## Poll

8 votes (53.33%) | |||

6 votes (40%) | |||

3 votes (20%) | |||

2 votes (13.33%) | |||

6 votes (40%) | |||

1 vote (6.66%) | |||

3 votes (20%) | |||

2 votes (13.33%) | |||

8 votes (53.33%) | |||

5 votes (33.33%) |

**15 members have voted**

April 2nd, 2020 at 2:39:22 PM
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Max volume = π

Call h the height of the rectangle and d its width, which is also the cylinder's diameter. The volume of the cylinder is:

V = π(d/2)

Given the perimeter of the rectangle as 6, the perimeter equation looks like:

2d + 2h = 6

or, solving for h, would be

h = 3 - d

Substitute h into the volume equation:

V = π(d/2)

multiply and you get:

V = 3π/4 x d

To get a maximum, take the first derivative, dV/dd, and set it equal to 0:

3π/2d - 3π/4d

Now, solve for d and you get:

d = 2, which means h = 1.

Plug d and h into the volume equation, and you get:

V = π(2/2)

which yields:

V = π

Call h the height of the rectangle and d its width, which is also the cylinder's diameter. The volume of the cylinder is:

V = π(d/2)

^{2}x hGiven the perimeter of the rectangle as 6, the perimeter equation looks like:

2d + 2h = 6

or, solving for h, would be

h = 3 - d

Substitute h into the volume equation:

V = π(d/2)

^{2}x (3 - d)multiply and you get:

V = 3π/4 x d

^{2}- π/3 x d^{3}To get a maximum, take the first derivative, dV/dd, and set it equal to 0:

3π/2d - 3π/4d

^{2}= 0Now, solve for d and you get:

d = 2, which means h = 1.

Plug d and h into the volume equation, and you get:

V = π(2/2)

^{2}x 1which yields:

V = π

"Dealer has 'rock'... Pay 'paper!'"

April 2nd, 2020 at 3:14:09 PM
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Quote:JoemanMax volume = π

Call h the height of the rectangle and d its width, which is also the cylinder's diameter. The volume of the cylinder is:

V = π(d/2)^{2}x h

Given the perimeter of the rectangle as 6, the perimeter equation looks like:

2d + 2h = 6

or, solving for h, would be

h = 3 - d

Substitute h into the volume equation:

V = π(d/2)^{2}x (3 - d)

multiply and you get:

V = 3π/4 x d^{2}- π/3 x d^{3}

To get a maximum, take the first derivative, dV/dd, and set it equal to 0:

3π/2d - 3π/4d^{2}= 0

Now, solve for d and you get:

d = 2, which means h = 1.

Plug d and h into the volume equation, and you get:

V = π(2/2)^{2}x 1

which yields:

V = π

"V = 3π/4 x d

^{2}- π/3 x d

^{3}

To get a maximum, take the first derivative, dV/dd, and set it equal to 0:

3π/2d - 3π/4d

^{2}= 0"

Isn't dV/dd = 3π/2 x d - π x d

^{2}?

Also, pardon me for being pedantic, but you didn't show that this makes V a maximum as opposed to a minimum or an inflection point.

April 2nd, 2020 at 4:06:27 PM
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I agree with above...

Using r saves a lot of 1/4 appearing as follows.

Area of top of cyclinder = Pi r^2.

Vol = h * Pi r^2.

Vol = (3 - 2r) * Pi r^2.

Vol = 3 Pi r^2 - 2 Pi r^3.

dV/dr = 6 Pi r - 6 Pi r^2 = 6 Pi r *(1 - r)

This is 0 when r=0 or r=1.

Clearly when r=0, v=0; and r = 1.5 v=0. Hence at r=1 this is a maximum.

r=1 gets d=2, h=1, Vol = Pi.

Area of top of cyclinder = Pi r^2.

Vol = h * Pi r^2.

Vol = (3 - 2r) * Pi r^2.

Vol = 3 Pi r^2 - 2 Pi r^3.

dV/dr = 6 Pi r - 6 Pi r^2 = 6 Pi r *(1 - r)

This is 0 when r=0 or r=1.

Clearly when r=0, v=0; and r = 1.5 v=0. Hence at r=1 this is a maximum.

r=1 gets d=2, h=1, Vol = Pi.

