How to solve this system of equations in Lagrange Multiplier problem

Question

Find the maximum and minimum values of ${x^{2} + y^{2} + z^{2}}$ subject to the conditions ${\frac{x^{2}}{4} + \frac{y^{2}}{5} + \frac{z^{2}}{25} = 1}$ and ${x + y - z = 0}$.

Using Lagrange multiplier method, I got following equations: $$ {2x = \frac{\lambda_{1} x}{2} + \lambda_{2}}$$ $$ {2y = \frac{2 \lambda_{1} y}{5} + \lambda_{2}}$$ $$ {2z = \frac{2 \lambda_{1} z}{25} - \lambda_{2}}$$ $${\frac{x^{2}}{4} + \frac{y^{2}}{5} + \frac{z^{2}}{25} = 1}$$ $${x + y - z = 0}$$

I'm stuck after this. I've tried to solve this system of equations to get critical point many times. Any help will be greatly appreciated. Also is there any other way to approach this problem?

Answer 1

I think Lagrange multipliers method is not necessary here.

We need to find a maximal and a minimal value of $x^2+y^2+(x+y)^2$, where $\frac{x^2}{4}+\frac{y^2}{5}+\frac{(x+y)^2}{25}=1.$

Let $2(x^2+xy+y^2)=k$.

Hence, the condition gives $29x^2+8xy+24y^2=100$ or $$k(29x^2+8xy+24y^2)=200(x^2+xy+y^2),$$ which says that the equation $$(29k-200)x^2+(8k-200)xy+(24k-200)y^2=0$$ has real solutions.

If $k=\frac{200}{29}$ we has solutions.

Let $k\neq\frac{200}{29}$.

Hence, $$(4k-100)^2-(29k-200)(24k-200)\geq0,$$ which gives $$\frac{75}{17}\leq k\leq10$$

It's obvious that the equality occurs in both cases and we are done!

Answer 2

The geometrical interpretation of this problem is that we wish to find the extremal values of the "distance-squared-from-the-origin" function $ \ f(x,y,z) \ = \ x^2+y^2+z^2 \ $ on the intersection curve of the ellipsoid $ \ 25x^2 + 20y^2 + 4z^2 = 100 \ $ and the plane $ \ x + y - z = 0 \ \ . $ The plot at left presents a view "looking down" the $ \ x-$ axis toward the origin.

Because the function and the equation of the ellipsoid have symmetry about the origin, we expect some sort of "mirroring" of extremum points. The equation of the intersection plane $ \ z = x + y \ $ "breaks" this symmetry somewhat. If the coordinates $ \ x \ $ and $ \ y \ $ have the same sign, then $ \ z \ $ should also have that sign, i.e., the points will be $ \ (x,y,z) \ $ and $ \ (-x,-y,-z) \ $ . If $ \ x \ $ and $ \ y \ $ have opposite signs, the sign of $ \ z \ $ will depend upon the sum of those coordinates, but the relative signs for the points will be $ \ (x,-y,z) \ $ and $ \ (-x,y,-z) \ $ .

It is reasonably convenient to set up systems of Lagrange equations using one or two multipliers. For a one-multiplier system, we insert $ \ z = x + y \ $ into our function and the ellipsoid equation to obtain

$$ f(x,y,z) \ = \ x^2 + y^2 + z^2 \ \ \rightarrow \ \ \phi(x,y) \ = \ 2·(x^2 + y^2 + xy) $$ and $$ 25x^2 + 20y^2 + 4z^2 \ = \ 100 \ \ \rightarrow \ \ 29x^2 + 24y^2 + 8xy \ = \ 100 \ \ . $$

The Lagrange equations are then

$$ 4x + 2y \ = \ \lambda · (58x + 8y) \ \ , \ \ 4y + 2x \ = \ \lambda · (48y + 8x) \ \ . $$

Solving these for $ \ \lambda \ $ yields $$ \lambda \ \ = \ \ \frac{2x \ + \ y}{29x \ + \ 4y} \ \ = \ \ \frac{2y \ + \ x}{24y \ + \ 4x} \ \ \Rightarrow \ \ 8x^2 + 52xy + 24y^2 \ = \ 29x^2 + 62xy + 8y^2 $$ $$ \Rightarrow \ \ 21x^2 \ + \ 10xy \ - \ 16y^2 \ = \ 0 \ \ . $$

At first glance, this doesn't seem all that helpful, but in fact it is the equation of a "degenerate hyperbola", a pair of intersecting lines. (This may be found from plotting the equation or applying the appropriate criterion for degenerate conics, $$ \ \det \left[ \begin{array}{cc} 21 & 10 \\ 10 & -16 \end{array} \right] \ < \ 0 \ \ . ) $$

The "curve" equation may be factored as $ \ (3x - 2y) · (7x + 8y) \ = \ 0 \ \ , $ which gives us two lines along which the extremal points may be found, $ \ y \ = \ \frac32 x \ $ and $ \ y \ = \ -\frac78 x \ \ . $ We thus have two possibilities to consider: inserting these into the ellipsoid equation produces

$$ \mathbf{ y \ = \ \frac32 x \ :} \quad 29x^2 \ + \ 24·\left(\frac32 x\right)^2 \ + \ 8·x·\left(\frac32 x\right) \ = \ 95x^2 \ = \ 100 $$ $$ \Rightarrow \ \ x \ = \ \pm \frac{10}{\sqrt{95}} \ = \ \pm 1.026 \ \ , \ \ y \ = \ \pm \frac{15}{\sqrt{95}} \ = \ \pm 1.539 \ \ , \ \ z \ = \ \pm \frac{25}{\sqrt{95}} \ = \ \pm 2.565 \ \ ; $$

$$ \mathbf{ y \ = \ -\frac78 x \ :} \quad 29x^2 \ + \ 24·\left(-\frac78 x\right)^2 \ + \ 8·x·\left(-\frac78 x\right) \ = \ \frac{323}{8} \ x^2 \ = \ 100 $$ $$ \Rightarrow \ \ x \ = \ \pm \frac{20\sqrt{2}}{\sqrt{323}} \ = \ \pm 1.574 \ \ , \ \ y \ = \ \mp \frac{35\sqrt{2}}{2\sqrt{323}} \ = \ \mp 1.377 \ \ , \ \ z \ = \ \pm \frac{5\sqrt{2}}{2\sqrt{323}} \ = \ \pm 0.197 \ \ . $$

