On the other hand, you don't actually need Lagrange multipliers!
Writing *v* for *g*(*x*, *y*),
if the constraint is *v* = 0 (or *v* equals any constant),
then differentiate this to get d*v* = 0.
(In fact, you could take any equation and just differentiate both sides.)
Then if you try to solve the system of equations
consisting of d*u* = 0 and d*v* = 0
for the differentials d*x* and d*y*,
you should immediately see that d*x* = 0 and d*y* = 0
is a solution.
However, if you actually go through the steps
of solving this as a system of linear equations
(which you can always do because differentials are always linear
in the differentials of the independent variables),
you'll find that at some point you need to divide by some quantity
involving *x* and *y*,
which is invalid if that quantity is zero!
So, setting whatever you divide by to zero
and combining that with the constraint equation *v* = 0,
you get two equations to solve for the two variables *x* and *y*.
This will give you the other critical points to check for extreme values.

Here is a typical problem:
The hypotenuse of a right triangle (maybe it's a ladder leaning against a wall)
is fixed at 20 feet,
but the other two sides of the triangle could be anything.
Still, since it's a right triangle,
we know that *x*^{2} + *y*^{2} =
20^{2},
where *x* and *y* are the lengths of legs of the triangle.
(If we think of *x* and *y* as independent variables,
then this equation is our constraint.)
Differentiating this,
2*x* d*x* + 2*y* d*y* = 0.
Now suppose that we want to maximize or minimize the area of this triangle.
Since it's a right triangle,
the area is *A* = ½*x**y*,
so d*A* =
½*y* d*x* + ½*x* d*y*.
If this is zero,
then
½*y* d*x* +
½*x* d*y* =
0,
to go along with the other equation
2*x* d*x* + 2*y* d*y* = 0.

The equations at this point are linear in the differentials
(and they always must be),
so think of this is a system of linear equations
in the variables d*x* and d*y*.
There are various methods for solving systems of linear equations;
I'll use the method of addition aka elimination,
but any other method should work just as well.
So ½*y* d*x* +
½*x* d*y* =
0
becomes
2*x**y* d*x* +
2*x*^{2} d*y* =
0
(multiplying both sides by 4*x*),
while
2*x* d*x* + 2*y* d*y* = 0
becomes
2*x**y* d*x* +
2*y*^{2} d*y* =
0
(multiplying both sides by *y*).
Subtracting these equations gives
(2*x*^{2} −
2*y*^{2}) d*y* =
0,
so either d*y* = 0
or *x*^{2} = *y*^{2}.
Now, *x* and *y* can change freely as long as they're positive,
but we have limiting cases:
*x* → 0^{+} and *y* → 0^{+}.
Since *x*^{2} + *y*^{2} = 400,
we see that *x*^{2} → 400, so *x* → 20,
as *y* → 0.
Similarly, *y* → 20 as *x* → 0.
In those cases, *A* = ½*x**y* → 0.
On the other hand, if *x*^{2} = *y*^{2},
then *x* = *y*,
so *x*, *y* =
10√2,
since *x*^{2} + *y*^{2} = 400.
In that case, *A* = ½*x**y* = 100.

So the largest area is 100 square feet, and while there is no smallest area, the area can get arbitrarily small with a limit of 0.

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This web page was written in 2015 and 2016 by Toby Bartels, last edited on 2016 April 26. Toby reserves no legal rights to it.

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