Math Help - Applications to Partial Differentiation

1. Applications to Partial Differentiation

I have a few questions that I need help with.

1. Find the critical points of F(x,y) = e^(2x)cos(y)

Fx = 2e^(2x)cos(y)
Fy = -e^(2x)sin(y)

2e^(2x)cos(y)= 0
-e^(2x)sin(y) = 0

e^2x will never be 0 so you can divide by that and you will get cos(y) = 0 and sin(y) = 0 but since their is no point where both sin and cos are 0 then their are no critical points? Did I do this right?

2. Use Lagrange Multipliers to find the max and min values of the function subject to the given constraint.

f(x,y,z) = xyz; x^2 + y^2 + z^2 = 12

The gradiant of f(x,y,z) = <yz, xz, xy>
The gradient of the constraint = <2x, 2y, 2z>

2x = λyz
2y = λxz
2z = λxy
x^2 + y^2 + z^2 = 12

x = 1/2λyz
y = 1/2λxz
z = 1/2λxy

x^2 + y^2 + z^2 = 12
(1/2λyz)^2 + (1/2λxz)^2 + (1/2λxy)^2 = 12
1/4λ^2y^2z^2 + 1/4λ^2x^2z^2 + 1/4λ^2x^2y^2 = 12
1/4λ^2(y^2z^2 + x^2z^2 + x^2y^2) = 12

I have no clue where to go from here.

3. Find the Volume of the largest rectangular box in the first octant with three faces in the coordinates and one vertex in the plane 2x+y + 3z = 6.

V(x,y,z) = xyz; 2x+y+3z = 6

The gradiant of V(x,y,z) = <yz, xz, xy>
The gradient of the constraint = <2, 1, 3>

2λ = yz
λ = xz
3λ = xy
x^2 + y^2 + z^2 = 12

Given the second equation the first and third equations becmoe
2xz = yz and 3xz = xy respectively.

Solving the second equation you get
2xz - yz = 0
z(2x-y) = 0
z = 0 2x - y = 0
y = 2x

plugging this into the constraint you get:
2x + 2x = 6
4x = 6
x = 3/2

this gets you y = 3 so the first point is (3/2,3,0)

Solving the third equation you get
3xz - xy = 0
x(3z-y) = 0
x = 0 3z - y = 0
y = 3z

plugging this into the constraint you get:
3z+3z = 6
6z = 6
z = 1

this gets you y = 3 so you get a point of (0, 3, 1)

V(3/2,3,0) = 0
v(0,3,1) = 0

what did I do wrong?

2. 2. Use Lagrange Multipliers to find the max and min values of the function subject to the given constraint.

f(x,y,z) = xyz; x^2 + y^2 + z^2 = 12

The gradiant of f(x,y,z) = <yz, xz, xy>
The gradient of the constraint = <2x, 2y, 2z>

2x = λyz
2y = λxz
2z = λxy
x^2 + y^2 + z^2 = 12

x = 1/2λyz
y = 1/2λxz
z = 1/2λxy

x^2 + y^2 + z^2 = 12
(1/2λyz)^2 + (1/2λxz)^2 + (1/2λxy)^2 = 12
1/4λ^2y^2z^2 + 1/4λ^2x^2z^2 + 1/4λ^2x^2y^2 = 12
1/4λ^2(y^2z^2 + x^2z^2 + x^2y^2) = 12

I have no clue where to go from here.
I will start from the beginning.

$f(x,y,z)=xyz$

$g(x,y,z)=x^{2}+y^{2}+z^{2}=12$

${\nabla}f(x,y,z)={\lambda}{\nabla}g(x,y,z)$

${\nabla}f=yzi+xzj+xyk$

${\nabla}g=2xi+2yj+2zk$

Equate:

$yz=2x{\lambda}..[1], \;\ xz=2y{\lambda}...[2], \;\ xy=2z{\lambda}...[3]$

${\lambda}=\frac{yz}{2x}, \;\ {\lambda}=\frac{xz}{2y}, \;\ {\lambda}=\frac{xy}{2z}$

From equations [1] and [2]:

$\frac{yz}{2x}=\frac{xz}{2y}\rightarrow y=x, \;\ y=-x$

From [2] and [3]:

$\frac{xz}{2y}=\frac{xy}{2z}\rightarrow z=y, \;\ z=-y$

Therefore, $x=y=z$

Sub into the constraint and we get $3x^{2}=12$

$x=2$

Of course, that means $y=2, \;\ z=2$

3. 3. Find the Volume of the largest rectangular box in the first octant with three faces in the coordinates and one vertex in the plane 2x+y + 3z = 6.

V(x,y,z) = xyz; 2x+y+3z = 6

The gradiant of V(x,y,z) = <yz, xz, xy>
The gradient of the constraint = <2, 1, 3>

2λ = yz
λ = xz
3λ = xy
x^2 + y^2 + z^2 = 12

Given the second equation the first and third equations becmoe
2xz = yz and 3xz = xy respectively.

Solving the second equation you get
2xz - yz = 0
z(2x-y) = 0
z = 0 2x - y = 0
y = 2x

plugging this into the constraint you get:
2x + 2x = 6
4x = 6
x = 3/2

this gets you y = 3 so the first point is (3/2,3,0)

Solving the third equation you get
3xz - xy = 0
x(3z-y) = 0
x = 0 3z - y = 0
y = 3z

plugging this into the constraint you get:
3z+3z = 6
6z = 6
z = 1

this gets you y = 3 so you get a point of (0, 3, 1)

V(3/2,3,0) = 0
v(0,3,1) = 0

what did I do wrong?
Maximize $V=xyz$ subject to $2x+y+3z=6$

$z=\frac{-(2x+y-6)}{3}$

$V=\frac{-2x^{2}y}{3}-\frac{xy^{2}}{3}+2xy$

$V_{x}=\frac{-4xy}{3}-\frac{y^{2}}{3}+2y=\frac{-(4x+y-6)y}{3}=0$

$V_{y}=\frac{-2xy}{3}-\frac{2x^{2}}{3}+2x=\frac{-2x(x+y-3)}{3}=0$

$x+y-3=0, \;\ and \;\ 4x+y-6=0$

Solving we see $x=1, \;\ y=2$

Now, do the second derivative test to see:

$V_{xx}V_{yy}-V_{xy}^{2}$

$V_{xx}=\frac{-4y}{3}$

$V_{yy}=\frac{-2x}{3}$

$V_{xy}^{2}=\frac{4(2x+y-3)^{2}}{9}$

$V_{xx}V_{yy}-V_{xy}^{2}=\frac{-4(4x^{2}+2x(y-6)+(y-3)^{2})}{9}$

Subbing in x=1 and y=2, we get 4/3.

Since 4/3>0, then we have a relative minimum at (1,2)

Check me out. Make sure I didn't make a mistake.