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avatar+583 

hi good people!,

 

I am given a cylinder which has a capacity of 2000 cubic cm. First they ask to show that the height of the cylinder can be expressed as \(h={2000 \over{pi x^2}}\)

 

this was easy to do...

 

the second question is to determine a formula for the outer surface of the cylinder in terms of x...

the third question is to determine the measurements which would minimize the material needed to make this cylinder..

 

please help me with questions 2 and 3...Thank you all very much!

 Jul 17, 2019
edited by juriemagic  Jul 17, 2019
edited by juriemagic  Jul 17, 2019
 #1
avatar+19797 
+2

X is the radius of the cylinder.

 

Surface area =   pi (2x) h     (if we do not include the endcaps)     

                           pi 2x h + 2 pi x^2  (if we include the ends)

 Jul 17, 2019
edited by Guest  Jul 17, 2019
 #2
avatar+105538 
+3

Juriemagic, it is good to see you here   laugh

 

 

\(S=2\pi x*h+2*\pi x^2\\ S=2\pi x*\frac{2000}{\pi x^2}+2*\pi x^2\\ S= \frac{4000}{x}+2\pi x^2\\ S= 4000x^{-1}+2\pi x^2\\ \frac{dS}{dx}= -4000x^{-2}+4\pi x\\ \frac{d^2S}{dx^2}= 8000x^{-3}+4\pi\\ \frac{d^2S}{dx^2}>0 \qquad \text{since x>0, concave up} \\ \text{So any turning point where x>0 will be a minimum}\\ \text{find minimum}\\ \frac{dS}{dx}= -4000x^{-2}+4\pi x=0\qquad x>0\\ -4000+4\pi x^3=0\\ \pi x^3=1000\\ x=\frac{10}{\sqrt[3]{\pi}} \)

 

so for minimum surface area  the radius is    \( \frac{10}{\sqrt[3]{\pi}}\;\;cm\)        and the height is   

 

\(h=  \frac{2000}{\pi x^2}\\ h=  2000 \div( \pi x^2)\\ h=  2000 \div[ \pi (\frac{10}{\pi ^{1/3}})^2]\\ h=  2000 \div[ \frac{100\pi}{\pi ^{2/3}}]\\ h=  2000 \div[ 100\pi^{1/3}]\\ h=\frac{20}{\sqrt[3]{\pi}}\)

 

 

 

 

\(min SA=4000x^{-1}+2\pi x^2\\ min SA=4000(\frac{10}{\pi^{1/3}})^{-1}+2\pi (\frac{10}{\pi^{1/3}})^2\\ min SA=4000(\frac{\pi^{1/3}}{10})+2\pi (\frac{100}{\pi^{2/3}})\\ min SA=400\pi^{1/3}+(\frac{200\pi}{\pi^{2/3}})\\ min SA=600\pi^{1/3}\\\)

 

 Jul 17, 2019
edited by Melody  Jul 18, 2019
 #3
avatar+583 
+1

Melody!!!!!!...thank you for this very complicated answer!!...smileysmiley....I will spend some time with it for sure, trust all is well with you?

juriemagic  Jul 19, 2019
 #4
avatar+105538 
0

Hi Juriemagic,

Yes thanks I am fine thanks :)

 

This technique takes a while to assimilate (to fully understand)

But once you get it there are many questions that will use this technique.

If you do not understand it from some particular line please tell me and I (or someone else) will try to explain better.

If i do no respond send me a private message because I may just not have seen your question.    laugh

Melody  Jul 19, 2019

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