1. ## Lagrange multiplier

1)Use Lagrange multipliers to find shortest distance from (10,-4,9) to the plane 7x+8y-10z=4
h=lambda(backwards h)

- I tried to use d=sqrt((x-10)^2+(y+4)^2+(z-9)^2) then split each one up to 2(x-10)=7h , 2(y+4)=8h , 2(z-9)=-10h
- I then solved for x,y,z and plugged into the original equation of 7x+8y-10z=4 got I got h=112/213 which is not correct and I'm not sure where I messed up.

2)Use lagrange multipliers to find volume of largest rectangular box in first octant with 3 faces in coordinate planes and one vertex in plane x+5y+2z=30.

-I use V=xyz and got yz=h , xz=5h , xy=2h , x+5y+2z=30
- Not sure where to go after this.

2. For the first one, you can easily check your answer by using the distance from a point to a plane formula. But first....the long way.

$f(x,y,z)=(x-10)^{2}+(y+4)^{2}+(z-9)^{2}$

${\nabla}f(x,y,z)=2(x-10)i+2(y+4)j+2(z-9)k$

Subject to the constraint $7x+8y-10z-4=0$

${\lambda}{\nabla}g(x,y,z)={\lambda}(7i+8j-10k)$

So, $2(x-10)i+2(y+4)j+2(z-9)k={\lambda}(7i+9j-10k)$

$2(x-10)=7{\lambda}, \;\ 2(y+4)=8{\lambda}, \;\ 2(z-9) = -10{\lambda}$

So, we have:

${\lambda}=\frac{2(x-10)}{7}, \;\ {\lambda}=\frac{y+4}{4}, \;\ {\lambda}=\frac{-(z-9)}{5}$

From the first two equations we get:

$\frac{2(x-10)}{7}=\frac{y+4}{4}$

where $y=\frac{8x}{7}-\frac{108}{7}$

From the first and third we get:

$\frac{2(x-10)}{7}=\frac{-(z-9)}{5}$

where $z=\frac{-10x}{7}+\frac{163}{7}$

Sub these into the constraint and get:

$\frac{213x}{7}-\frac{2522}{7}=0$

Solving for x we get $x=\frac{2522}{213}$

Which leads to $y=\frac{-404}{213}, \;\ z=\frac{1357}{213}$

Sub these into g(x,y,z):

$\sqrt{(x-10)^{2}+(y+4)^{2}+(z-9)^{2}}$

We get $\boxed{\frac{56}{\sqrt{213}}}$

Check it with the distance formula:

$\frac{|7(10)+8(-4)-10(9)-4|}{\sqrt{7^{2}+8^{2}+(-10)^{2}}}=\boxed{\frac{56}{\sqrt{213}}}$

Check....

3. worked out perfectly. When I plugged x,y,z into distance I got (56*sqrt(213)) / 213. Thanks for the steps that helped me a lot.