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?