- #1

Hyperreality

- 202

- 0

Now some of you might have seen this problem in Fundamentals of Physics Extended Fifth Edition. Please give me some hints.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Hyperreality
- Start date

- #1

Hyperreality

- 202

- 0

Now some of you might have seen this problem in Fundamentals of Physics Extended Fifth Edition. Please give me some hints.

- #2

HallsofIvy

Science Advisor

Homework Helper

- 43,021

- 970

A cylinder of radius r and height h has three surfaces: the top and bottom, each of area [pi]r

V= [pi]r

the surface area as A= 2[pi](r

Differentiate with respect to r, set equal to 0 and solve for r.

You will get a formula for r that depends on V. Substitute for V with the formula above and see what happens.

- #3

Hyperreality

- 202

- 0

First solution:

A minimum area is obtained when h = r for A = [pi] r^2 + 2[pi] rh.

Second solution:

A minimum surface area is obtained when h = 2r.

Can anyone please explain the result to me?

- #4

Hyperreality

- 202

- 0

Second solution: when A = 2[pi] r^2 + 2[pi] rh

- #5

HallsofIvy

Science Advisor

Homework Helper

- 43,021

- 970

the "lateral surface" area is 2[pi]rh and each end has area [pi]r

A= [pi]r

Share:

- Last Post

- Replies
- 6

- Views
- 815

- Last Post

- Replies
- 11

- Views
- 571

- Last Post

- Replies
- 2

- Views
- 469

- Replies
- 1

- Views
- 398

MHB
Graphic Help

- Last Post

- Replies
- 2

- Views
- 330

MHB
Help

- Last Post

- Replies
- 4

- Views
- 597

MHB
Please Help!

- Last Post

- Replies
- 1

- Views
- 365

MHB
Please help

- Last Post

- Replies
- 2

- Views
- 336

MHB
Help

- Last Post

- Replies
- 5

- Views
- 554

MHB
Help

- Last Post

- Replies
- 3

- Views
- 541