"By using Lagrange's method, find the points on the curve $10x^2 + 12xy + 10y^2 = 1$ that are nearest and farthest from the origin."

I've used $f(x,y) = x^2+y^2 $from the distance formula $d$ $=$ $\sqrt{(x-x_1)^2+(y-y_1)^2}$ $F(x,y) = λ G (x,y)$ where $G$ is the curve from the question.

And I've ended up with these 3 equations after partial derivating and etc.

  1. $2x = λ(20x+12y)$

  2. $2y = λ(12x+20y)$

  3. $10x^2 +12xy + 10y^2 -1 = 0$

And from there I've solved for $x$ and ended up with $x^2 = y^2$ and $x = \pm \sqrt{\frac{1}{32}}$, which gives $y = \pm \sqrt{\frac{1}{32}}$.

But that only solves for the closest distance from origin and I can't figure out how to find the point furthest away and also how do I know which combinations of $x,y$ are valid?


2 Answers 2


You've done fine to conclude that $x^2=y^2.$ That means $x=y$ or that $x=-y.$ Plug each of those into your curve's equation. You should find four distinct points of two distinct distances from the origin.


Note that this method relies heavily on the AM-GM inequality,but you can connect it with your Lagrage Multiplier method to sort out the points that lie furthest from the origin. For the farthest, observe that $(x+y)^2 \ge 0 \implies xy \ge -\dfrac{x^2+y^2}{2}\implies 1=10x^2+12xy + 10y^2 \ge 10(x^2+y^2)-6(x^2+y^2)=4(x^2+y^2)\implies x^2+y^2 \le \dfrac{1}{4}$ with $=$ occurs at $x+y = 0$ or $x = -y\implies 20x^2 - 12x^2 = 1\implies 8x^2 = 1\implies x = \pm \dfrac{1}{2\sqrt{2}}, y = \mp \dfrac{1}{2\sqrt{2}}$. For the nearest, note that $(x-y)^2 \ge 0 \implies 2xy \le x^2+y^2\implies 1 = 10(x^2+y^2)+12xy \le 10(x^2+y^2) +6(x^2+y^2)=16(x^2+y^2)\implies x^2+y^2 \ge \dfrac{1}{16}$. $=$ occurs at $x = y = \pm \dfrac{1}{4\sqrt{2}}$.


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