# Applications to Partial Differentiation

• Jul 9th 2008, 11:22 AM
algebrapro18
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?
• Jul 9th 2008, 01:00 PM
galactus
Quote:

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.

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

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

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

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

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

Equate:

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

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

From equations [1] and [2]:

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

From [2] and [3]:

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

Therefore, $\displaystyle x=y=z$

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

$\displaystyle x=2$

Of course, that means $\displaystyle y=2, \;\ z=2$
• Jul 9th 2008, 01:38 PM
galactus
Quote:

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 $\displaystyle V=xyz$ subject to $\displaystyle 2x+y+3z=6$

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

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

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

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

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

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

Now, do the second derivative test to see:

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

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

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

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

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

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.