, we solve for x:
y + z = λyz y = λyz – z x + y = λxy y = λxy – x
y = (λy – 1)z y = (λy – 1)x
y
=z
λy – 1
54. Lagrange Multipliers
Example: Find the largest volume possibe of a box with a
fixed total surface area.
The volume is V = xyz.
The constraint is 2xy + 2xz + 2yz = c or xy + xz + yz = c/2
Hence g(x, y, z) = xy + xz + yz – c/2 = 0. Set up the system:
gx = λfx y + z = λyz eq1
gy = λfy x + z = λxz eq2
gz = λfz x + y = λxy eq3
xy + xz + yz = c/2 eq4
g(x, y, z) = 0
From eq1, we solve for z: From eq3, we solve for x:
y + z = λyz y = λyz – z x + y = λxy y = λxy – x
y = (λy – 1)z y = (λy – 1)x
y y
=z =x
λy – 1 λy – 1
55. Lagrange Multipliers
Example: Find the largest volume possibe of a box with a
fixed total surface area.
The volume is V = xyz.
The constraint is 2xy + 2xz + 2yz = c or xy + xz + yz = c/2
Hence g(x, y, z) = xy + xz + yz – c/2 = 0. Set up the system:
gx = λfx y + z = λyz eq1
gy = λfy x + z = λxz eq2
gz = λfz x + y = λxy eq3
xy + xz + yz = c/2 eq4
g(x, y, z) = 0
From eq1, we solve for z: From eq3, we solve for x:
y + z = λyz y = λyz – z x + y = λxy y = λxy – x
y = (λy – 1)z y = (λy – 1)x
y y
=z =x
λy – 1 λy – 1
Hence either x = z or λy – 1 = 0.
58. Lagrange Multipliers
Assume λy – 1 = 0, then from y = (λy – 1)z, we get y = 0.
Put this in eq1 and eq3, then x and z must be 0 also,
59. Lagrange Multipliers
Assume λy – 1 = 0, then from y = (λy – 1)z, we get y = 0.
Put this in eq1 and eq3, then x and z must be 0 also,
which is impossibe because of eq4 .
60. Lagrange Multipliers
Assume λy – 1 = 0, then from y = (λy – 1)z, we get y = 0.
Put this in eq1 and eq3, then x and z must be 0 also,
which is impossibe because of eq4 .
Therefore x = z.
61. Lagrange Multipliers
Assume λy – 1 = 0, then from y = (λy – 1)z, we get y = 0.
Put this in eq1 and eq3, then x and z must be 0 also,
which is impossibe because of eq4 .
Therefore x = z.
The system is symmetric so similar algebra yields x = y
and y = z.
62. Lagrange Multipliers
Assume λy – 1 = 0, then from y = (λy – 1)z, we get y = 0.
Put this in eq1 and eq3, then x and z must be 0 also,
which is impossibe because of eq4 .
Therefore x = z.
The system is symmetric so similar algebra yields x = y
and y = z.
Therefore x = y = z and the box must be a cube.
63. Lagrange Multipliers
Assume λy – 1 = 0, then from y = (λy – 1)z, we get y = 0.
Put this in eq1 and eq3, then x and z must be 0 also,
which is impossibe because of eq4 .
Therefore x = z.
The system is symmetric so similar algebra yields x = y
and y = z.
Therefore x = y = z and the box must be a cube.
