1. ## Lagrange Multipliers

F(x,y,z)=4x^2+y^2+5z^2. Use Lagrange Multipliers to find the point on the plane 2x+3y+4z=12 at which F(x,y,z) is the least value.

Δf(x,y,z)= 8xi+2yj+10zk
γΔg(x,y,z)= 2λ+3λ+4λ

8x=2λ
3y=3λ
10z=4λ

Im stuck

2. Originally Posted by JJ007
F(x,y,z)=4x^2+y^2+5z^2. Use Lagrange Multipliers to find the point on the plane 2x+3y+4z=12 at which F(x,y,z) is the least value.

Δf(x,y,z)= 8xi+2yj+10zk
γΔg(x,y,z)= 2λ+3λ+4λ

8x=2λ
3y=3λ
10z=4λ

Im stuck
use your fourth equation and try to work out by multiplying each equation by the missing variables.

3. Originally Posted by Warrenx
use your fourth equation and try to work out by multiplying each equation by the missing variables.
Fourth equation, you mean γΔg(x,y,z)= 2λ+3λ+4λ? How do I solve for λ?

4. Originally Posted by JJ007
F(x,y,z)=4x^2+y^2+5z^2. Use Lagrange Multipliers to find the point on the plane 2x+3y+4z=12 at which F(x,y,z) is the least value.

Δf(x,y,z)= 8xi+2yj+10zk
γΔg(x,y,z)= 2λ+3λ+4λ

8x=2λ
3y=3λ
10z=4λ

Im stuck
might be wrong, but here goes:
g= lambda
since 8x = 2g
that means that 8 = 2g/x
which means that 12 = 6g/2x
so 6g/2x = 2x +3y + 4z
6g = 2x(2x + 3y + 4z)
6g = 1/2g(1/2g + 3g + 4z)
12g^2 -3.5g = 4z
10z = 4g
z = 2/5g
8/5g + 3.5g =12g^2
1.6g + 3.5g = 5.1g
5.1g/g
5.1 = 12g

go from there.. i think

5. Hello, JJ007!

$f(x,y,z)\:=\:4x^2+y^2+5z^2$

Use Lagrange Multipliers to find the point on the plane $2x+3y+4z\:=\:12$
at which $f(x,y,z)$ is a minimum.

We have: . $F(x,y,z,\lambda) \;=\;(4x^2 + y^2 + 5z^2) + \lambda(2x + 3y + 4z - 12)$

Set the partial derivatives equal to zero.

. . $\begin{array}{ccccccc}F_x \:=\:8x + 2\lambda \:=\:0 & \Rightarrow & x \:=\:-\dfrac{1}{4}\lambda & [1] \\ \\[-3mm]

F_y \:=\:2y + 3\lambda \:=\:0 & \Rightarrow & y \:=\:-\dfrac{3}{2}\lambda & [2]\\ \\[-3mm]

F_z \:=\:10z + 4\lambda \:=\:0 & \Rightarrow & z \:=\:-\dfrac{2}{5}\lambda & [3]\end{array}$

. . $F_{\lambda} \:=\:2x + 3y + 4z - 12 \:=\:0 \qquad \quad\;\;\; [4]$

Substitute [1], [2], [3] into [4]:

. . $2\left(-\frac{1}{4}\lambda\right) + 3\left(-\frac{3}{2}\lambda\right) + 4\left(-\frac{2}{5}\lambda\right) -12\:=\: 0$

. . $-\frac{66}{10}\lambda \:=\:12 \quad\Rightarrow\quad \lambda \:=\:-\frac{20}{11}$

Substitute into [1], [2], [3]:

. . $\begin{Bmatrix}x &=& \left(-\frac{1}{4}\right)\left(-\frac{20}{11}\right) &=& \dfrac{5}{11} \\ \\[-3mm]
y &=& \left(-\frac{3}{2}\right)\left(-\frac{20}{11}\right) &=& \dfrac{30}{11} \\ \\[-3mm]
z &=& \left(-\frac{2}{5}\right)\left(-\frac{20}{11}\right) &=& \dfrac{8}{11} \end{Bmatrix}$