1. ## Maxima problem

The variables $\displaystyle u$ and $\displaystyle v$ take positive values such that $\displaystyle \frac{1}{u}+\frac{1}{v}=\frac{1}{20}$. If $\displaystyle s=u+v$, find the least value of $\displaystyle s$ as $\displaystyle u$ varies.

Since they want the least value of s, my guess is that differentiating s and equating that equation to zero will get me the answer. but i cant figure of anything to differentiate s with respect to

2. Well, you will have to clearly make use of the first equation. Notice that

$\displaystyle \frac{1}{u} + \frac{1}{v} = \frac{u + v}{uv} = \frac{s}{uv}$

Now you want to find the least value of s as u varies. Stated differently, I would interpret that as finding the least value of s while v is held constant (partial differentiation?). We already have s as a function of u and v. So I think this translation is sound, no? See if that helps you figure out the problem.

3. Originally Posted by Punch
The variables $\displaystyle u$ and $\displaystyle v$ take positive values such that $\displaystyle \frac{1}{u}+\frac{1}{v}=\frac{1}{20}$. If $\displaystyle s=u+v$, find the least value of $\displaystyle s$ as $\displaystyle u$ varies.

Since they want the least value of s, my guess is that differentiating s and equating that equation to zero will get me the answer. but i cant figure of anything to differentiate s with respect to
I would set up this problem by solving the first equation for v.

$\displaystyle \frac{1}{u}+\frac{1}{v}=\frac{1}{20} \iff 20v+20u=uv \iff v=\frac{20u}{u-20}$

Now plug this into the equation for s to get

$\displaystyle s=u+\frac{20u}{u-20}$

Now you can use single variable calculus to minimize the function.

4. On second thought, I would think about solving the first equation (the restrictions on u and v) in terms of v. This should help a lot more!

Late Edit: Yeah, like he said lol

5. so with that equation, all i have to do now is ds/du=0, right?

6. Originally Posted by Punch
so with that equation, all i have to do now is ds/du=0, right?
Yes that is correct, but watch out for extraneous solutions.

7. I managed to form the equation $\displaystyle s=u+\frac{20u}{20-u}$ with your help.

Then, $\displaystyle \frac{ds}{du}=1+\frac{(20-u)(20)-(20u)(-1)}{(20-u)^2}$

$\displaystyle =1+\frac{20(20-u)+20u}{(20-u)^2}$

$\displaystyle \frac{ds}{du}=0, 1+\frac{20(20-u)+20u}{(20-u)^2}=0$

$\displaystyle \frac{400-20u+20u}{(20-u)^2}=-1$

$\displaystyle 400=-(20-u)^2$

$\displaystyle 400=-400+40u-u^2$

$\displaystyle u^2-40u+800=0$

but i couldn't find u because u^2-40u+800 has no roots.

8. So you have

$\displaystyle s=u+\frac{20u}{u-20}=\frac{u^2}{u-20} \right)$

Now if you use the product rule you get or quotient rule

$\displaystyle s'=\frac{2u}{(u-20)}-\frac{u^2}{(u-20)^2}=\frac{u}{u-20}\left(2-\frac{u}{u-20} \right)$

Can you finish from here?

9. This is how I would do it

Plug in values, find the minimum s
s = u + v
1/u + 1/v = 1/20

1/40 + 1/40 = 1/20
s=80

1/30 + 1/60 = 1/20
s=90

S keeps growing, the v will be > 80 (your smallest value of s so far)

1/25 + 1/100 = 1/20
s=125

1/24 + 1/120 = 1/20
s=144

1/22 + 1/220 = 1/20
s=242

Etc. Your min s is where u=v so 80.

10. Originally Posted by TheEmptySet
So you have

$\displaystyle s=u+\frac{20u}{u-20}=\frac{u^2}{u-20} \right)$

Now if you use the product rule you get or quotient rule

$\displaystyle s'=\frac{2u}{(u-20)}-\frac{u^2}{(u-20)^2}=\frac{u}{u-20}\left(2-\frac{u}{u-20} \right)$

Can you finish from here?
$\displaystyle \frac{ds}{du}=0, 1+\frac{20(20-u)+20u}{(20-u)^2}=0$

$\displaystyle \frac{400-20u+20u}{(20-u)^2}=-1$

$\displaystyle 400=-(20-u)^2$

$\displaystyle 400=-400+40u-u^2$

$\displaystyle u^2-40u+800=0$

But was what i did wrong?

