# Finding the Maximum and Minimum Distance by Lagrange's Method of Multipliers

Q. Find the maximum and minimum distance of a point from origin such that the point lies in the curve $3x^2+4xy+6y^2=140$

I am unable to solve these three equations simultaneously for $(x,y)$

$2x+\lambda(6x+4y)=0$

$2y+\lambda(4x+12y)=0$

$3x^2+4xy+6y^2=140$

• why are you unable to solve them ? – what'sup Oct 13 '13 at 18:29
• i don't know :( i tried substituting for $x$ from first equation to the second, this gives a complex number. I tried equating $\lambda$ from first two equation, that doesn;t get me anywhere. Can you please point in the right direction ? i know i am just missing it ! @what'sup – Aman Mittal Oct 13 '13 at 18:32
• Exactly what did you get when you "tried equating λ from first two equation[s]"? Show your work here. – Steve Kass Oct 13 '13 at 18:33
• @SteveKass Equating the two $\lambda$, i end up getting this $2x^2-2y^2+3xy=0$ – Aman Mittal Oct 13 '13 at 18:35
• Looks good then, $(2x-y)(x+2y)=0$. – André Nicolas Oct 13 '13 at 18:37

Another related approach that can be taken makes use of the symmetry of the geometric figure and the function to be extremized. The conic section, although rotated, still retains symmetry about the origin, that is, if the point $\ (x,y) \$ lies on the curve, $\ (-x,-y) \$ does as well. The distance-from-the-origin function $\ s(x,y) \ = \ \sqrt{x^2 \ + \ y^2} \$ has "radial" symmetry about the origin. Since the points $\ (x,y) \$ and $\ (-x,-y) \$ are equidistant from the origin, the points on the curves at extremal distances from the origin lie on lines $\ y \ = \ mx \$ (or possibly the vertical line $\ x \ = \ c \$ ) .

We can thus search for such points by inserting $\ y \ = \ mx \$ into the equation for the conic section and using the result in finding critical points of the "distance-squared" function

$$s^2 \ = \ x^2 \ + \ y^2 \ = \ x^2 \ + \ (mx)^2 \ = \ ( \ 1 \ + \ m^2 \ ) \ x^2 \ \ .$$

From the equation for our rotated ellipse, we thus obtain

$$3x^2 \ + \ 4xy \ + \ 6y^2 \ = \ 140 \ \ \Rightarrow \ \ 3x^2 \ + \ 4mx^2 \ + \ 6m^2x^2 \ = \ 140$$

$$x^2 \ = \ \frac{140}{6m^2 \ + \ 4m \ + \ 3} \ \ \Rightarrow \ \ s^2 \ = \ \frac{140 \ ( \ 1 \ + \ m^2 \ )}{6m^2 \ + \ 4m \ + \ 3} \ \ .$$

By implicit differentiation, we locate critical values of $\ m \$ from

$$2s \ \frac{ds}{dm} \ = \ \frac{[ \ (6m^2 \ + \ 4m \ + \ 3) \cdot 140 \cdot 2m \ ] \ - \ [ \ ( \ 1 \ + \ m^2 \ ) \cdot 140 \cdot (12m \ + \ 4) \ ]}{(6m^2 \ + \ 4m \ + \ 3)^2} \ = \ 0$$

$$\Rightarrow \ \ [ \ (6m^2 \ + \ 4m \ + \ 3) \ m \ ] \ - \ [ \ ( \ 1 \ + \ m^2 \ ) \cdot (6m \ + \ 2) \ ] \ = \ 0$$

[we can just investigate the numerator, as the denominator has no real zeroes]

$$\Rightarrow \ \ 6m^3 \ + \ 4m^2 \ + \ 3m \ - \ 6m^3 \ - \ 2m^2 \ - \ 6m \ - \ 2 \ = \ 0 \ \ \Rightarrow \ \ 2m^2 \ - \ 3m \ - \ 2 \ = \ 0 \ \ .$$

[Note that we can also arrive here by inserting $\ y \ = \ mx \$ into the result from the Lagrange equations, $\ 2x^2 \ + \ 3xy \ - \ 2y^2 \ = \ 0 \$ . ]

By factoring or applying the quadratic formula, we find the solutions $\ m \ -\frac{1}{2} \ , \ 2 \$ , so these are the slopes of the lines through the origin on which the extremal-distance points lie (perhaps unsurprisingly, the lines are perpendicular).

For the question at hand, we don't need to know the coordinates of those points (although they can be determined readily), so we compute the extremal distances as

$$\mathbf{m \ = \ 2 \ : } \quad s^2 \ = \ \frac{140 \ ( \ 1 \ + \ 2^2 \ )}{6 \cdot 2^2 \ + \ 4 \cdot 2 \ + \ 3} \ = \ \frac{140 \cdot 5}{35} \ = \ 20 \ \ \Rightarrow \ \ s \ = \ 2 \ \sqrt{5} \ \ ;$$

$$\mathbf{m \ = \ -\frac{1}{2} \ : } \quad s^2 \ = \ \frac{140 \ ( \ 1 \ + \ \left[-\frac{1}{2}\right]^2 \ )}{6 \cdot \left[-\frac{1}{2}\right]^2 \ + \ 4 \cdot \left[-\frac{1}{2}\right] \ + \ 3} \ = \ \frac{140 \cdot \left(\frac{5}{4}\right)}{\left(\frac{5}{2}\right)} \ = \ 70$$ $$\Rightarrow \ \ s \ = \ \sqrt{70} \ \ .$$

Here is a graph illustrating the solution:

*The ellipse in question is marked in blue; the points closest to the origin lie on the red line, tangent to the red circle of radius $\ 2 \sqrt{5} \$ ; those farther from the origin are on the green line, tangent to the green circle of radius $\ \sqrt{70} \$ .

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[And I apologize, but as an American of a not insignificant age, I just couldn't look at that graph without being put in mind of this: ]