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

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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

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this is a tricky one....
are thos answers? or formulas to apply?
formulas it's a related rates question

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Other answers:

yes, optimizations and such
yes they want D, H, and Amin
V = 1/2 cup, soo; .5 = (pi) (r^2) (h) right?
Yes
A = 2 circles and a rectangle :)
haha :)
A = 2pi (r^2) + h(2pi r)
use the volume to calibrate either h or r h = .5/pi r^2 seems easiest right?
substitute that "value" into the Area formula and then derive
which value the 946cc?
the .5 cups...worry about conversions afterwards
oh okay gotcha
so...\[h = \frac{1}{2\pi r^2}\] right?
A = 2pi (r^2) + h(2pi r) sooo \[A = 2\pi r^2 + \frac{2\pi r}{2\pi r^2}\]
I did get that for A
\[A = 2\pi r^2 + {1 \over r}\] correct?
yes
now derive :) you know how to derive?
yes i do
A = 4pi r - 1/r^2 then correct?
A' that is lol
Yes that is what I just got
if we combine these with like denoms we get: \[A'=\frac{4\pi r^3 - 1}{r^2}\] correct?
Yes that looks good
\[A'=0=4pir^3 -1\] so what are the zeros?
r = ? the easiest way is to plug this into wolframs site and itll tell us right away ;)
http://www.wolframalpha.com/
okay and the equation that I would type in would be \[\sqrt[3]{1/4\pi}\] right
4pi r^3 - 1 :) and i got abt .4 as the only root
oh okay you don't have to get r by itself
not with wolframs site; itll do it for you
r = aprrox. .430127
okay and r^2 is the diameter and I can use that to solve for Amin
Yes; or simply use it to solve 'h'
Yeah I guess that would be easier lol
how many cups to a quart? 8 or 4?
4 cups
4 cups to 1 quart and we need half a cup so 1/8th of a quart right?
yes
1/2 cup = 946/8 = 118.25 centimeters squared (cc) then
2 = abt 2.1 centimeters then
r = abt. 2.1 centimeters
A = 2pi(2.1)^2 + 1/(2.1)
Total min area = 28.18 cc
wouldn't it be 2.1^4 since it was d^2
oh never mind I see it now
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
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!
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...
Yes it is it matches up to the parts that I had figured out
cool :)
Thanks again!

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