maximize the volume, from a given surface area...

@katie00

Question

maximize the volume, from a given surface area...

@katie00

anonymous · Answer

??

danjs · Answer

that second practice prob you liked me

danjs · Answer

What is the maximum volume for a sealed cylinder, if the surface area is set at 600pi

danjs · Answer

for a cylinder , volume and surface area
$$V = \pi*r^2*h$$
$$S.A = 2*\pi * r^2 + 2*\pi*r*h$$

danjs · Answer

given S.A is 600pi,  what is the max possible value for the Volume...

danjs · Answer

One thing to think of,  the graph of the function for A ' ,  the first derivative of Volume

will show you when it is increasing/decreasing, depending on the sign of the derivative

danjs · Answer

Volume  and S.A.  are both functions of   height, and radius, cant do the derivative of A with that

danjs · Answer

$$S.A = 600\pi  = 2*\pi * r^2 + 2*\pi*r*h $$

you can solve that surface area  either for r or h,   solving for h will be easier, since not roots to deal with


h = (300/r)  - r

danjs · Answer

put that h in terms of r,  intto the Volume Function, substitute

$$V = \pi*r^2*[\frac{ 300 }{ r } - r]$$

danjs · Answer

now you can differentiate that and get the rate of change of the Volume with respect to the radius


$$V(r) = 300*\pi*r  -  \pi*r^3$$
and
$$V '(r) = 300\pi - 3*\pi*r^2$$

danjs · Answer

V ' will = 0  at   r = 10

danjs · Answer

before  r=10,   0<r<10   the Area is increasing

after r=10,  starts decreasing


local max there

danjs · Answer

Overall - 

You want maximum Volume, that keeps in line with the surface area staying at 600pi

'optimization probs'

danjs · Answer

1 - put the volume in terms of one variable, we did r

2 - differentiate

3 - the maximums change from increasing functions to decreasing,  that occurs at r = 10

danjs · Answer

most probs are like that process, just different things,  boxes, and whjatever

anonymous · Answer

Ty I think I understadn this one much better than the previous one you did. I was working on the previous one and it's still kind of confusing.

danjs · Answer

oh, can you link me back there

danjs · Answer

or we can start again new on it

anonymous · Answer

let's start a new one on that one plz!

anonymous · Answer

@DanJS would there be a grpah to this problem?