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## Lagrange multipliers and constrained optimization

Current time:0:00Total duration:7:50

# Lagrange multiplier example, part 1

## Video transcript

- [Instructor] So let's say you're running some kind of company and
you guys produce widgets. You produce some little trinket
that people enjoy buying and the main costs that
you have are labor, the workers that you have
creating these and steel. And let's just say that your labor costs are $20 per hour, $20 each hour and then your steel costs are $2,000. Keep the numbers kind of
related to each other. $2,000 for every ton of steel. And then you've had your
analysts work a little bit on trying to model the
revenues you can make with your widgets as a function of hours of labor and tons of steel. Now let's say the revenue model
that they've come up with. The revenue as a function
of hours of labor and then S for steel, let's say, is equal to about 100
times the hours of labor to the power of 2/3 multiplied by the tons of steel to the power of 1/3. If you put in a given amount of labor and a given amount of steel,
this is about how much money you're gonna expect to earn. And of course you wanna
earn as much as you can plus you actually have
a budget for how much you're able to spend on all these things. And your budget is $20,000. You're willing to spend $20,000 and you wanna make as
much money as you can, according to this model based on that. Now this is exactly the kind of problem that the Lagrange multiplier
technique is made for. We're trying to maximize
some kind of function and we have a constraint. Now right now the constraint
isn't written as a formula, but we can pretty easily
write as a formula. Because what makes up our budget? Well, it's gonna be the number of hours of labor multiplied by 20, so that's gonna be $20 per hour multiplied by the number
of hours you put in plus $2000 per tons of steel times the tons of steel that you put in. So the constraint is basically that you have to have these guys equal $20,000. I mean you could say less than, you could say you're not willing
to go any more than that, but intuitively and in reality, it's gonna be the case that in order to maximize your revenues, you
should squeeze every dollar that you have available and
actually hit this constraint. So this right here is the
constraint of our problem. And let's go ahead and
give this guy a name, the function that we're
dealing with a name. And I'm gonna call it g of h,
s, which is gonna be that guy. And now if you'll remember
in the last few videos, the way we visualize something like this, is to think about the set
of all possible input, so in this case, you might be
thinking about the h s plane, the number of hours of labor on one axis, the number of tons of steel on another. And this constraint, well, in this case, it's a linear function, so this
constraint is gonna give us some kind of line that tells
us which pairs of s and h are gonna achieve that constraint. And then the revenue function that we're dealing with will
have certain contours. Maybe revenues of $10,000 have a certain contour that looks like this and revenues of $100,000
have a certain contour that looks like this. But what we want is to find which value is barely touching the constraint curve. Just tangent to it at a given point cause that's gonna be the contour line, where if you up the value
by just a little bit, it would no longer
intersect with that curve, there would no longer be values of h and s that satisfy this constraint. And the way to think about
finding that tangency is to consider the vector perpendicular to the tangent line to
the curve at that point, which fortunately is
represented by... (mumbles) Make some room for myself here. Represented by the gradient,
the gradient of our r function, the function whose contour's
this is, the revenue. And what it means for this to be tangent to the constraint line is that there's gonna be another vector, the gradient of g, of
our constraint function, that points in the same direction, that's proportional to that. And typically the way you write this is to say that the
gradient of this function is proportional to the gradient of g and this proportionality constant is called our Lagrange multiplier. It's called the Lagrange multiplier. So, let's go ahead and
start working it out. Let's first compute the gradient of r. So the gradient of r is gonna
be the partial derivative of r with respect to its first
variable, which is h, so partial derivative with respect to h. And the second component
is its partial derivative with respect to that second
variable, s, with respect to s. And in this case, that
first partial derivative, if we treat h as a variable
and s as a constant, then that 2/3 gets brought down, so that'll be 100 times 2/3
times h to the power of, well, we've gotta subtract one from 2/3 when we bring it down, so
that'll be negative 1/3 multiplied by s to the 1/3. And then the second component here, the partial derivative with respect to s, is gonna be 100 times, well now, by treating s as the variable, we take down that 1/3, so that's 1/3, h to the 2/3 just looks like a constant as far as s is concerned
and then we take s to the 1/3 minus 1, which is negative 2/3. Great, so that's the gradient of R. And now we need the gradient of g. And that one's a lot easier, actually, cause g is just a linear function. So when we take the gradient of g, which is its partial derivative, with respect to h, partial
h, and its partial derivative with respect to s, partial s. Well, the partial with
respect to h is just 20. The function looks like
20 times h plus something that's a constant so
that ends up being 20. And then the partial with respect to s, likewise is just 2,000,
cause it's just some constant multiplied by s plus
a bunch of other stuff that looks like constants. So that's great, and this means when we set the gradient of
r equal to the gradient of g, the pair of equations that we get, here let me just write it all out again, is we have this top one,
which I'll call 200/3 times, and let's go ahead and
do a little simplifying while I'm rewriting things here. So h to the 1/3 is really one
over, h to the negative 1/3, sorry, is one over h to the
1/3, that's s to the 1/3. So all of this, that first component, is being set equal to the first component of the gradient of g, which is 20 times lambda, times this
Lagrange multiplier, cause we're not setting the
gradients equal to each other, we're just setting them
proportional to each other. So that's the first equation
and then the second one, I'll go ahead and do some simplifying while I rewrite that one also. That's gonna be 100/3
and then h to the 2/3 so times h to the 2/3
divided by s to the 2/3 cause s to the negative 2/3 is the same as 1 over s to the 2/3. All of that is equal
to 2,000 times lambda. And the important thing
it's that same lambda, because the entire vector
has to be proportional. And I think right here's
probably a pretty good point to stop and in the next video, I'll go ahead and work through the details and we'll land on a solution.