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I'm learning single variable calculus; I finished a section on related rates several weeks ago. I'm sure the novelty of related rates and simple optimization problems will wear off eventually, but right now I'm having a lot of fun solving these kinds of problems and creating my own. My question is about the following problem:

Consider the function $f(x)={x}\sqrt{a-x}$, $a∈ \mathbb{R}^{+}$. If the value of $a$ is increasing at some rate $R$, how fast is the maximum value of $f$ increasing when $a=k?$

Am I right in thinking that to solve this, you would:

(1) Determine the maximum value of $f$ in terms of $a$. Since $f'(x)= \frac{2a-3x}{2\sqrt{a-x}},$ the global maximum occurs when $x=\frac{2a}{3}.$ Thus the maximum functional value is $f(\frac{2a}{3})=\frac{2\sqrt{3}{a}^{\frac 32}}{9}$.

(2) Treat this maximum value as a function, $F(a)=\frac{2\sqrt{3}{a}^{\frac 32}}{9}$, which gives the maximum value of $f$ for any given $a$.

(3) Differentiate $F(a)$ using the chain rule, yielding $F'(a)=R\cdot \frac{\sqrt{3}{k}^{\frac 12}}{3}?$

Is this correct?

BTW, I just want to say how much I appreciate those who are willing to share their expertise on this site. I found out about this site a couple of days ago. I asked my first question about how to prove that certain bounded functions must have at least one inflection point, and within minutes--minutes!--an emeritus professor showed my how this could be done. He even took the time to guide me through a little hurdle in his proof. Just incredible. This site rocks.

Ryan
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  • Always remember to show some of your work or anything that you have attempted, that way people are more likely to help you –  Jul 25 '13 at 03:28
  • @DannyCheuk This user has done a pretty good job at showing his work and his attempt. In fact, it looks like he's mostly solved it. – rurouniwallace Jul 25 '13 at 03:36
  • @ZettaSuro, you got me wrong here, it's just a comment to his "BTW, I just ..." opinion about this site, I agree that he showed his work on this question. Apologies for the confusion, should have stated that on my previous comment –  Jul 25 '13 at 03:39
  • @Danny Cheuk Thanks for your suggestion. I've made a few changes, showing more work. I wasn't sure at first how many steps I should include. – Ryan Jul 25 '13 at 03:46

1 Answers1

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First determine the maximum value of $f(x)$ as a function of a:

$$\begin{align} f'(x)=&-\frac{x}{2\sqrt{a-x}}+\sqrt{a-x}=\frac{-x+2a-2x}{2\sqrt{a-x}} \\ \end{align}$$ Equate the first derivative to zero: $$\begin{align} 0=&-3x+2a \implies x=\frac{2}{3}a \end{align}$$

Since the domain of $a$ is an open interval, the absolute maximum of $f(x)$ is when $x=\frac{2}{3}a$.

$$m(a)=\frac{2}{3}a$$

What is $\frac{d}{da}f(m(a))$ when $a=k$?

Starlight
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rurouniwallace
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  • The maximum value of the function is $f(\frac{2a}{3})$, right? – Ryan Jul 25 '13 at 03:50
  • @Ryan yes, now you need to find what the rate of change of the maximum is when $a=k$. – rurouniwallace Jul 25 '13 at 03:51
  • $f(\frac{2a}{3})= \frac{2a}{3} \cdot \sqrt{ a -\frac{2a}{3}}=\frac{2a}{3} \cdot \sqrt{\frac {a}{3}}=\frac{2\sqrt{3}{a}^{\frac 32}}{9}.$ Correct? – Ryan Jul 25 '13 at 04:01
  • And then we treat this last expression, $\frac{2\sqrt{3}{a}^{\frac 32}}{9},$ as a function itself--the function that returns the maximum value of $f$ in terms of a. Call this function $F(a)=\frac{2\sqrt{3}{a}^{\frac 32}}{9}.$ Then we differentiate $F(a)$, and finally plug in the value $k$ into $F'(a).$ Is that all correct? – Ryan Jul 25 '13 at 04:04
  • I guess my question is: Is the above correct? I've detailed in the rest of the steps in my original question. I just wanted to know if I did it all correctly. – Ryan Jul 25 '13 at 04:13
  • @Ryan Yes, that is correct :) – rurouniwallace Jul 25 '13 at 04:15