These points are picked out in the graph at right above by violet and red arrows, respectively. (The view "looks upward" along the $ \ z-$ axis toward the origin.)

The extremal values of the "distance-squared" function are then given by

$$ \mathbf{ y \ = \ \frac32 x \ :} \quad 2·\frac{100}{95} \ + \ 2 · \left(\frac32 \right)^2 · \frac{100}{95} \ + \ 2·\left( \frac{10}{\sqrt{95}} \right) · \left( \frac32 · \frac{10}{\sqrt{95}} \right) \ = \ \frac{200+450+300}{95} \ \ = \ \ 10 $$ [absolute maximum]

and

$$ \mathbf{ y \ = \ -\frac78 x \ :} \quad 2·\frac{800}{323} \ + \ 2 · \left(-\frac78 \right)^2 · \frac{800}{323} \ + \ 2·\left( \frac{20 \sqrt{2}}{\sqrt{323}} \right) · \left( -\frac78 · \frac{20 \sqrt{2}}{\sqrt{323}} \right) $$ $$ = \ \frac{1600+1225-1400}{323} \ \ = \ \ \frac{1425}{323} \ \ = \ \ \frac{3·5·5·19}{17·19} \ \ = \ \ \frac{75}{17} \ \ . $$ [absolute minimum]

This agrees with the work by Michael Rosenberg and the calculations from WolframAlpha.

$$ \ \ $$

If we set up a system with two multipliers as you did (though I prefer using $ \ \mu \ $ rather than a subscripted letter), we obtain

$$ 2x \ = \ \lambda·50x \ + \ \mu \ \ , \ \ 2y \ = \ \lambda·40y \ + \ \mu \ \ , \ \ 2z \ = \ \lambda · 8z \ - \mu \ \ , $$

using $ \ 25x^2 + 20y^2 + 4z^2 = 100 \ $ as the equation for the ellipsoid. We may start here by eliminating $ \ \mu \ $ among the equations:

$$ \mu \ \ = \ \ 2x \ - \ 50·\lambda·x \ \ = \ \ 2y \ - \ 40·\lambda·y \ \ = \ \ 8·\lambda·z \ - \ 2z \ \ ; $$

solving each pair of equations for $ \ \lambda \ $ then yields

$$ \lambda \ \ = \ \ \frac{x \ - \ y}{5·(5x \ - \ 4y)} \ \ = \ \ \frac{x \ + \ y}{4z \ + \ 25x} \ \ = \ \ \frac{y \ + \ z}{4·(z \ + \ 5y)} \ \ ; $$

if we now cross-multiply each pair of ratios in turn (none of the denominators will turn out to equal zero) and simplify the resulting equations, we obtain each time $ \ 5xy + 21xz - 16yz \ = \ 0 \ \ ! $ We now solve this for $ \ z \ $ and combine the result with the plane constraint equation to find

$$ z \ \ = \ \ -\frac{5xy}{21x \ - \ 16y} \ \ = \ \ x \ + \ y \ \ \Rightarrow \ \ 21x^2 \ + \ 10xy \ - \ 16y^2 \ = \ 0 \ \ , $$

the degenerate hyperbola equation we found above (cross-multiplication is again "safe" here). The two solutions are consequently $ \ \ ( \ x \ , \ -\frac78 \ x \ , \ \frac18 \ x \ ) \ \ $ and $ \ \ ( \ x \ , \ \frac32 \ x \ , \ \frac52 \ x \ ) \ \ $ with the rest of the calculations proceeding as before.

Answer 3

With the Lagrange multipliers

$$ L(x,y,z,\lambda,\mu) = x^2+y^2+z^2+\lambda\left(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}-1\right)+\mu(x+y-z) $$

The stationary points are located at

$$ \nabla L = 0 = \left\{ \begin{array}{l} \frac{2\lambda x}{a}+\mu +2 x \\ \frac{2 \lambda y}{b}+\mu +2 y \\ \frac{2 \lambda z}{c}-\mu +2 z \\ \frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}-1 \\ x+y-z \\ \end{array} \right. $$

taking the first three rows we have

$$ \left(\frac{2\lambda x}{a}+\mu +2 x\right)x+\left(\frac{2 \lambda y}{b}+\mu +2 y\right)y+\left(\frac{2 \lambda z}{c}-\mu +2 z\right)z=2\lambda+2(x^2+y^2+z^2)=0 $$

and also

$$ \cases{ x = -\frac{\mu}{2+2\lambda/a}\\ y = -\frac{\mu}{2+2\lambda/b}\\ z = \frac{\mu}{2+2\lambda/c}\\ } $$

or

$$ \mu\left(\frac{1}{2+2\lambda/a}+\frac{1}{2+2\lambda/b}+\frac{1}{2+2\lambda/c}\right)=0 $$

Solving for $\lambda$ we have the values for $x^2+y^2+z^2$ at the stationary points.