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Find the maximum and minimum distances from the origin to the curve $5x^3+6xy+5y^2-8=0$

My attempt:

We have to maximise and minimise the following function $x^2+y^2$ with the constraint that $5x^3+6xy+5y^2-8=0$.

Let $$F(x,y)=x^2+y^2+\lambda(5x^2+6xy+5y^2-8)$$ $$\frac{\delta F(x,y)}{\delta x}=2x+\lambda(10x+6y)$$ and $$\frac{\delta F(x,y)}{\delta y}=2y+\lambda(6x+10y)$$ Multiplying the 2 equations by y,x respectively and subtracting I get $$\lambda(y^2-x^2)=0$$ Hence $$y=x$$ Substituting $x=y$ in $5x^3+6xy+5y^2-8=0$, I get the $x=\pm \frac{1}{\sqrt2}$ and $y=\pm \frac{1}{\sqrt2}$ . Now I am stuck. Both the points corresponds to only one distance. Did I do something wrong?

GTX OC
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4 Answers4

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if the case is $5x^2+6xy+5y^2-8=0$, it could be solved in a easy way:

$5x^2+6xy+5y^2=8,2xy \le x^2+y^2 \implies 5x^2+5y^2 +3(x^2+y^2) \ge 8 \implies x^2+y^2 \ge 1$ when $x=y$

$2xy \ge -(x^2+y^2) \implies 2(x^2+y^2) \le 8 \implies x^2+y^2 \le 4$ when $x=-y$

chenbai
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This answer is just for the sake of the alternative solution:

Pass to the polar coordinates: $$x=r\cos(a)$$

$$y=r\sin(a)$$ Now we get $$5r^2+6r^2\cos(a)\sin(a)=8$$ $$5r^2+3r^2\sin(2a)=8$$ $$r^2(5+3\sin(2a))=8$$ Here $r$ attains its minimum when the value in the parentheses attains its maximum and vice versa. So it suffices to plug in values $\pm1$ instead of $\sin(2a)$. So the corresponding extremums for $r$ are $\sqrt{\frac8{5-3}}=2$ and $\sqrt{\frac{8}{5+3}}=1$

Mihail
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You wrote:

Multiplying the 2 equations by y,x respectively and subtracting I get

λ(y^2−x^2)=0

Hence

y=x

This is true, but not the entirety of the truth. λ(y^2−x^2)=0 is a difference of squares, and may be written as λ(y+x)(y-x)=0, so this condition is also met with y=-x

Lagrange Multipliers do not optimize for you, but report conditions for optimization. It is up to you to test and make sense of your method's output.

Jordan A
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You have found the points that, under the contraint $5x^2+6xy+5y^2−8=0$, minimise $x^2+y^2$and therefore minimise the euclidean distance between the origin and the point $(x,y)$ that is

$$d_{euclidean}\left( (0,0), (x,y) \right) \doteq \sqrt{(x-0)^2+(y-0)^2} = \sqrt{x^2+y^2}$$

So your minimum distance is the distance of $\left(+\frac{1}{\sqrt{2}}, +\frac{1}{\sqrt{2}}\right)$ from the origin (the same with $\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)$. You have found the minimum distance: $$\sqrt{x^2+y^2}=\sqrt{\left( -\frac{1}{\sqrt{2}} \right)^2+\left( -\frac{1}{\sqrt{2}} \right) ^2}=\sqrt{\frac{1}{2}+\frac{1}{2}}=\sqrt{1}= 1$$

As lsp said, from $\lambda (y^2-x^2)$ you find an other possibility: $y=-x$ from which you get the points $\left(-\sqrt{2}, +\sqrt{2}\right)$ and $\left(+\sqrt{2}, -\sqrt{2}\right)$. Those points are the furthest from the origin as you will see when computing the distance to the origin: $$\sqrt{x^2+y^2}=\sqrt{\sqrt{2}^2+\sqrt{2}^2}=\sqrt{2+2}=\sqrt{4}= 2$$

Luc M
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  • Yeah I got it. Distance is $x^2+y^2$. Hence the second point $\left(-\sqrt{2}, +\sqrt{2}\right)$ is at a distance of 4 units from the origin, not 2. – GTX OC Jan 29 '14 at 13:22
  • @GTXOC Yes, but you have to take the square root (for the euclidean distance ), so the distance is given by $\sqrt{x^2+y^2}$. I will edit my answer to make it clearer. – Luc M Jan 29 '14 at 13:33