We're asked to find the absolute extrema
of f of x comma y equals four x plus five y
within the domain x squared plus y squared
less than or equal to four squared.
Looking at the notes below,
the first step is to find the critical points
that lie in the bounded region
and determine the function values at these points.
And then step two is to find the extrema
of the function on the boundary.
We already solved this problem
using the steps outlined here,
but in this video, for step two,
to find the extrema of the function on the boundary,
we'll be using Lagrange multipliers
instead of Calculus I techniques.
To begin, though, we need to find the critical points
of the function f of x comma y in the bounded region.
The critical points occur
where both first order partial derivatives
are equal to zero or undefined,
which means the first step is to find
the partial of f with respect to x
and the partial of f with respect to y.
To find the partial of f with respect to x,
we differentiate f with respect to x,
treating y as a constant, which gives us four.
To find the partial of f with respect to y,
we differentiate f with respect to y,
treating x as a constant, which gives us five.
Now we set these equal to zero and solve
as a system of equations.
We notice here both are constants,
which are never equal to zero and always defined,
which means f of x comma y has no critical points
in the bounded region.
The next step is to find the extrema
of the function on the boundary,
and, again, for this video, we will do this
using the method of Lagrange multipliers.
To review the method of Lagrange multipliers,
it's given that f and g are differentiable
on a specified region,
and the gradient of g
does not equal the zero vector on the curve
g of x comma y equals zero.
To locate the max and mins of f
subject to the constraint of g,
we follow steps one and two.
For the first step, we determine x, y, and lambda
that satisfy the equations,
the gradient of f equals lambda
times the gradient of g,
and g of x comma y equals zero.
Looking at this equation here, this tells us
the max and mins occur when the two gradients
are parallel to each other,
and this occurs where
the level curves of f and g are tangent,
which we'll take a look at graphically in just a moment.
These two equations result in this system of equations,
which we solve for x, y, and lambda.
And in step two we determine which solutions produce
the maximum and minimum function value.
So going back to our question,
the objective function is f of x comma y.
The constraint is g of x comma y,
which on the boundary is x squared plus y squared equals 16.
Because we need g of x comma y to be equal to zero,
we'll say g of x comma y equals x squared
plus y squared minus 16 equals zero.
Before we set up the system of equations, though,
let's look at this graphically
by looking at the level curves of f and g.
The level curves of f of x comma y are blue,
the level curve of the constraint is the red circle.
Notice how the level curves are tangent at this point,
and these two vectors here
are the gradient of g and the gradient of f.
Notice how they are parallel,
again because the level curves are tangent at this point,
which means at this point
and this point down here
is where we will find the max and mins
of the objective function, f of x comma y,
using the method of Lagrange multipliers.
So let's go and set up our system of equations.
We already know the partial of f with respect to x
is equal to four, and the partial of f with respect to y
is equal to five.
We also need to find the partial of g with respect to x
and the partial of g with respect to y.
The partial of g with respect to x is two x,
the partial of g with respect to y is two y,
which means we need to solve the system of equations,
four equals lambda times two x,
five equals lambda times two y,
and the constraint x squared
plus y squared minus 16 equals zero.
Let's solve this first equation for lambda
by dividing both sides by two x,
which gives us lambda
equals four divided by two x.
Simplifying, we have two divided by x.
Solving the second equation for lambda,
we have lambda
equals five divided by two y.
Because these are both equal to lambda, it follows
that two divided by x
must equal five divided by two y.
Let's solve this equation for y.
Because we have a proportion, we can cross multiply.
Two times two y equals five times x,
which means four y equals five x.
Dividing both sides by four, we know y is equal to 5/4 x.
Now we'll use this equation
and perform a substitution into the constraint,
which will give us one equation with one variable.
Substituting 5/4 x for y, we have the equation x squared
plus 5/4 x squared
minus 16 equals zero.
Let's go ahead and solve this equation for x
on the next slide.
We have x squared
plus here we'll have 25/16 x squared.
Let's add 16 to both sides of the equation.
And then one x squared plus 25/16 x squared
is equal to 41/16 x squared.
Now let's multiply both sides of the equation
by the reciprocal of 41/16,
which is 16/41.
This gives us x squared
equals, 16 times 16 is equal to 256,
we have 256/41.
And now we take the square root
of both sides of the equation.
We already know the square root of 256 is equal to 16,
which means x is equal to
plus or minus 16 divided by the square root of 41.
Now we'll use the equation y equals 5/4 x
to determine the corresponding y-coordinates.
When x is equal to positive
16 divided by the square root of 41,
we have y equals 5/4
times 16 divided by the square root of 41.
Simplifying before multiplying,
there's one four in four and four fours in 16,
this gives us
y equals 20 divided by the square root of 41.
And then when x is equal to
negative 16 divided by the square root of 41,
y is equal to 5/4
times negative 16 divided by the square root of 41,
which is negative 20
divided by the square root of 41.
To determine the extrema of f on the boundary,
we need to evaluate f
at the point 16 divided by square root of 41
comma 20 divided by square root of 41,
and negative 16 divided by square root of 41
comma negative 20 divided by square root of 41.
And let's say some time I've already done this.
When both the x and y coordinates are positive,
we get four square root of 41,
which is approximate 25.6125.
When both coordinates are negative,
we get a function value of negative four square root of 41,
which is approximately negative 25.6125.
So we know the greatest value is the absolute max
and the least value is the absolute minimum,
which means this is the absolute max
and this is the absolute min.
Going back to the graph one last time,
this is the point where we find the absolute maximum
of f of x comma y,
and this is the point where we find
the absolute minimum of f of x comma y.
Notice that both points,
the gradient of g and the gradient of f
are parallel and the level curves are tangent.
To record the results,
we'll use the exact values,
and we'll say the absolute minimum
is negative four square root of 41
and the absolute max is positive four square root of 41.
I hope you found this helpful.