Calculus optimization question

[carlosruiz]

find the greatest value of the sum of r+t on the circle with equation r^2+t^2=9. Any help would be greatly appreciated, specifically if it could be broken down into step by step pieces with an explanation as I'm completely lost on this homework. Thanks!

 

[galactus]

The max is achieved if the values are the same. That is r=t

So, we have \(\displaystyle r^{2}+r^{2}=9\Rightarrow 2r^{2}=9\)

solve for r. t is the same. find r, then double it to find the max.

 

[carlosruiz]

Thank you! I found out that the sum was 2(sq.rt. of (9/2). Is there any way I can prove this using calculus?

 

[galactus]

Yes, there is.

We want to maximize \(\displaystyle M=r+t\)............[1]

But, \(\displaystyle r^{2}+t^{2}=9\)

\(\displaystyle t=\sqrt{9-r^{2}}\)

Sub into [1]:

\(\displaystyle r+\sqrt{9-r^{2}}\)

differentiate:

\(\displaystyle \frac{dM}{dr}=\frac{\sqrt{9-r^{2}}-r}{\sqrt{9-r^{2}}}\)

Set to 0 and solve for r:

\(\displaystyle \sqrt{9-r^{2}}-r=0\)

Solve for r. I bet you get the same answer.

 

[carlosruiz]

Once again thank you. You made this topic much clearer to me. My final question (sorry to keep bothering you) is how you got the equation that was differentiated?

 

[galactus]

How did I differentiate \(\displaystyle r+\sqrt{9-r^{2}}\)?.

\(\displaystyle 1+\underbrace{\frac{1}{2}(9-r^{2})^{\frac{-1}{2}}(-2r)}_{\text{chain rule}}\)

simplify:

\(\displaystyle 1-\frac{r}{\sqrt{9-r^{2}}}\)

Common denominator. Just like fractions. Multiply the left side, 1, by \(\displaystyle \frac{\sqrt{9-r^{2}}}{\sqrt{9-r^{2}}}\)

\(\displaystyle \frac{\sqrt{9-r^{2}}-r}{\sqrt{9-r^{2}}}\)

 

[carlosruiz]

Oh I see now, thank you