A few questions... please help :)

[jane1234]

1. A can of food has a surface area given by A(r)=2(pi)r^2 + 1000/r

a) Draw graph over a suitable domain.
Would this "suitable domain" be r E (0,infinity) ... as r can't be negative or zero??

But another question says:

b) What would be a suitable domain for the radius (r) ? Justify your answer.
I'm guessing it wouldn't be the same domain as part a) ... does anyone know what this domain would best be restricted to and why??

c) Another can has SA given by A(r) = 2(pi)r^2 + 1500/r. Why is the minimun TP for the second greater than the first?
I know it should be pretty obvious... because the number over r is bigger? Having a hard time putting this into words that sound halfway sophisticated...

d) Comment on the size of the cans. Which one would have a better shape? Justify.
Confused. They both have various shapes given by the changing SA. Anyone have ideas as to which one would be considered "better"??

Any help would be much appreciated. Thanks guys!!

[Mao]

a) and b) are as you described.

c), you can show that A2(r) > A1(r) for all r, thus the minimum turning point must be greater also.

do not understand what d is asking.

[pi]

lol, part d) makes no mathematical sense. I'd just (for lols btw) make a comment on which I would find more aesthetically pleasing. But seriously, what a useless question.

[jane1234]

a) and b) are as you described.

So would b) be the same as a)??
Or should it be restricted further, say to r E (2,10)


And @Rohitpi I know. My teacher loves making up pointless questions...   

[BubbleWrapMan]

For c

1500 > 1000
=> 1500/r > 1000/r for r > 0
=> 2(pi)r^2 + 1500/r > 2(pi)r^2 + 1000/r for r > 0

So the turning point must be higher.

[jasoN-]

for b, heres a fun opportunity to go to ur pantry and measure the radius of ur smallest and largest can, approximate radius would be considered suitable
u can also measure the TSA of a couple of ur cans and compare with the equation for part d i guess.

but yeah clearly theres no exact, proper answer. stupid question is stupid!

[jane1234]

for b, heres a fun opportunity to go to ur pantry and measure the radius of ur smallest and largest can, approximate radius would be considered suitable
u can also measure the TSA of a couple of ur cans and compare with the equation for part d i guess.

but yeah clearly theres no exact, proper answer. stupid question is stupid!

Hmm ... I think we have cans of cat food and that's about it 

[BubbleWrapMan]

Since the surface area of the cylindrical part is 1000/r
And it is, in general, measured by pi x height x diameter = 2(pi)rh
Then 2(pi)(r)(h) = 1000/r
h = 500/(pi)r^2 or alternatively 750/(pi)r^2

(I think?)

Based on this I notice two things:

1) Height can be more easily calculated from the radius in order to find a value of r for which the dimensions of the can are realistic.
2) The volume -- base x height -- is constant for all r > 0. That is, (pi)r^2 x 500/(pi)r^2 = 500, or 750, cubic units.

Not sure if this is enough to answer d? Weird question. I guess it depends how much you want to put in the can.

[jane1234]

Since the surface area of the cylindrical part is 1000/r
And it is, in general, measured by pi x height x diameter = 2(pi)rh
Then 2(pi)(r)(h) = 1000/r
h = 500/(pi)r^2 or alternatively 750/(pi)r^2

(I think?)

Based on this I notice two things:

1) Height can be more easily calculated from the radius in order to find a value of r for which the dimensions of the can are realistic.
2) The volume -- base x height -- is constant for all r > 0. That is, (pi)r^2 x 500/(pi)r^2 = 500, or 750, cubic units.

Not sure if this is enough to answer d? Weird question. I guess it depends how much you want to put in the can.

Yeah volumes are constant. There is also a cost factor (min SA = lowest cost) so I thought maybe the 500 one as it is lower... but what stumped me was how the question asks for which can has the best shape... and the shape is variable with both...

[BubbleWrapMan]

Perhaps compare what the cans look like at minimum surface area?