# Maximum value of $p+q$ when $pq=r^2$

Question : If $$pq=r^2$$, then the maximum value of $$(p+q)$$ is :

$$1.)$$ I calculated the minimum value of $$(p+q)$$ which is $$2r$$ by AM-GM inequality. But, how do I calculate the maximum value using the same inequality?

$$2.$$ I did follow an alternative method of using the first and second derivatives to obtain critical values and then calculate the maximum value of $$p+q$$.

Following is the calculation executed:

Since, $$pq = r^2$$, this implies $$q = \frac{r^2}{p}$$

Substituting the value of $$q$$ in the function $$f(p, q) = p + q$$ , we get,

$$F(p) = p + \frac{r^2}{p}$$

To find the critical values, we need to calculate the derivative of $$F(p)$$ with respect to $$p$$ , i.e. $$F '(p)$$:

$$F '(p) = 1 - \frac{r^2}{p^ 2}$$

Now, setting $$F '(p) = 0$$ provides the critical values:

$$1 - \frac{r^2}{p^2} = 0$$

Solving for $$p$$, we get $$p = \pm r$$

Now, check the sign of $$F''(p)$$ for $$p = r$$ and $$p = -r$$.

$$F''(p)$$ = $$\frac{r^2}{p^3}$$

For $$p = r$$, $$F''(p)$$ = $$\frac{\large 2}{\large r}$$ $$>0$$ (assuming $$r > 0$$) and thus $$p= r$$ has a minimum

For $$p = -r$$, $$F''(p)$$ = $$\frac{-2}{r}\space\space <0$$ (assuming $$r > 0$$) and thus $$p= -r$$ has a maximum

Now, let's find the corresponding values of $$q$$ by substituting the values of $$p$$ in $$q = \frac{r^2}{p}$$:

For $$p = r$$, $$q = r$$ and $$p+q = 2r$$ (Minimum Value)

For $$p = -r$$, $$q = -r$$ and $$p+q = -2r$$ (Maximum Value)

So, is the maximum value $$-2r$$? I am skeptical about using the assumption ($$r>0$$) as it is not mentioned in the question and thus think that $$-2r$$ cannot be said as the maximum value of $$p+q$$.

• Why are the exponents $2$ and $3$ typeset so large? I know they can get hard to read when typeset in a fraction, but if you just use display mode they should be large enough. For an equation on a line by itself, put it between $$ and $$, and for something in line with text, you can use \dfrac instead of \frac. Aug 6, 2023 at 5:06
• This question is more like a test of "checking plausibility before taking action". You can take arbitrarily large values for $p,$ as long as you take appropriately small values for $q$ so their product remains fixed. Therefore there's no reason to think that $p+q$ attains a maximum value, it gets as large as you want.
– Edd
Aug 6, 2023 at 5:16
• Most of your fractions had a "huge" tag. I removed them. The fonts now seem a little small on your fractions for some reason, but it looks like an improvement to me. Aug 6, 2023 at 6:20

$$F(p) = p + \frac{r^2}p$$ notice that $$F \rightarrow \infty$$ as $$p \rightarrow \infty$$.

So here is your answer. It can get as big as you wish to.

Maybe you want more rigor, let's take $$p>0$$, so:

$$F(p) = p +\frac{r^2}p > p$$

what if $$p <0$$? Then $$\frac{r^2}p < 0 \implies F(p) < p$$ so the function can actually go to $$-\infty$$ if you allow any sign for $$p$$ and $$q$$

My primary point begins after the horizontal line.

First, please consider a note on $$r>0$$. There is a phrase used more commonly in undergraduate level proofs, "Without Loss of Generality", abbreviated WLOG. This means we can make an assumption, which does not significantly restrict the exercise or change the results. In this case, we may assume WLOG, $$r>0$$. This affects the sign of p and later on the sign of q. If you have any doubt about assuming $$r>0$$, work out both options, $$r>0\text{ and }r<0$$. I think you'll find that p and q are simply reversed.

Technically, we should also deal with the possibility, $$r = 0$$. This means at least one of p or q also $$=0$$. Of course, we may need to use the original $$F(p, q)$$ and be careful not to divide by $$q=0$$. For simplicity, let $$p=0$$. Then q is totally free. In this case q has no bound whatsoever, $$q\rightarrow\infty$$. Or the reverse is an alternate solution, $$q=0\text{ and }p\rightarrow\infty$$.

You have correctly found local extrema. A quick glance at the graph of $$F(p) = p + \frac{r^2}{p}$$ explains your counterintuitive results.

For the graph, I used $$r = 1$$, but any value for r will demonstrate similar results. WolframAlpha graph This confirms @HelloFriends results.

$F(p) = p + \frac1p$" />

When the exercise simply asks for maximum value this is usually interpreted as a global maximum. F(p) has no global maximum.

Be aware, that as $$p\rightarrow \infty \text{ or as }p\rightarrow 0^+, F\rightarrow \infty$$ in addition to the case @HelloFriends mentioned.

Remember, absolute maxima are only proven to exist for continuous functions on closed domains like [1,6], not open domains, like $$(-\infty, +\infty)$$

It can be easily seen that maximum is infinity (i.e. we can achieve any large value as we want)… just take $$p = M >0$$ and $$q = \dfrac{r^2}{M}$$ then we have $$pq= r^2$$ and $$p+q = M + \dfrac{r^2}{M}$$ which is not bounded above as we can choose large value of $$M$$.

If $$\,pq=r^2\,,\,$$ then there does not exist the maximum value of $$\,(p+q)\,,\,$$ in fact $$\;\sup\left\{p+q:pq=r^2\right\}=+\infty\,.$$

Proof :
For any $$\,M>0\,$$ there exist $$\,p=M\,$$ and $$\,q=\dfrac{r^2}M\,$$ such that
$$pq=r^2\;$$ and $$\;p+q>M\,.$$