Welcome to another example of lagrange multipliers.
This time with three independent variables.
In order to determine the maximum or minimum function values of f.
Under the constraint of g.
We're gonna set the gradient of f equal to lambda times the gradient of g.
That will give us the first three equations as we see here.
Then the fourth equation will be the constraint g equal to zero.
Once we determine the values of x, y, z, and lambda.
Which determine which of these produce a max and min function value.
Let's go and take a look at our example.
So here we want to determine the minimum function value of f.
Subject to two x minus three y minus four z equal to 49.
So the first thing we should do is probably recognize that g of x, y, z.
Would have to be equal to two x
minus three y
minus four z
minus 49
equal zero.
So now we'll set the gradient of f.
Equal to lambda times the gradient of g.
That will give us these first three equations.
So the partial derivative of f with respect to x. That's gonna give us four x.
Equals lambda times the partial
of g with respect to x that would be two.
Second equation will be the partials respect to y.
So the partial of f respect to y would be two y.
Equals lambda times the partial
of g with respect to y
that's gonna be negative three.
And then we'll have the partial of f respect to z that will be six z.
Equals lambda times the partial of g with respect to z.
That will be negative four.
And there are constraint is the fourth equation.
This will be equation one, equation two, equation three, and equation four.
So what we do here is solve
equations one, two, and three for x, y, and z.
And then perform substitution into equation four.
So for equation one if we divide both sides by four.
We would have x equals one half lambda.
In equation two if we divide both sides by two we would have
y equals negative three halves lambda.
And then for equation three if we divide both sides by six
we would have z equals negative two third
lambda.
So what were gonna do now is use equation four replace x, y, and z.
In terms of lambda
Let's do that on the next slide.
Okay so here's x, y, and z in terms of lambda.
Here's equation four.
And then here's the substitution notice everything is in terms of lambda now.
So we have one equation with one unknown.
So now we can solve for lambda.
Here we would have lambda.
Plus
nine halves lambda.
Plus eight third.
Lambda.
Let's go and add 49 to both sides.
Let's go and get a common denominator of six.
So this would be
six plus
multiply this by three over three that will be 27 over six.
Multiply this by two over two that's 16.
49 six lambda.
Go ahead and multiply by the reciprocal.
And lambda equal six.
Now that we have lambda let's go back to the previous slide and determine x, y, and z.
So now that we know that lambda is equal to six.
One half time six that means x would be three.
Negative three halves time six that would give us
y equals negative nine.
And for z negative two third times six that will give us negative four.
So notice we only have one point to consider.
Which means it would be our minimum value.
Go ahead and determine what that function value would be.
So here's our original function.
Here the values that we found for x, y, and z.
We sub these values into f were gonna get 147
as our minimum function value.
And that's going to do it for this video. I hope you found it helpful.