I am working on this Maxima and Minima Problem :

Determine the Max and Min distance of origin from the curve $3x^2+4xy+6y^2=140$

I tried it solving using the lagrange's method of multipliers. I get the following equations





So i get $z=0$

Please suggest how to go about solving for $x$ and $y$ ?

  • $\begingroup$ Shouldn't that be $6y\theta$? $\endgroup$ – Gerry Myerson Jul 11 '13 at 10:07
  • $\begingroup$ Which one ?.... $\endgroup$ – user85860 Jul 11 '13 at 10:08
  • $\begingroup$ You could eliminate $\theta$ from the first two equations; then you have two equations in the two variables $x$ and $y$. $\endgroup$ – Gerry Myerson Jul 11 '13 at 10:08
  • $\begingroup$ Putting value of $\theta$ from the first equation in second, i get $x^2+y^2={-7 \over 2} xy$ $\endgroup$ – user85860 Jul 11 '13 at 10:17
  • $\begingroup$ I'll take your word for it. That has two solutions of the form $y=\alpha x$, which you can stick into the other quadratic. $\endgroup$ – Gerry Myerson Jul 11 '13 at 10:25

Another approach. Draw a circle centered at the origin. You want the radii which intersect the ellipse in exactly two points. So apply the equation $x^2+y^2=r^2$. Define $A=\frac{140}{r^2}$. Then $3x^2+4xy+6x^2=A(x^2+y^2)$. Divide through by $x^2$, and define $z=\frac{y}{x}$, yielding a quadratic in z. We want the discriminat to equal zero, which occurs when $A=7$ or $2$. Thus $r^2=20$ or $70$.


\begin{align} \min_{x,y}&x^2+y^2\\s.t.\\ 3x^2+6y^2+4xy&=140\\ L(x,\lambda)&=x^2+y^2-\lambda(3x^2+6y^2+4xy-140)\\ \nabla L(x,\lambda)&=\binom{2x}{2y}-\lambda\binom{6x+4y}{12y+4x}=0\\ \text{Necessary Conditions:}\\ 2x-\lambda(6x+4y)&=0\\ 2y-\lambda(12y+4x)&=0\\ 3x^2+6y^2+4xy&=140\\ \end{align} Solve these 3 equations and you have your answer. \begin{align} 2x&=\lambda(6x+4y)\\ \lambda&=\frac{2x}{6x+4y}\\ 2y-\lambda(12y+4x)&=0\\ 2y-\frac{2x}{6x+4y}(12y+4x)&=0\\ y&=\frac{x}{6x+4y}(12y+4x)\\ (6x+4y)y&=x(12y+4x)\\ 6xy+4y^2&=12xy+4x^2\\ 2y^2&=3xy+2x^2\\ 3x^2+6y^2+4xy&=140\\ \text{Solving you get,}\\ x=\pm2,y=\pm4\\ d=20 \end{align}

  • $\begingroup$ I wonder whether OP can get from those two quadratics to the solutions. $\endgroup$ – Gerry Myerson Jul 12 '13 at 13:24

Let $x^2+y^2=k$.

Hence, $$k(3x^2+4xy+6y^2)=140(x^2+y^2)$$ or $$(3x-140)x^2+4kxy+(6x-140)y^2=0.$$ If $k=\frac{140}{3}$ so we get one of distances point on our curve from the origin.

Let $k\neq\frac{140}{3}$. Hence, $$4k^2-(3k-140)(6k-140)\geq0$$ or $$k^2-90k+1400\leq0$$ or $$20\leq x^2+y^2\leq70.$$ It's obvious that the equality occurs, which gives the answer:

Let $O$ is the origin and $A$ is a point on the curve $3x^2+4xy+6y^2=140$.

Hence, $\max{AO}=\sqrt{70}$ and $\min AO=\sqrt{20}$.


A third approach. Rotate the ellipse to bring the major and minor axes coincident with the coordinate axes. Then the equation becomes $\lambda_1x^2+\lambda_2y^2=140$, where $\lambda_1$ and $\lambda_2$ are the eigenvalues of the symmetric matrix which generates the quadratic form on the left side of the equation. In this case the eigenvalues are 7 and 2. It is not necessary to calculate the eigenvectors.


An approach with a bit of geometry:

$3x^2 + 4xy + 6y^2 = 140$ is an ellipse rotated about the origin.


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

$2x^2 + (x+2y)^2 + 2y^2 = 140$;

$(x^2 + y^2) = 140/2 - [(x + 2y)^2]/2$.

Let $r$ be the distance from the origin,

$r^2 = (x^2 + y^2) = 70 - $ $[(x + 2y)^2 ]/2$.


Since the bracket on the r.h.s., a square, is $\\ \geq 0$, we get for the maximum:

$r^2 = 70$, at $ 2y + x = 0$.

The major axis of the ellipse, call it $a$, lies along the line $2y + x = 0$, and has squared length:

$a^2 = 70$.


The major axis lies along $2y +x = 0$, or $y = - (1/2)x$ , a straight line with slope $- 1/2$.

The line $y = 2x$ is perpendicular to $y = - (1/2) x$, the orientation of the major axis, I.e. a line along the minor axis.

Intersection of $y = 2x$ with the ellipse:

$3x^2 + 4x(2x) + 6(2x)^2 = 140$; $35x^2 = 140$; $\\x^2 = 4$, and using $y= 2x$, we find $y^2 = 4x^2 = 16$.


$r^2 = min ( x^2 +y^2) = 20$;

For the squared minor axis, call it $b$, we have

$b^2 = 20$.


It appears to be a straightforward problem using Lagrange multiplier.

$$F(x,y)= 3x^2+4 xy +6y -140 ; \quad G(x,y)=x^2+y^2; \tag1 $$

Partially differentiate the Lagrangian

$$\frac{F_x}{F_y}=\frac{G_x}{G_y}=\lambda \tag2 $$

$$\frac{6x+ 4y}{4 x}=\frac{3x+2y}{2 x}=\frac{x}{y}=\lambda \tag3$$

Cross-multiply , let $\dfrac{y}{x}=t,$ and simplify

$$ 3 xy - 2 y^2 +2 x^2=0,\quad t^2-3t+1=0 \tag4$$

Solve this quadratic to find slopes

$$ t= \frac{3 \pm \sqrt{13}}{2} = \frac{y}{x} \tag5$$

which are a pair of perpendicular straight lines passing through the origin.

To find touching points at minimum distance solve the given rotated ellipse and pair of straight lines (exercise for student..)

Verify graphically.

Aug 16, 3019


Those called homogeneous problems can be easily solved

Calling $d^2 = x^2+y^2$ and making $y = \lambda x$ we have

$$ d^2 = x^2(1+\lambda^2) $$

but from $3x^2+4 x y +6 y^2 = 140$ we obtain

$$ x^2(3+4\lambda+6\lambda^2) = 140 $$


$$ d^2 = \frac{140(1+\lambda^2)}{3+4\lambda+6\lambda^2} $$

now deriving

$$ \frac{d}{d\lambda}d^2 = \frac{200(2\lambda^2-3\lambda-2)}{(3+4\lambda+6\lambda^2)^2} $$

and the condition for local minimum/maximum is

$$ 2\lambda^2-3\lambda-2 = 0 $$

obtaining $\lambda = \{-\frac 12, 2\}$ and

$$ \min d^2 = 20\Rightarrow d = \sqrt{20}\\ \max d^2 = 70\Rightarrow d = \sqrt{70} $$


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