April 2nd, 2020 at 5:03:52 PM
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Quote:ThatDonGuyQuote:JoemanMax volume = π

Call h the height of the rectangle and d its width, which is also the cylinder's diameter. The volume of the cylinder is:

V = π(d/2)^{2}x h

Given the perimeter of the rectangle as 6, the perimeter equation looks like:

2d + 2h = 6

or, solving for h, would be

h = 3 - d

Substitute h into the volume equation:

V = π(d/2)^{2}x (3 - d)

multiply and you get:

V = 3π/4 x d^{2}- π/3 x d^{3}

To get a maximum, take the first derivative, dV/dd, and set it equal to 0:

3π/2d - 3π/4d^{2}= 0

Now, solve for d and you get:

d = 2, which means h = 1.

Plug d and h into the volume equation, and you get:

V = π(2/2)^{2}x 1

which yields:

V = π

"V = 3π/4 x d^{2}- π/3 x d^{3}

To get a maximum, take the first derivative, dV/dd, and set it equal to 0:

3π/2d - 3π/4d^{2}= 0"

Isn't dV/dd = 3π/2 x d - π x d^{2}?

Also, pardon me for being pedantic, but you didn't show that this makes V a maximum as opposed to a minimum or an inflection point.

You are correct! Not sure how that happened. I did this by hand and then copied it over. Must have lost (or gained!) something in translation!

And yes, I did not demonstrate that it was a maximum. I did check my answer before posting, and the volume decreased using both d = 2.01 and 1.99. I figured that was good enough for an internet message board! :)

d = 0 is also a solution, but obviously not a meaningful one for a maximum volume.

And yes, I did not demonstrate that it was a maximum. I did check my answer before posting, and the volume decreased using both d = 2.01 and 1.99. I figured that was good enough for an internet message board! :)

d = 0 is also a solution, but obviously not a meaningful one for a maximum volume.

"Dealer has 'rock'... Pay 'paper!'"

April 2nd, 2020 at 6:33:47 PM
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I agree that pi is the correct answer. Congratulations to all those who solved it.

It's not whether you win or lose; it's whether or not you had a good bet.

April 2nd, 2020 at 11:07:10 PM
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Quote:WizardI agree that pi is the correct answer. Congratulations to all those who solved it.

Just use 2 formulas and differentiation( dy/dx) will get it.

r = radius of cylinder

h= cylinder height

dV/dr = PI(6r - 6r^2) = 0

r=1

V = PI *(1)^1 *1= PI

April 3rd, 2020 at 6:30:51 AM
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Shouldn’t the volume be expressed as cubic units? Am I missing something here? If so, sorry for being not math literate.

NO KILL I

April 3rd, 2020 at 6:53:09 AM
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Quote:MoscaShouldn’t the volume be expressed as cubic units? Am I missing something here? If so, sorry for being not math literate.

The units in the volume are cubic units - not the number. If the perimeter is, say, 6 cm, then the maximum volume is π cm

^{3}.

April 3rd, 2020 at 7:03:12 AM
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Quote:ThatDonGuyThe units in the volume are cubic units - not the number. If the perimeter is, say, 6 cm, then the maximum volume is π cm

^{3}.

Thanks, that’s what I thought.

NO KILL I

April 3rd, 2020 at 12:59:39 PM
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New puzzle:

Imagine the following game, which costs $1.00 to play. At the beginning of each round, a random number between 0 and 1 is generated and then multiplied by $1.00. You get paid a portion of that as follows. If, for instance, you’re on round 3 and the random number is 0.8169 then you get paid 0.8169 / (3 * 3.8169) = $0.07.

You can play as many rounds as you want and you keep winnings from all rounds. Total winnings are paid when you decide to stop playing, and that is the only time they are rounded to the nearest cent.

For calculation purposes, you can assume players will play forever. What is the house edge of this game?

Imagine the following game, which costs $1.00 to play. At the beginning of each round, a random number between 0 and 1 is generated and then multiplied by $1.00. You get paid a portion of that as follows. If, for instance, you’re on round 3 and the random number is 0.8169 then you get paid 0.8169 / (3 * 3.8169) = $0.07.

You can play as many rounds as you want and you keep winnings from all rounds. Total winnings are paid when you decide to stop playing, and that is the only time they are rounded to the nearest cent.

For calculation purposes, you can assume players will play forever. What is the house edge of this game?

Last edited by: Ace2 on Apr 3, 2020

It’s all about making that GTA