300. Lagrange's multiplier. Examples and problems.
by Virgil Nicula, Jul 22, 2011, 12:27 AM
PP1. Find the maximum value of the expression 
, where 
and 
.
Proof. We introduce a new variable
(Lagrange's multiplier) so we study the function 
.
The partial derivatives with respect to
and 
are 
, 
, 
and 
respectively.
Solving this system for
, 
and 
yields 
and substituting this into the expression we want to maximize gives 
.
An easy extension. Consider the numbers
such that 
can be the sidelengths of a triangle. Prove that for any
we have 
. For 
obtain that 
.
Proof. Denote
and apply the method of Lagrange's multipliers. Consider the function 
.
From the system
obtain that 
and
. Prove easily the identity 
. In conclusion,
, i.e. 
.
Remark. Apply the identity
in 
.
PP2. If
and 
, then find the minimum value of 
(British Math Olympiad).
Proof 1 (with the Lagrange's multipliers). Denote
. Since 
consider
for any . Now can take all of the partial derivatives (in terms of , , and )
and set them to zero to get the minimum (or maximum). Taking them in terms of , , and respectively we get
. Take the first minus, the second and the second minus the third to get ![$\left\{\begin{array}{c}
(x - y)\left[2 + 3\lambda (x + y + z)\right] = 0\\\\
(y - z)\left[2 + 3\lambda (x + y + z)\right] = 0\end{array}\right\|$](//latex.artofproblemsolving.com/2/a/1/2a1798744c4ebd1a538f6ce5bc69e57b1a35200d.png)
.
Taking cases from these equations, we find that the solutions of this system are
, and permutations.
They all give
, so 
or 
. But notice that 
, 
and ![$z = \dfrac{\sqrt[3]{7}}{4}$](//latex.artofproblemsolving.com/0/e/1/0e14db59e28a1fa0547aff8e3576d659f05f249d.png)
for the problem conditions, and they give 
. Thus 
,
at
, and permutations.
Proof 2.![$x^3 + y^3 + z^3 - 3xyz = 1 \Leftrightarrow (x + y + z)[x^2 + y^2 + z^2 - (xy + yz + xz)] = 1$](//latex.artofproblemsolving.com/5/e/4/5e4342198ff9d9b8f82bb6bea2f478964d92e889.png)
. So 
.
Note that
, so 
. Plugging this back in we find
. Let 
. Now we have the task of minimizing 
. I'm sure there are other ways, but I took the
derivative of this to find
, or 
. Note 
gives non-real solutions, so 
and 
. Thus the answer is 
.
, where 
and 
.Proof. We introduce a new variable
(Lagrange's multiplier) so we study the function 
.The partial derivatives with respect to
and 
are 
, 
, 
and 
respectively.Solving this system for
, 
and 
yields 
and substituting this into the expression we want to maximize gives 
.An easy extension. Consider the numbers
such that 
can be the sidelengths of a triangle. Prove that for any we have 
. For 
obtain that 
.Proof. Denote
and apply the method of Lagrange's multipliers. Consider the function 
.From the system
obtain that 
and
. Prove easily the identity 
. In conclusion,
, i.e. 
.Remark. Apply the identity
in 
. PP2. If
and 
, then find the minimum value of 
(British Math Olympiad).Proof 1 (with the Lagrange's multipliers). Denote
. Since 
consider
for any . Now can take all of the partial derivatives (in terms of , , and )and set them to zero to get the minimum (or maximum). Taking them in terms of
, , and respectively we get
. Take the first minus, the second and the second minus the third to get ![$\left\{\begin{array}{c}
(x - y)\left[2 + 3\lambda (x + y + z)\right] = 0\\\\
(y - z)\left[2 + 3\lambda (x + y + z)\right] = 0\end{array}\right\|$](http://latex.artofproblemsolving.com/2/a/1/2a1798744c4ebd1a538f6ce5bc69e57b1a35200d.png)
.Taking cases from these equations, we find that the solutions of this system are
, and permutations.They all give
, so 
or 
. But notice that 
, 
and ![$z = \dfrac{\sqrt[3]{7}}{4}$](http://latex.artofproblemsolving.com/0/e/1/0e14db59e28a1fa0547aff8e3576d659f05f249d.png)
for the problem conditions, and they give 
. Thus 
,at
, and permutations.Proof 2.
![$x^3 + y^3 + z^3 - 3xyz = 1 \Leftrightarrow (x + y + z)[x^2 + y^2 + z^2 - (xy + yz + xz)] = 1$](http://latex.artofproblemsolving.com/5/e/4/5e4342198ff9d9b8f82bb6bea2f478964d92e889.png)
. So 
.Note that
, so 
. Plugging this back in we find
. Let 
. Now we have the task of minimizing 
. I'm sure there are other ways, but I took thederivative of this to find
, or 
. Note 
gives non-real solutions, so 
and 
. Thus the answer is 
.This post has been edited 52 times. Last edited by Virgil Nicula, Nov 21, 2015, 7:46 AM
August 2018
January 2018