A cylindrical can is to hold 1/2 cup of tomato sauce. What are the dimensions of its diameter and height in centimeters, to minimize the amount of metal to make the can?
V=(pi*d^2*h)/4
A=(pi*d^2)+pi*d*h
1qt=946cc

- anonymous

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- schrodinger

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- anonymous

this is a tricky one....

- amistre64

are thos answers? or formulas to apply?

- anonymous

formulas it's a related rates question

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## More answers

- amistre64

yes, optimizations and such

- anonymous

yes they want D, H, and Amin

- amistre64

V = 1/2 cup, soo;
.5 = (pi) (r^2) (h) right?

- anonymous

Yes

- amistre64

A = 2 circles and a rectangle :)

- anonymous

haha :)

- amistre64

A = 2pi (r^2) + h(2pi r)

- amistre64

use the volume to calibrate either h or r
h = .5/pi r^2 seems easiest right?

- amistre64

substitute that "value" into the Area formula and then derive

- anonymous

which value the 946cc?

- amistre64

the .5 cups...worry about conversions afterwards

- anonymous

oh okay gotcha

- amistre64

so...\[h = \frac{1}{2\pi r^2}\] right?

- amistre64

A = 2pi (r^2) + h(2pi r) sooo
\[A = 2\pi r^2 + \frac{2\pi r}{2\pi r^2}\]

- anonymous

I did get that for A

- amistre64

\[A = 2\pi r^2 + {1 \over r}\] correct?

- anonymous

yes

- amistre64

now derive :) you know how to derive?

- anonymous

yes i do

- amistre64

A = 4pi r - 1/r^2 then correct?

- amistre64

A' that is lol

- anonymous

Yes that is what I just got

- amistre64

if we combine these with like denoms we get:
\[A'=\frac{4\pi r^3 - 1}{r^2}\] correct?

- anonymous

Yes that looks good

- amistre64

\[A'=0=4pir^3 -1\] so what are the zeros?

- amistre64

r = ? the easiest way is to plug this into wolframs site and itll tell us right away ;)

- amistre64

http://www.wolframalpha.com/

- anonymous

okay and the equation that I would type in would be \[\sqrt[3]{1/4\pi}\]
right

- amistre64

4pi r^3 - 1 :) and i got abt .4 as the only root

- anonymous

oh okay you don't have to get r by itself

- amistre64

not with wolframs site; itll do it for you

- amistre64

r = aprrox. .430127

- anonymous

okay and r^2 is the diameter and I can use that to solve for Amin

- amistre64

Yes; or simply use it to solve 'h'

- anonymous

Yeah I guess that would be easier lol

- amistre64

how many cups to a quart? 8 or 4?

- anonymous

4 cups

- amistre64

4 cups to 1 quart and we need half a cup
so 1/8th of a quart right?

- anonymous

yes

- amistre64

1/2 cup = 946/8 = 118.25 centimeters squared (cc) then

- amistre64

2 = abt 2.1 centimeters then

- amistre64

r = abt. 2.1 centimeters

- amistre64

A = 2pi(2.1)^2 + 1/(2.1)

- amistre64

Total min area = 28.18 cc

- anonymous

wouldn't it be 2.1^4 since it was d^2

- anonymous

oh never mind I see it now

- amistre64

diameter = 2r = 2(2.1) = 4.2
V = 128.25 = 2pi(2.1)^2 * h
128.25
---------- = h = abt. 4.63
2pi (2.1)^2

- anonymous

Thank you very much I've been working on this problem for about 3 hours and you figured it out in 10 minutes lol! You're a lifesaver!

- amistre64

is it right? cuase i might see an error in it; i allowed the area of the material to be 2 circles, top and bottom of a can; and a height...

- anonymous

Yes it is it matches up to the parts that I had figured out

- amistre64

cool :)

- anonymous

Thanks again!

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