Let $g(x,y)=\frac{3x^2+16xy+15y^2}{x^2+y^2}$. If $y=0$ then $g(x,y)=3$. Assume otherwise, that $y\ne0$. Dehomogenizing, we may assume that $y=1$. Then: \begin{align*} f(x)&=g(x,y)\\ &=\frac{3x^2+16x+15}{x^2+1}\\ &=3+\frac{16x+12}{x^2+1}\end{align*}We have: $$f'(x)=-\frac{8(2x-1)(x+2)}{(x^2+1)^2}.$$Then since $\lim_{x\to-\infty}f(x)=3$, $f(-2)=-1$, $f\left(\frac12\right)=19$, and $f$ is decreasing on $\left(\frac12,\infty\right)$ we have a maximum of $\boxed{19}$.

Let $g(x,y)=x^2+y^2=c,$ so our function $f(x,y) = \frac{3c+16xy+12y^2}{c} = 3 + \frac{16xy+12y^2}{c}$ Now it's time to pull out LM! Maximize $f(x,y)$ subject to $g(x,y) = c$ So we have $\nabla f = \lambda \nabla g$ ie. $\left\langle \frac{16y}{c}, \frac{16x+24y}{c} \right\rangle = \lambda \langle 2x, 2y \rangle$ ie. $8y = \lambda c x$ and $8x+12y = \lambda c y$ so $8x = (\lambda c - 12)y$ and $x^2+y^2=c$ $y = \frac{\lambda c x}{8},$ giving us $8x = (\lambda c - 12) \lambda cx / 8$ ie. $8 = (\lambda c - 12) \lambda c / 8$ then $64 = (\lambda c)(\lambda c - 12),$ or $(\lambda c)^2 - 12(\lambda c) - 64 = 0,$ ie. $\lambda c = -4, 16.$ If $\lambda c = 16,$ then $8y=16x,$ so $y=2x,$ and $8x=4y,$ so $y=2x.$ Thus, $x^2+4x^2=c,$ ie. $x = \pm \sqrt{c / 5}$ and $y = \pm 2 \sqrt{c / 5}$ which gives us maximum $g(x,y) = 3 + \frac{16 \cdot 2c/5 + 12 \cdot 4c/5}{c}=19,$ If $\lambda c = -4,$ then $8y=-4x,$ so $x=-2y,$ so $x=\pm 2 \sqrt{c/5}$ and $y= \mp \sqrt{c/5}$ which yields the minimum $g(x,y) = 3 + \frac{-16 \cdot 2c/5 + 12 c / 5}{c} = -1.$ so the maximum is indeed $\boxed{19}.$

Common mistake with LM, you haven't proved that a maximum exists.

The region we're considering is a fairly common constraint in LM problems (the disk $x^2+y^2=c$), is obviously closed and bounded so maxima and minima both exist by extreme value theorem. It's pretty clear and you don't have to explicitly state that every time you use LM for standard problem constraints like this so i don't think this is necessarily a "common mistake." Fair point to be precise ig

Elementary solution: $\frac{3{{x}^{2}}+16xy+15{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}=m\Leftrightarrow (3-m){{x}^{2}}+13yx+(15-m){{y}^{2}}=0,x\in \mathbb{R}\Rightarrow {{\Delta }_{x}}\ge 0$ So ${{m}^{2}}-18m-19\le 0\Rightarrow m\le 19$

beautiful solution!

parmenides51 wrote: Find the largest possible value in the real numbers of the term $$\frac{3x^2 + 16xy + 15y^2}{x^2 + y^2}$$with $x^2 + y^2 \ne 0$. https://artofproblemsolving.com/community/c4h319781p1852351

jasperE3 wrote: Common mistake with LM, you haven't proved that a maximum exists. This is a common mistake with minima maxima solutions in general, not just Lagrange multipliers. Proving the possible existence of a maximum is not the same as showing that it is actually there. As an example, #2 actually did this correctly (and it isn’t hard to do!).

