anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
this is a tricky one....
amistre64
  • amistre64
are thos answers? or formulas to apply?
anonymous
  • anonymous
formulas it's a related rates question

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amistre64
  • amistre64
yes, optimizations and such
anonymous
  • anonymous
yes they want D, H, and Amin
amistre64
  • amistre64
V = 1/2 cup, soo; .5 = (pi) (r^2) (h) right?
anonymous
  • anonymous
Yes
amistre64
  • amistre64
A = 2 circles and a rectangle :)
anonymous
  • anonymous
haha :)
amistre64
  • amistre64
A = 2pi (r^2) + h(2pi r)
amistre64
  • amistre64
use the volume to calibrate either h or r h = .5/pi r^2 seems easiest right?
amistre64
  • amistre64
substitute that "value" into the Area formula and then derive
anonymous
  • anonymous
which value the 946cc?
amistre64
  • amistre64
the .5 cups...worry about conversions afterwards
anonymous
  • anonymous
oh okay gotcha
amistre64
  • amistre64
so...\[h = \frac{1}{2\pi r^2}\] right?
amistre64
  • amistre64
A = 2pi (r^2) + h(2pi r) sooo \[A = 2\pi r^2 + \frac{2\pi r}{2\pi r^2}\]
anonymous
  • anonymous
I did get that for A
amistre64
  • amistre64
\[A = 2\pi r^2 + {1 \over r}\] correct?
anonymous
  • anonymous
yes
amistre64
  • amistre64
now derive :) you know how to derive?
anonymous
  • anonymous
yes i do
amistre64
  • amistre64
A = 4pi r - 1/r^2 then correct?
amistre64
  • amistre64
A' that is lol
anonymous
  • anonymous
Yes that is what I just got
amistre64
  • amistre64
if we combine these with like denoms we get: \[A'=\frac{4\pi r^3 - 1}{r^2}\] correct?
anonymous
  • anonymous
Yes that looks good
amistre64
  • amistre64
\[A'=0=4pir^3 -1\] so what are the zeros?
amistre64
  • amistre64
r = ? the easiest way is to plug this into wolframs site and itll tell us right away ;)
amistre64
  • amistre64
http://www.wolframalpha.com/
anonymous
  • anonymous
okay and the equation that I would type in would be \[\sqrt[3]{1/4\pi}\] right
amistre64
  • amistre64
4pi r^3 - 1 :) and i got abt .4 as the only root
anonymous
  • anonymous
oh okay you don't have to get r by itself
amistre64
  • amistre64
not with wolframs site; itll do it for you
amistre64
  • amistre64
r = aprrox. .430127
anonymous
  • anonymous
okay and r^2 is the diameter and I can use that to solve for Amin
amistre64
  • amistre64
Yes; or simply use it to solve 'h'
anonymous
  • anonymous
Yeah I guess that would be easier lol
amistre64
  • amistre64
how many cups to a quart? 8 or 4?
anonymous
  • anonymous
4 cups
amistre64
  • amistre64
4 cups to 1 quart and we need half a cup so 1/8th of a quart right?
anonymous
  • anonymous
yes
amistre64
  • amistre64
1/2 cup = 946/8 = 118.25 centimeters squared (cc) then
amistre64
  • amistre64
2 = abt 2.1 centimeters then
amistre64
  • amistre64
r = abt. 2.1 centimeters
amistre64
  • amistre64
A = 2pi(2.1)^2 + 1/(2.1)
amistre64
  • amistre64
Total min area = 28.18 cc
anonymous
  • anonymous
wouldn't it be 2.1^4 since it was d^2
anonymous
  • anonymous
oh never mind I see it now
amistre64
  • 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
  • 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
  • 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
  • anonymous
Yes it is it matches up to the parts that I had figured out
amistre64
  • amistre64
cool :)
anonymous
  • anonymous
Thanks again!

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