Find $b$ so that $f(x,y) = y^3+3x^2y-15y-12bx$ has some critical point

Question

I am trying to solve this excersice but I can't seem to get to anything but dead ends.

Let $b\gt 0$ and $f(x,y) = y^3+3x^2y-15y-12bx$, find all possible values of $b$ so that $f$ has at least one critical point.

I started by deriving:

$f_x(x,y) = 6xy-12b$
$f_y(x,y) = 3y^2+3x^2-15$

So, if $(x,y)$ is a critical point:

$xy = 2b$
$x^2+y^2 = 5$

That would say that the possible values of $b$ are those that define an homographic function $y = \frac{1}{x}2b$ such that its graphic intersects with the border of the circle centered at the origin with radius equal to $\sqrt{5}$.

That is my intuition, but I can't seem to find a way of getting those values in numbers. I don't know if it would help, or if it is even correct, but I also attempted using Lagrange multipliers but that route didn't yield any valid result.

Thank you very much.

score 1 · Answer 1 · answered Nov 24 '14 at 23:33

1

Find the minimum squared distance from the origin to the graph of $y=\frac{2b}x.$ If this minimum squared distance is greater than $5,$ it will not intersect the circle; otherwise, it will.

answered Nov 24 '14 at 23:33

Cameron Buie

102,994

You mean like the function $d(x) = x^2 + \frac{4b^2}{x^2}$? – John Nov 25 '14 at 00:00
Yes! Can you determine when this achieves a minimum? – Cameron Buie Nov 25 '14 at 00:03
I seem to get that $d$ has a critical point when $x=\sqrt{2b}$ but that doesn't seem right. – John Nov 25 '14 at 00:14
That is nearly correct. It should actually be $x=\pm\sqrt{2b}.$ Regardless, plugging either of those values of $x$ in gives us $d(x)=2b+2b=4b.$ Hence, it will not intersect the circle when $4b>5,$ but will otherwise. – Cameron Buie Nov 25 '14 at 00:17
Got it. I wasn't plugging it back in $d$. Thank you very much, Cameron! – John Nov 25 '14 at 00:18
You are very welcome. – Cameron Buie Nov 25 '14 at 00:23

score 1 · Answer 2 · answered Nov 24 '14 at 23:33

1

Hint: $x^2 + y^2 \geq 2xy $ gives you the condition on $b$.

answered Nov 24 '14 at 23:33

DeepSea

77,651

Wow, that was a good one! I get that $b \leq \frac{5}{4}$ – John Nov 24 '14 at 23:46

score 1 · Answer 3 · answered Nov 24 '14 at 23:45

1

Recall that $x^2+y^2=(x+y)^2-2xy$, so your system becomes $$ \begin{cases} (x+y)^2=5+4b\\ xy=2b \end{cases} $$ that forces $5+4b\ge0$ or $b\le-5/4$.

Once this condition is granted, you have two possibilities: $$ \begin{cases} x+y=\sqrt{5+4b}\\ xy=2b \end{cases} \qquad \begin{cases} x+y=-\sqrt{5+4b}\\ xy=2b \end{cases} $$ The first corresponds to the roots of $$ z^2-z\sqrt{5+4b}+2b=0 $$ which has discriminant $5-4b$, and this requires $b\le5/4$. The same for the second system. So, finally, you get $$ -\frac{5}{4}\le b\le\frac{5}{4} $$

answered Nov 24 '14 at 23:45

egreg

238,574

Cool. I didn't see that "trick". As the instructions said that b is positive I can discard the negative values. Thanks a lot, egreg! – John Nov 25 '14 at 00:01
@John: Also, you can discard $0$. – Cameron Buie Nov 25 '14 at 00:03
@John Well, that restriction seems unnatural, because if you change $b$ into $-b$ and $x$ into $-x$ the function remains the same. – egreg Nov 25 '14 at 00:03
@CameronBuie The way with the minimum distance can be handy in other cases. – egreg Nov 25 '14 at 00:06
I will have both approaches in mind for the future. Thank you. – John Nov 25 '14 at 00:20

Find $b$ so that $f(x,y) = y^3+3x^2y-15y-12bx$ has some critical point

3 Answers3