- Thread starter
- #1

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

- Thread starter MoneyKing
- Start date

- Thread starter
- #1

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

- Feb 5, 2012

- 1,621

Hi MoneyKing,

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

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.

- Thread starter
- #3

Hey Sudharaka,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.

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

- Feb 5, 2012

- 1,621

Hi MoneyKing,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

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.

- Thread starter
- #5

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 chapterHi 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.

- Admin
- #6

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.

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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