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Finding surface area with volume

MoneyKing

New member
Jan 31, 2012
4
Ok, so I did a test today for advance functions and there was a question: the volume of a pop can is 350ml, find the minimum surface area and determine the dimensions.
Where I got stuck:


350=pi(r)^2*h
h=350/pi(r)^2


SA= 2pi(r)^2+2pi(r)( h)
SA= 2pi(r)^2+2pi(r)(350/pi(r)^2)


=(2(pi(r)^3+350))/(pi(r))


I'm stuck here :S
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Ok, so I did a test today for advance functions and there was a question: the volume of a pop can is 350ml, find the minimum surface area and determine the dimensions.
Where I got stuck:


350=pi(r)^2*h
h=350/pi(r)^2


SA= 2pi(r)^2+2pi(r)( h)
SA= 2pi(r)^2+2pi(r)(350/pi(r)^2)


=(2(pi(r)^3+350))/(pi(r))


I'm stuck here :S
Hi MoneyKing, :)

You should get,

\[A_{s}=\frac{2(\pi r^3+350)}{r}\]

Use, the first derivative test or the second derivative test to find out the value of \(r\) which minimizes \(A_{s}\).

Kind Regards,
Sudharaka.
 

MoneyKing

New member
Jan 31, 2012
4
Hi MoneyKing, :)

You should get,

\[A_{s}=\frac{2(\pi r^3+350)}{r}\]

Use, the first derivative test or the second derivative test to find out the value of \(r\) which minimizes \(A_{s}\).

Kind Regards,
Sudharaka.
Hey Sudharaka,
How did you get \[A_{s}=\frac{2(\pi r^3+350)}{r}\]
and I didn't learn how to do the first derivative test or the second, can you show me, or tell me where I can learn how to do that?
please and thanks
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Hey Sudharaka,
How did you get \[A_{s}=\frac{2(\pi r^3+350)}{r}\]
and I didn't learn how to do the first derivative test or the second, can you show me, or tell me where I can learn how to do that?
please and thanks
Hi MoneyKing, :)

You have made a mistake in the highlighted part in post #1.

\[350=\pi r^2 h\mbox{ and }A_{S}=2\pi r^2+2\pi r h\]

\[\Rightarrow A_{S}=2\pi r^2+2\pi r\left(\frac{350}{\pi r^2}\right)\]

\[\Rightarrow A_{S}=\frac{2(\pi r^3+350)}{r}\]

There are loads of resources on the internet that you can learn about the first and second derivative tests. Here are two videos that you will find useful.

First Derivative Test: Calculus: 3.2 - Extrema and the First-Derivative Test - YouTube


Second Derivative Test: Calculus: Concavity and the Second-Derivative Test - YouTube

Kind Regards,
Sudharaka.
 

MoneyKing

New member
Jan 31, 2012
4
Hi MoneyKing, :)

You have made a mistake in the highlighted part in post #1.

\[350=\pi r^2 h\mbox{ and }A_{S}=2\pi r^2+2\pi r h\]

\[\Rightarrow A_{S}=2\pi r^2+2\pi r\left(\frac{350}{\pi r^2}\right)\]

\[\Rightarrow A_{S}=\frac{2(\pi r^3+350)}{r}\]

There are loads of resources on the internet that you can learn about the first and second derivative tests. Here are two videos that you will find useful.

First Derivative Test: Calculus: 3.2 - Extrema and the First-Derivative Test - YouTube


Second Derivative Test: Calculus: Concavity and the Second-Derivative Test - YouTube

Kind Regards,
Sudharaka.
thanks I see my mistake, umm if possible can you show me how to do this question? I still didn't take calculus, I just started Advance Functions, just finished first chapter
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I don't think you would gain much from seeing differential calculus applied to find the extrema. Since you have not had calculus yet, I suspect you are to use graphical means to approximate the value of r for which the surface area is minimized.

If you graph the function, your calculator may even have a feature which allows you to find the minimum.

edit: It is also possible you may have been given the fact that for a given volume, the cylinder with minimum surface area has its height equal to its diameter. This would allow you to find the dimensions directly.
 
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