Arr0w wrote: jasperE3 wrote: Common mistake with LM, you haven't proved that a maximum exists. This is a common mistake with minima maxima solutions in general, not just Lagrange multipliers. Proving the existence of a maximum is not the same as showing that it is actually there. As an example, #2 actually did this correctly (and it isn’t hard to do!). Not really. For LM once you establish that the region is compact then it is known that maxima/minima exist; and if $\nabla g(x,y) \neq 0$ at the critical points then LM finds them / shows that the maxima/minima will satisfy those conditions - that's the whole point of LM. So my solution is correct and I've showed that the maxima and minima are as claimed.

MelonGirl wrote: Uh no, this is false. For LM once you establish that the region is compact then it is known that maxima/minima exist Why should I assume that in your solution? What you say is true, but you need to use the Heine-Borel Theorem/Extreme Value Theorem link to prove this true (which is literally why that's true, compactness shows extrema exist), which you didn't do in #3. All you did was use LM which only proves existence. So yes, what I said was true. See this which says verbatim what I said in #9. Also, your solution isn't wrong, your solution is right in the sense that you got the right answer. If you took the Puerto Rico TST from 2010, you would have been fine (except for the reasons I and JasperE3 both stated to you). The way you got there was fine, but the way you proved it was wrong.

Bro, this is an online math forum, not a TST. You don't have to necessarily go over every detail in a solution post. (funnily enough, the TST might be where a point could be docked for not paying attention to this detail, sure.) The region being closed and bounded -> maxima/minima exists by extreme value theorem is something I literally mentioned in #5, can you read? That's the only valid criticism you have which I addressed. Also that's something you learn in standard multivar classes when doing LM. It's not a terribly "advanced" argumentation. And I already pointed out that since this is such a standard LM problem with a very common condition that you don't have to explicitly say "Because the region is closed and bounded maxima and minima exist..." since it's clear that this is the case with $g(x,y) = x^2+y^2=c.$

MelonGirl wrote: Can you read? No

Arr0w wrote: MelonGirl wrote: Can you read? No Understandable. Have a good day

JasperE3's point is correct; Arro0w's is partially correct. The post by greenturtle that Arr0w doesn't say anything new. "All you did was use LM which proves existence." Not quite. Here, EVT proves existence; LM actually finds the correct maximum once you know that it exists. If you look at greenturtle's post all they did was make sure to cite compactness/EVT to prove that the maximum does exist. Once you do that, applying LM will find the optimum values.

Using the nice technique outlined in this post, slightly adapted: We write $3x^2+16xy+15y^2 = a(x^2+y^2)+b(mx+ny)^2$, and solving gives that we can write it as $19(x^2+y^2)-(4x-2y)^2$ and $-(x^2+y^2)+(2x+4y)^2$. The first one is the one that will give us a bound for the maximum. We have that \[\frac{3x^2+16xy+15y^2}{19(x^2+y^2)-(4x-2y)^2} = 1 \implies \frac{3x^2+16xy+15y^2}{19(x^2+y^2)} \leq 1 \implies \frac{3x^2+16xy+15y^2}{x^2+y^2} \leq 19\] Equality holds when $4x=2y$, for example at $(1,2)$.

We claim that $\frac{3x^2+16xy+15y^2}{x^2+y^2}\le19$ with equality when $(x,y)=(2021,4042)$. Note that $3x^2+16xy+15y^2\le19x^2+19y^2\Leftrightarrow(2x-y)^2\ge0$, and we're done.

Similar to the above two. so happy that there was a cauchy solution $\frac{3x^2+16xy+15y^2}{x^2+y^2} = \frac{(2x+4y)^2 - (x^2+y^2)}{x^2+y^2} = \frac{(2x+4y)^2}{x^2+y^2}-1$ By Cauchy, $(2x+4y)^2 \leq (x^2+y^2)(2^2+4^2) = 20(x^2+y^2),$ so $\frac{(2x+4y)^2}{x^2+y^2} \leq 20$ Thus maximum of original fraction is $\boxed{19},$ with equality at $4x=2y, x \neq y.$