Demand $z=x+y$ and $x^2/4 + y^2/5 + z^2/25 = 1$. What is the maximum value of $f(x,y,z) = x^2+y^2+z^2$?
I've been attempting this with Lagrange multipliers in a few different ways. However, the resulting system of equations with two lagrangians has so many variables that it becomes very complicated. Can someone show how this is to be done manually?
I also attempted turning it into two unknowns by replacing $z$ with $x+y$. However, this also led nowhere.
On
You have two equations in three unknowns, so just have to choose one variable to maximize over. When you eliminate $z$ you have to do it from the second constraint as well as the objective function. Your problem becomes to maximize $2x^2+2y^2+2xy$ subject to $x^2/4+y^2/5+(x+y)^2/25=1=\frac {29}{100}x^2+\frac 6{25}y^2+\frac 2{25}xy$ Now solve the second constraint for one of the variables using the quadratic formula, plug that into the objective, and the objective is a function of one variable. Differentiate, set to zero.....
On
A geometric interpretation of this problem is to locate the points of the given triaxial ellipsoid $ \ \frac{x^2}{4} \ + \ \frac{y^2}{5} \ + \ \frac{z^2}{25} \ = \ 1 \ $ , which also lie in the plane $ \ z \ = \ x + y \ $ which lie at the greatest distance from the origin (which we shall find by maximizing the "distance-squared" function $ \ f(x, \ y, \ z) \ $ in $ \ \mathbb{R}^3 \ $ ) . While the function and the ellipsoid have symmetry about the origin, and the choice of the intersecting plane would seem to also permit some symmetry for the solution, the triaxiality unfortunately "breaks" that symmetry, so determining the solution is made a bit complicated. What we do expect is that the points at maximum and at minimum distance are located symmetrically about the origin.
The first diagram presents the geometrical arrangement; the second is a "cut-away" showing the intersection ellipse on which the solutions lie.
The "Lagrange-multiplier" method is very powerful and can reduce certain sorts of extremization problems almost to triviality, while leading to somewhat "nightmarish" algebra for others (particularly when non-linear equations are produced). One finds that one must be something of an opportunist in order to attempt to reduce the level of complication: reluctant mathematician shows what can arise for this problem if all of the variables are kept. If we eliminate $ \ z \ $ as Ross Millikan describes, the amount of effort is reduced somewhat. By "imbedding" one of the constraints into the expressions for the function $ \ f \ $ and the equation of the ellipsoid, we can work with
$$ \ f(x, \ y) \ = \ 2x^2 \ + \ 2xy \ + \ 2y^2 \ \ , \ \ g(x, \ y) \ = \ 25x^2 \ + \ 4xy \ + \ 24y^2 \ - \ 100 $$
and a single Lagrange-multiplier. As such, the "Lagrange equations" don't inspire much hope:
$$ 4x \ + \ 2y \ = \ \lambda \ \left( \ \frac{29x \ + \ 4y}{50} \ \right) \ \ , \ \ 2x \ + \ 4y \ = \ \lambda \ \left( \ \frac{4x \ + \ 24y}{50} \ \right) \ \ . $$
But if we re-arrange these equations into the linear system
$$ \begin{array}{c} (200 \ - \ 29 \lambda) \ x \ \ + \ \ (100 \ - \ 4 \lambda) \ y \ \ = 0 \\ (25 \ - \ \lambda) \ x \ \ + \ \ (50 \ - \ 6 \lambda) \ y \ \ = 0 \end{array} \ \ , $$
we know that we can find non-trivial results for $ \ x \ $ and $ \ y \ $ when
$$ \left| \begin{array}{cc} (200 \ - \ 29 \lambda) & (100 \ - \ 4 \lambda) \\ (25 \ - \ \lambda) & (50 \ - \ 6 \lambda) \end{array} \right| \ \ = \ \ 0 \ \ $$
(and we know that $ \ (0, \ 0, \ 0 + 0 = 0 ) \ $ can't be a solution to our problem). With some patience (or a little help from a convenient computer-algebra system), we find that this determinant equation is satisfied by $ \ \lambda \ = \ 10 \ $ and $ \ \lambda \ = \ \frac{75}{17} \ $ .
For each value of the multiplier $ \ \lambda \ $ , the system of equations becomes
$ \ \mathbf{\lambda \ = \ 10 \ \ --} \ $
$$ \begin{array}{c} -90 \ x \ \ + \ \ 60 \ y \ \ = 0 \\ 15 \ x \ \ - \ \ 10 \ y \ \ = 0 \end{array} \ \ \Rightarrow \ \ y \ = \ \frac{3}{2} \ x \ \ \Rightarrow \ \ z \ = \ \frac{5}{2} \ x \ \ ; $$
$ \ \mathbf{\lambda \ = \ \frac{75}{17} \ \ --} \ $
$$ \begin{array}{c} \frac{1225}{17} \ x \ \ + \ \ \frac{1400}{17} \ y \ \ = 0 \\ \frac{350}{17} \ x \ \ + \ \ \frac{400}{17} \ y \ \ = 0 \end{array} \ \ \Rightarrow \ \ y \ = \ -\frac{7}{8} \ x \ \ \Rightarrow \ \ z \ = \ \frac{1}{8} \ x \ \ . $$
[One learns not to try to work directly with expressions for $ \ \lambda \ $ itself... except when that is the easy thing to do! (That is to say, there isn't any general method for approaching these problems.)]
