# Lagrange multiplier example, part 2

- [Instructor] So where we left off we have these two
different equations that we wanna solve
and there's three unknowns.
There's s, the tons of steel that you're using,
h the hours of labor, and then lambda,
this Lagrange Multiplier we introduced
that's basically a proportionality constant
between the gradient vectors of the revenue function
and the constraint function.
And always the third equation that we're dealing with here
to solve this, is the constraint itself.
That gives us another equation that'll help solve for
h and s, and ultimately lambda,
if that's something that you want as well.
So it's kind of a first pass here,
I'm gonna do a little more simplifying
but I'm gonna make a substitution
that'll make this easier for us.
So I see s over h here, and they're both to the same power,
so I feel it might be a little bit easier
if I just substitute u in for s divided by h.
And what that'll let me do, is rewrite
this first equation here as 200 thirds,
200 thirds times u to the power of one third.
And that's equal to 20 times lambda.
And then likewise, what that means for this guy,
is, well, this is h over s, not s over h,
so that one's gonna be 100 thirds,
not times u to the power of two thirds,
but times u to the negative two thirds
because we swapped the h and s here.
So that's u to the negative two thirds
and this is just to make it a little bit cleaner, I think.
We kind of want to treat h and s in the same package.
Now I'm gonna go ahead and put all the constants together
and then I'm gonna take this guy
and multiply it by three divided by 200,
multiply both sides of that just to cancel out
what's on the left side here.
And what that's gonna give me, and I'll go ahead and
write it over here, kind of all over the place,
u to the one third is equal to,
let's see three over 200, so that 20
almost cancels out with the 200, it just leaves a 10,
so that's gonna give me three tenths
of lambda.
And then similarly over here,
I'm gonna take this full equation
and multiply it by three over 100
and what that's gonna leave me with
is that u to the negative two thirds,
u to the negative two thirds is equal to,
let's see this 2,000 when we divide it by 100
becomes 20, and that 20 times three is 60,
so that'll be 60 times lambda.
Alright, so now I want a way to simplify each of these,
and what I notice is they look pretty similar on each side,
you know it's something related to u equal to lambda,
so if I can get this in a form
where I'm really isolating u, that would be great.
The way I'm gonna do this is I'm gonna multiply
each one of them by u to the two thirds,
so I'm gonna multiply it into this guy
and I'm gonna multiply it into that guy,
because on the top, it's gonna turn this into just u,
which will be nice, and on the bottom
it'll cancel out that u entirely.
So it feels like it'll make both of these nicer,
even though it might make the right side a little uglier,
those right sides will still kind of be the same,
and we'll take advantage of that.
So, when I do that to the top part,
like I said, that u to the one third
times u to the two thirds ends up being u,
and then on the right side we have
our three tenths lambda,
but now u to the two thirds.
And then on the bottom here, we,
when I multiply it by u to the two thirds,
the right side becomes one, 'cause it cancels out
with u to the negative two thirds,
and the right side is 60 times lambda
times u to the two thirds.
Now these right sides look very similar,
and the left sides are quite simple.
So I'm gonna multiply this top one
by whatever it takes to make it look exactly
like that right side.
So in this case I'm gonna multiply that top by 10,
which will get it to three, and then by another 20
to make that constant 60, so I'm gonna multiply this
entire top equation by 200,
and what that gives me is that 200 times u
is equal to 60 times lambda
times u to the two thirds.
And now these two equations, these two equations
have the same right side.
So this is the same as saying,
200 times u is equal to, well, one.
Because each one of those expressions
equals the same complicated thing.
And now 200 times u, well that's s divided by h.
So this is the same thing as saying 200
times s over h equals one,
which we can write much more simply as h
is equal to 200
times s.
Great, so I'm gonna go ahead and circle that,
h is equal to 200 times s.
And now what we apply that to
is the constraint, is the 200 times h
plus 2,000 times s equals our budget.
We'll go ahead and kind of write that down again.
That are 20 times h, I think, 20 times the hours of labor
plus \$2,000 per ton of steel
is equal to our budget
of \$20,000, and now we can just substitute in.
Instead of h I'm gonna write 200 s, so that's 200,
sorry, 20 times 200 s,
200 s,
plus 2,000
times s
is equal to 20,000.
And now this right side, 20 times 200 is equal to 4,000,
and I'm just gonna go ahead and kind of write
so this here
is 4,000 s, so the entire right side of the equation
simplifies to 6,000,
6,000 times s
is equal to 20,000
and when those cancel out, what that gives us
is s is equal to 20 divided by six,
which is the same as 10 divided by three.
So that's how many tons of steel we should get.
S is 10 over three, then when we apply that
to the fact that h is 200 times s,
that's gonna mean that h is equal
to 200 times that value,
10 over three, which is equal to
2,000,
2,000 thirds,
2,000 thirds, that's how many hours of labor we want.
So, evidently in our original problem,
where we have this model for our revenue function
for our Widgets with \$20 per hour of labor,
and \$2,000 per ton of steel,
with a budget of \$20,000, the allocation
that you should make is to buy 10 thirds of a ton of steel
and 2,000 thirds hours of labor.