11. Originally Posted by Punch
$\displaystyle \frac{ds}{du}=0, 1+\frac{20(20-u)+20u}{(20-u)^2}=0$

$\displaystyle \frac{400-20u+20u}{(20-u)^2}=-1$

$\displaystyle 400=-(20-u)^2$

$\displaystyle 400=-400+40u-u^2$

$\displaystyle u^2-40u+800=0$

But was what i did wrong?
$\displaystyle 20-u \ne u-20$

$\displaystyle 1+\frac{20(u-20)-20u}{(u-20)^2}=0$

$\displaystyle \frac{(u-20)^2-400}{(u-20)^2}=\frac{u(u-40)}{(u-20)^2}$

12. Starting with $\displaystyle s=\frac{u^2}{u-20}$

$\displaystyle s'=\frac{(u-20)(2u)-u^2}{(u-20)^2}$

$\displaystyle =\frac{u^2-40u}{(u-20)^2}$

$\displaystyle s'=0, \frac{u^2-40u}{(u-20)^2}=0$

$\displaystyle u(u-40)=0$

$\displaystyle u=0(reject) or u=40$

but i dont have the value of v, so how am i to find s?

13. Originally Posted by Punch
Starting with $\displaystyle s=\frac{u^2}{u-20}$

$\displaystyle s'=\frac{(u-20)(2u)-u^2}{(u-20)^2}$

$\displaystyle =\frac{u^2-40u}{(u-20)^2}$

$\displaystyle s'=0, \frac{u^2-40u}{(u-20)^2}=0$

$\displaystyle u(u-40)=0$

$\displaystyle u=0(reject) or u=40$

but i dont have the value of v, so how am i to find s?
Use the constraint equation.

$\displaystyle \frac{1}{u}+\frac{1}{v}=\frac{1}{20}$

$\displaystyle \frac{1}{40}+\frac{1}{v}=\frac{1}{20}$

Now solve for v.

14. For a completed proof, we were given

$\displaystyle s = u + v,\ u>0,\ v>0,\ \frac{1}{u} + \frac{1}{v} = \frac{1}{20}$

By solving the constraint for v we get

Spoiler:
$\displaystyle u + v = \frac{uv}{20} \iff 20u + 20v = uv \iff v = \frac{20u}{u - 20}$

We can then restate s with this substitution:

$\displaystyle s = u + \frac{20u}{u - 20} = \frac{u^2}{u-20}$

Let us take the derivative of this with respect to u. This is a straight application of the product rule with two functions:

$\displaystyle f(x) = x^2\ \mathrm{and}\ g(x) = (x-20)^{-1}$

Each of which has derivatives:

$\displaystyle f'(x) = 2x\ \mathrm{and}\ g'(x) = -(x-2)^{-2}\times 1$

Using the product rule we get,

Spoiler:
$\displaystyle s'= f' (u)g(u) + f(u)g' (u) = 2u(u-20)^{-1} - u^2(u-20)^{-2}$

Did you get this result? Now let us put it back into a more manageable form.

$\displaystyle s' = \frac{2u}{u-20} - \frac{u^2}{(u-20)^2} = \frac{2u(u-20) - u^2}{(u-20)^2}=\frac{u^2 - 40u}{(u-20)^2}$

Now this is where you need to simplify, set it equal to 0, notice that you can eliminate the denominator, and see the easy solution.

Spoiler:
$\displaystyle s' = \frac{u(u-40)}{(u-20)^2} \stackrel{set}{=} 0 \iff u(u-40) = 0 \iff u = 0 \mathrm{\ or\ } u = 40$

But as emptyset pointed out, we have to be careful of our solutions. If u = 0 we have a problem with our original constraint that u > 0. Therefore, we have found our single extrema. We can also find the value of v under this value of u because we have a statement of v in terms of u. However, we have s in terms of u, so we can find s by directly substituting u.

Spoiler:
$\displaystyle s = \frac{u^2}{u-20} = \frac{40^2}{40 - 20} = \frac{1600}{20} = 80,\ \ v = \frac{20u}{40-20} = u = 40$