The result for each Lagrange-multiplier value is then
$ \ \mathbf{\lambda \ = \ 10 \ \ --} \ $
$$ 25x^2 \ + \ 20 \ (\frac{3}{2} \ x)^2 \ + \ 4(\frac{5}{2} \ x)^2 \ = \ 100 \ \ \Rightarrow \ \ 95 \ x^2 \ = \ 100 \ \ \Rightarrow \ \ x^2 \ = \ \frac{20}{19} \ \ $$
$$ \Rightarrow \ \ f(x, \ y, \ z) \ = \ x^2 \ + \ (\frac{3}{2} \ x)^2 \ + \ (\frac{5}{2} \ x)^2 \ = \ \frac{38}{4} \ x^2 \ = \ \frac{2 \cdot 19}{4} \ \cdot \ \frac{4 \cdot 5}{19} \ = \ 10 \ \ ; $$
$ \ \mathbf{\lambda \ = \ \frac{75}{17} \ \ --} \ $
$$ 25x^2 \ + \ 20 \ (-\frac{7}{8} \ x)^2 \ + \ 4(\frac{1}{8} \ x)^2 \ = \ 100 \ \ \Rightarrow \ \ \frac{2584}{64} \ x^2 \ = \ \frac{323}{8} \ x^2 \ = \ 100 $$ $$ \Rightarrow \ \ x^2 \ = \ \frac{800}{323} \ \ $$
$$ \Rightarrow \ \ f(x, \ y, \ z) \ = \ x^2 \ + \ (-\frac{7}{8} \ x)^2 \ + \ (\frac{1}{8} \ x)^2 \ = \ \frac{114}{64} \ x^2 $$ $$ = \ \frac{2 \cdot 3 \cdot 19}{2 \cdot 32} \ \cdot \ \frac{32 \cdot 25}{17 \cdot 19} \ = \ \frac{75}{17} \ \approx \ 4.412 \ \ . $$
So $ \ \lambda \ = \ 10 \ $ corresponds to the pair of points symmetric about the origin at "squared-distance" $ \ 10 \ $ [the maximal value], with the second multiplier value representing the symmetric points at "squared-distance" $ \ \frac{75}{17} \ $ [the minimal value] . The diagram below presents this situation.
Beginning with the equation for the Lagrange Multipliers, it is a matter of (tedious) algebraic manipulation with a goal to systematically eliminate parameters. We have $x+y-z=0$ and $\frac{x^2}{4}+\frac{y^2}{5}+\frac{z^2}{25}=1$, $f(x,y,z)=x^2+y^2+z^2$
$$\begin{pmatrix} 2x \\ 2y \\ 2z \end{pmatrix} = \lambda \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} + \gamma \begin{pmatrix} x/2 \\ 2y/5 \\ 2z/25 \end{pmatrix} \implies \begin{cases} 2x= \lambda+\frac{1}{2}\gamma x \\ 2y= \lambda+\frac{2}{5}\gamma y \\ 2z= -\lambda+\frac{2}{25}\gamma z \end{cases}$$
$$\begin{cases} \lambda = 2x-\frac{1}{2}\gamma x \\ \lambda = 2y-\frac{2}{5}\gamma y \\ \lambda = -2z + \frac{2}{25}\gamma z \end{cases} \implies \begin{cases} \frac{1}{2}\gamma x-2x = \frac{2}{5}\gamma y-2y \\ \frac{1}{2}\gamma x-2x = -\frac{2}{25}\gamma z +2z\end{cases} \implies \begin{cases} \gamma=\frac{2x-2y}{x/2-2y/5} \\ \gamma = \frac{2x+2z}{x/2+2z/25} \end{cases}$$ It then remains to solve the system of equations: $$\frac{2x-2y}{x/2-2y/5} = \frac{2x+2z}{x/2+2z/25}\ \ \ \ \wedge \ \ \ \ x+y-z=0 \ \ \ \ \wedge \ \ \ \ \frac{x^2}{4}+\frac{y^2}{5}+\frac{z^2}{25}=1$$
The solutions to this system may be any of the following: maxima, minima, or saddle points of f(x,y,z) under the given constraints.
edit: One of the resulting quadratics is factorable.
$$\frac{2x-2y}{x/2-2y/5} = \frac{2x+2z}{x/2+2z/25} \implies (x-y)(25x+4z)=(5x+5z)(5x-4y)$$ After expanding and eliminating the z variable (z=x+y), the quadratic reduces to: $$21x^2+10xy-16y^2=0$$ $$(3x-2y)(7x+8y)=0$$ Thus, we have $y=(3/2)x$ or $y=(-7/8)x$, hence, an explicit solution from the quadratic formula using the ellipsoid constraint. Plugging these back into f, we see a maximum at $f(2\sqrt{\frac{5}{19}},3\sqrt{\frac{5}{19}},5\sqrt{\frac{5}{19}})=10$.