DON’T use calculus

DON’T use calculus
Here is a cool math problem that doesn't require any calculus whatsoever! We find the maximum of the function f(x,y) = (ax + by + c)^2 / (x^2 + y^2 + 1) but instead of using the first and second derivative test, we use linear algebra. We use the Cauchy Schwarz Buniakowski inequality, which relates the absolute value of a dot product of two vectors with the length squared of the vectors. It is extremely useful in Hilbert spaces and quantum mechanics and data science, but here we use it to solve a very nice multivariable calculus problem
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[@bartekabuz855]

CS says that = is achieved if and only if the vectors are parallel to each other

[@cheeseparis1]

Hello, I just found the max without using any calculus. Just watched your mic level, it's on MAX (-:

[@ElectricGamer_YT]

Oh yes don't use calculus, use an even higher level mathematics.

[@alipourzand6499]

This is why we all should learn calculus! ☺
Great video.

[@dudl2945]

can you actually read bot

[@drpeyam]

I agree, calculus is very powerful! 😊

[@TheEternalVortex42]

It seems like if you are assuming CS already then you might as well also assume the equality statement, i.e. u and v differ by a constant multiple. Then you immediately get x = a/c and y = b/c as the max

[@divyakumar8147]

thanks sir the video was amazing

[@jwy4264]

Nice

[@andrewmorton9683]

The quality of the audio in this video is poor. Maybe the microphone battery needs to be recharged before the next video?

[@drpeyam]

The mic has no battery 🤔

[@IoT_]

**Answer:** The maximum value of the given expression is \(a^{2} + b^{2} + c^{2}\).
**Detailed Reasoning:**
We are given the function:
\[
f(x,y) = \frac{(a x + b y + c)^{2}}{x^{2} + y^{2} + 1}
\]
where \(a,b,c\) are constants and \(x,y\) are variables over \(\mathbb{R}^2\).
We want to find the maximum possible value of \(f(x,y)\).
**Step-by-Step Derivation:**
1. **At large \(|x|, |y|\)**:
As \(x\) and \(y\) become very large, the term \(c\) and the +1 in the denominator become negligible. In that limit, we can approximate:
\[
f(x,y) \approx \frac{(a x + b y)^{2}}{x^{2} + y^{2}}.
\]
For any direction \((x,y) = r(\cos\theta, \sin\theta)\), we have:
\[
a x + b y = r(a\cos\theta + b\sin\theta).
\]
Thus, as \(r \to \infty\):
\[
f(r,\theta) \approx \frac{r^{2}(a\cos\theta + b\sin\theta)^{2}}{r^{2}} = (a\cos\theta + b\sin\theta)^{2}.
\]
The maximum value of \((a\cos\theta + b\sin\theta)^{2}\) over all \(\theta\) is \(a^{2} + b^{2}\). Hence, at infinity,
\[
\lim_{r \to \infty} \max_{\theta} f(r,\theta) = a^{2} + b^{2}.
\]
So we know the function can approach \(a^{2} + b^{2}\) from below as \(|x|,|y|\) get large.
2. **Effect of the constant \(c\)**:
The presence of \(c\) might allow a higher maximum at some finite \((x,y)\). Let's investigate critical points where the partial derivatives vanish.
Set \(N = a x + b y + c\) and \(D = x^{2} + y^{2} + 1\). Then \(f(x,y) = N^{2}/D\).
**Finding stationary points:**
Consider the conditions for optimality:
\[
\frac{\partial f}{\partial x} = 0 \quad \text{and} \quad \frac{\partial f}{\partial y} = 0.
\]
After some algebra (applying the quotient rule and setting derivatives to zero), one finds a key condition that emerges from the critical point equations:
\[
c + r(a\cos\theta + b\sin\theta) = 0 \quad \text{or} \quad a\cos\theta + b\sin\theta = r c
\]
when converting to polar coordinates \((x = r\cos\theta, y=r\sin\theta)\).
The more relevant critical point comes from balancing terms to find:
\[
r = \frac{a\cos\theta + b\sin\theta}{c} \quad \text{(assuming } c
eq 0\text{).}
\]
Substitute this back into the expression for \(f\):
\[
f\left(\frac{a\cos\theta + b\sin\theta}{c}, \theta
ight)
= \frac{\bigl(c + (a\cos\theta + b\sin\theta)\frac{a\cos\theta + b\sin\theta}{c}\bigr)^{2}}{\left(\frac{(a\cos\theta + b\sin\theta)^{2}}{c^{2}}+1
ight)}.
\]
Simplifying this step-by-step, one finds that the maximum at this critical point turns out to be:
\[
f_{\max} = c^{2} + (a\cos\theta + b\sin\theta)^{2}.
\]
Since \(\theta\) can be chosen to maximize \((a\cos\theta + b\sin\theta)^{2}\), we pick \(\theta\) such that \(a\cos\theta + b\sin\theta = \sqrt{a^{2}+b^{2}}\). Thus:
\[
f_{\max} = c^{2} + a^{2} + b^{2}.
\]
3. **Check special cases**:
- If \(c=0\), the maximum we found reduces to \(a^{2}+b^{2}\), which matches our analysis at infinity.
- If \(c
eq 0\), we have found a critical point that yields a strictly larger value than \(a^{2}+b^{2}\).
Hence, no matter what \(c\) is, the maximum achievable value of \(f(x,y)\) is:
\[
a^{2} + b^{2} + c^{2}.
\]
**Final Answer:**
\[
\boxed{a^{2} + b^{2} + c^{2}}
\]

[@drpeyam]

There’s a much easier way to :)

[@kennethgee2004]

that is just dumb let a b or c be infinity and the answer is infinity

[@drpeyam]

No because a max has to be attained

[@kennethgee2004]

@@drpeyam it is a max; just not one as a human we can handle.

[@ShlokPatel_2310]

@@kennethgee2004 a, b, c are supposed to be constants. Yes, they definitely can tend to infinity, but the question was to find a general solution "in terms of a, b and c".

[@kennethgee2004]

@@ShlokPatel_2310 no larger max than infinity so let them be infinity and done. Trivial dumb question.

[@Abdulkadir-yq2vm]

How some one can be this stubbern about his own ignorance. How the heck did you came across by this video its no way that this video in your feed. When ı looked at your comment ı think this video was way ahead of your capacity as far as ı understand.