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Can anybody help me with the following question, Use the method of Lagrange multipliers to find $\max$ and $\min$ values for $f(x,y,z)=x^2+y^2+z^2$ subject to the constraint $4x^2 + y^2 + 9z^2=36$.

I have $2x=\lambda8x.$(1)

$2y=\lambda2y$ (2)

$2z=\lambda18z$(3)

$4x^2 + y^2 + 9z^2=36(4)$

From the the second equation (2) I got $\lambda =1$, but in equation (3)I got $\lambda = 9$ how can this be possible. Is lambda different in each equation ?, I assumed it it was equal for each equation.

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  • $\begingroup$ Can you please tell us what you have tried so far and/or where you are stuck? You can edit and add those details to the question! $\endgroup$ – Jesse P Francis May 10 '15 at 8:41
  • $\begingroup$ what is the exponant of $z$ in $4x^2+y^2+9z^$? $\endgroup$ – Dr. Sonnhard Graubner May 10 '15 at 8:42
  • $\begingroup$ Aren't you missing something in the constraint ? $\endgroup$ – Claude Leibovici May 10 '15 at 8:43
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    $\begingroup$ From $(2)$, $\lambda=1$ if $y \neq 0$. If you take that into account in all other possible places, you will find several solutions. $\endgroup$ – Claude Leibovici May 10 '15 at 9:14
  • $\begingroup$ I am getting all max points because of the function f(x,y,z) is all positive squares with no minus constant, is this possible? $\endgroup$ – Kevin May 10 '15 at 11:38
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It will perhaps clarify the situation if we make a geometric interpretation of the problem. The function $ \ f(x,y,z) \ = \ x^2 + y^2 + z^2 \ $ is the "distance-squared" function in $ \ \mathbb{R}^3 \ $ for distances of points from the origin. The "constraint surface" is a triaxial ellipsoid centered on the origin with its own axes aligned with the coordinate axes.

Your "Lagrange equations" factor as $ \ 2 \ x \ (1 - 4 \ \lambda) \ = \ 0 \ $ , $ \ 2 \ y \ (1 - \lambda) \ = \ 0 \ $ , and $ \ 2 \ z \ (1 - 9 \ \lambda) \ = \ 0 \ $ . While each of the variables has a zero solution, they cannot all be zero at the same time, since extremal points must lie on the constraint surface. Setting only one of the variables at a time equal to zero gives us cross-sections of the ellipsoid in each of the coordinate planes, which does not determine extremal values of $ \ f \ $ for us.

The three different values of $ \ \lambda \ $ are suggestive, since we are dealing with a triaxial ellipsoid. For $ \ \lambda \ = \ 1 \ $ , the equations become

$$ \ 2 \ x \ ( - 3) \ = \ 0 \ \ , \ \ 2 \ y \ \cdot \ 0 \ = \ 0 \ \ , \ \ 2 \ z \ ( - 8) \ = \ 0 \ \ , $$

which requires that $ \ x \ = \ z \ = \ 0 \ $ , but leaves $ \ y \ = \ 0 \ $ open to assume any value. The constraint equation, however, tells us that $ \ 4 \cdot 0^2 \ + \ y^2 \ + \ 9 \cdot 0^2 \ = \ 36 \ \Rightarrow \ y \ = \ \pm 6 $ . So this value of $ \ \lambda \ $ corresponds to the endpoints of the longest axis of the ellipsoid and to the value for our function $ \ f(0, \ \pm 6 , \ 0 ) \ = \ 36 \ $ [this proves to be the maximum value for $ \ f \ $ ] . By similar reasoning, we find that

$$ \lambda \ = \ \frac{1}{4} \ \ \Rightarrow \ \ y \ = \ z \ = \ 0 \ \ , \ \ x \ = \ \pm 3 \ \ , \ \ f( \pm 3 , \ 0, \ 0 ) \ = \ 9 $$

and

$$ \lambda \ = \ \frac{1}{9} \ \ \Rightarrow \ \ x \ = \ y \ = \ 0 \ \ , \ \ z \ = \ \pm 2 \ \ , \ \ f( 0 , \ 0, \ \pm 2 ) \ = \ 4 \ \ \text{[the minimum value]} . $$

The Lagrange-multiplier method locates points where level curves or surfaces of the function $ \ f \ $ are tangent to the constraint curve or surface (what it actually does is find points where the normal vectors are parallel or anti-parallel). Since our "distance-squared" function has "spherical symmetry", it can have level surfaces with tangent points to the constraint ellipsoid at three pairs of antipodal points. These are marked by the three values of the multiplier $ \ \lambda \ $ .

An analogous problem for an ellipse in the plane can be found here .

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Easier: cross multiply the two gradients using the determinant method and equate the result with 0. You get xy=0=yz=xz. The candidates are (x,0,0),(0,y,0) and (0,0,z). Plugging it into the constraint you get (3,0,0),(-3,0,0),(0,6,0),(0,-6,0),(0,2,0),(0,0,-2). The last two give min, and the the middle two give max or check.

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  • $\begingroup$ Thanks, the past exam paper says you have to use lagrange multipliers. $\endgroup$ – Kevin May 10 '15 at 12:57
  • $\begingroup$ I , for some reason thought that a min value needed to be < 0 $\endgroup$ – Kevin May 10 '15 at 12:58

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