Inequality with 2 Variables

[Jordan2]

I have the following problem. I have an inequality describing a plane and I need to sketch a graph of the plane. Consequently I need to simplify the inequality to the point of being able to enter an x and get a corresponding y. Any (quick) help or pointers would be hugely appreciated!! I'm desperate and have spent hours on this:

This is the plane's definition: \(\displaystyle $A:= \{{(x,y) \in R^2: 2x + 2y} \geq {(1 + \sqrt{3}) * \sqrt{x^2 + y^2}\}$\)


\(\displaystyle ${{2x + 2y}} \geq {(1 + \sqrt{3}) * \sqrt(x^2 + y^2)}$\)

\(\displaystyle $\frac{2x + 2y}{\sqrt{x^2 + y^2}} \geq {(1 + \sqrt{3})}$\)

\(\displaystyle ${\frac{4 * (x^2 + y^2) + 8xy}{(x^2 + y^2)}} \geq {4 + 2 * \sqrt{3}}$\)

\(\displaystyle ${4 + \frac{8xy}{(x^2 + y^2)} \geq {4 + 2 * \sqrt{3}} $\)

\(\displaystyle $\frac{8xy}{(x^2 + y^2)} \geq {2 * \sqrt{3}}$\)

\(\displaystyle ${\frac{xy}{(x^2 + y^2)}}\geq{\frac{2* \sqrt{3}}{8}}$\)

Simply can't work with the inequality from here on out ...

 

[stapel]

1) A:= {(x,y) e R^2: 2x + 2y >= (1 + \/3') * \/x^2 + y^2'}

I think you are using ":=" to mean "is defined as", R[sup:cs34pjcw]2[/sup:cs34pjcw] to be the real plane, and "e" to mean "is an element of". But I"m afraid I can't guess what you might mean by "\/3'"...? And what is the meaning of the notations after your lines of work? What do "\(\/x^2 + y^2')", "|^2", and ":8" mean...?

What are you supposed to be doing with these "inequalities" (which actually look like set definitions)...? What were the instructions...?

Please be complete. Thank you!

Eliz.

 

[Jordan2]

Dang it, I didn't even see this supports TeX. I've rewritten the headpost in a more legible fashion and taken out the second inequality (which I've figured out on my own). If someone can help with this, it would only be of use within the next hour.

 

[mmm4444bot]

Re: Inequalities with 2 Variables

... this [site] supports TeX.

This site renders some forms of TeX syntax, but not others (as you can see from your post).

-

\(\displaystyle \frac{(2x + 2y)}{\sqrt{x^2 + y^2}} \;\geq\; 1 + \sqrt{3}\)

\(\displaystyle 4 + \frac{(2x + 2y)^2 + 8xy}{(x^2 + y^2)} \;\geq\; 4 + 2 \cdot \sqrt{3}\)

\(\displaystyle 4 + \frac{8xy}{(x^2 + y^2)} \;\geq\; 4 + 2 \cdot \sqrt{3}\)

This does not look right because (2x + 2y)[sup:3bg4jsc0]2[/sup:3bg4jsc0] equals 4x[sup:3bg4jsc0]2[/sup:3bg4jsc0] + 4y[sup:3bg4jsc0]2[/sup:3bg4jsc0] + 8xy.

We could square both sides of the original inequality without first dividing.

\(\displaystyle (2x + 2y)^2 \;\geq\; (1 + \sqrt{3})^2 \cdot (\sqrt{x^2 + y^2})^2\)

\(\displaystyle 4x^2 + 8xy + 4y^2 \;\geq\; (4 + 2\cdot\sqrt{3}) \cdot (x^2 + y^2)\)

\(\displaystyle (4)(x^2 + 2xy + y^2) \;\geq\; (4)(1 + \frac{\sqrt{3}}{2}) \cdot (x^2 + y^2)\)

\(\displaystyle x^2 + 2xy + y^2 \;\geq\; x^2 + y^2 + \frac{\sqrt{3}}{2}x^2 + \frac{\sqrt{3}}{2}y^2\)

\(\displaystyle \frac{\sqrt{3}}{2}x^2 - 2xy + \frac{\sqrt{3}}{2}y^2 \;\leq\; 0\)

This looks like the equation for a rotated hyperbola (the discriminant is greater than 1). Are you familiar with rotating the coordinate system to eliminate the xy-term?

 

[Jordan2]

This does not look right because (2x + 2y)[sup:y6wja42c]2[/sup:y6wja42c] equals 4x[sup:y6wja42c]2[/sup:y6wja42c] + 4y[sup:y6wja42c]2[/sup:y6wja42c] + 8xy.

You must have replied right as I was editing the headpost. I had a couple TeX typos - all is now as it should be in the headpost.

We could square both sides of the original inequality without first dividing.

\(\displaystyle (2x + 2y)^2 \;\geq\; (1 + \sqrt{3})^2 \cdot (\sqrt{x^2 + y^2})^2\)

\(\displaystyle 4x^2 + 8xy + 4y^2 \;\geq\; (4 + 2\cdot\sqrt{3}) \cdot (x^2 + y^2)\)

\(\displaystyle (4)(x^2 + 2xy + y^2) \;\geq\; (4)(1 + \frac{\sqrt{3}}{2}) \cdot (x^2 + y^2)\)

\(\displaystyle x^2 + 2xy + y^2 \;\geq\; x^2 + y^2 + \frac{\sqrt{3}}{2}x^2 + \frac{\sqrt{3}}{2}y^2\)

\(\displaystyle \frac{\sqrt{3}}{2}x^2 - 2xy + \frac{\sqrt{3}}{2}y^2 \;\leq\; 0\)

Yeah, that makes sense - thanks. My last equation would feed into yours with the steps / \(\displaystyle $ * (x^2 + y^2)$; $- xy$; $* 2$\)

Is there really no other way but to rotate the coordinate system? I'm familiar with rotating a coordinate system but wouldn't know how to do so in order to get rid of the xy term. Anyhow, time's kind of up as I need to take off in a bit. But thanks a bunch for the replies - hugely appreciated.