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My daughter tackled this problem yesterday and her answer was out by one unit. We can't work out if it's just a rounding issue or if there was something fundamentally wrong with her process. Sadly, her textbook doesn't give worked solutions 😕

I'm wondering if anyone here is willing to give it a go and we can compare answers. @square_25 or @kiana perhaps? Thank you 🌻

 

You have a sphere of cheese with radius 8cm. What is the largest cylinder of cheese you can cut from this? Give the volume of this cylinder of cheese.

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I’m just taking a shot at this, while waiting for actual math wizards (or even math Muggles) to chime in.

I imagine a triangle with two 8cm sides, coming from the centre of the sphere. If half of that is a right triangle, then I want the two sides of the right triangle, with 8cm being the hypotenuse. One side is the radius of the cylinder, the other is half the length.

Edited by arctic_bunny
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Calculus, right? 

I'm going to mess this up because I'm too lazy to get a piece of paper, but here goes. Say that the bottom face of the cylinder has radius r. Then the distance from the center of the sphere to the face of the cylinder is sqrt(64 - r^2) by Pythagoras, so the height of the cylinder is 2*sqrt(64 - r^2) and the volume is 

V = pi*r^2 * 2*sqrt(64 - r^2)

Have I done anything silly so far? At this point, I'd get all the rs under the square root and maximize what's under there (so I don't have to deal with maximizing square roots.) 

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10 minutes ago, arctic_bunny said:

I’m just taking a shot at this, while waiting for actual math wizards (or even math Muggles) to chime in.

I imagine a triangle with two 8cm sides, coming from the centre of the sphere. If half of that is a right triangle, then I want the two sides of the right triangle, with 8cm being the hypotenuse. One side is the radius of the cylinder, the other is half the length.

Yes, this is the first step my daughter took too. 

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1 minute ago, square_25 said:

Calculus, right? 

I'm going to mess this up because I'm too lazy to get a piece of paper, but here goes. Say that the bottom face of the cylinder has radius r. Then the distance from the center of the sphere to the face of the cylinder is sqrt(64 - r^2) by Pythagoras, so the height of the cylinder is 2*sqrt(64 - r^2) and the volume is 

V = pi*r^2 * 2*sqrt(64 - r^2)

Have I done anything silly so far? At this point, I'd get all the rs under the square root and maximize what's under there (so I don't have to deal with maximizing square roots.) 

That's looking similar. My daughter got V in terms of h though, not r. I can't imagine that matters overall though.

And yes, it's in an applications of calculus chapter. Good fun 👍

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Nope, shouldn't matter. OK, let me keep going. Then we're maximizing sqrt(r^4*(64 - r^2)), so maximizing

64r^4 - r^6.

Taking the derivative gets me 256 r^3 - 6r^5, setting to 0 gets 

256 r^3 - 6r^5 = 0 , ----> r = 0 or r^2 = 256/6 = 128/3. 

Plugging in, that gives me 

V = pi*128/3*2*sqrt(64 - 128/3), which is about 1238.

Did you get anything like that? 

 

Edited by square_25
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That's exactly what my daughter got - 1238 cubic cm. 

The answer key says 1239.

Thanks heaps for your help. 🌻

I think we can now put this problem aside and be confident she's got it. Thanks so much.

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I think I've mentioned in some other thread that my daughter is really focused on mastery. So, near enough isn't good enough. If she misses a question, she needs to know why, and she won't move on. 

If I can tell her that an AoPS teacher got her answer, I think she'll be happy that the answer key may have been wrong this time. 🙂

Thanks heaps.

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8 minutes ago, chocolate-chip chooky said:

That's exactly what my daughter got - 1238 cubic cm. 

The answer key says 1239.

Thanks heaps for your help. 🌻

I think we can now put this problem aside and be confident she's got it. Thanks so much.

Yeah, looks like they just rounded wrong to me :-). Glad to help! 

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By the way, I totally skipped the steps which made it clear why setting the derivative to 0 is the right thing here... those are good to think about, too! 

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The only way that I get 1239 is if I round r^2 and h two decimal places prior to substituting into the final formula. 

Which is WRONG. ARGH. 

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13 hours ago, square_25 said:

By the way, I totally skipped the steps which made it clear why setting the derivative to 0 is the right thing here... those are good to think about, too! 

I think she's pretty clear on what she's doing and why. When she was showing me her working and trying to work out that lost unit in the final answer, she talked through all of the steps, including why she was differentiating (needing a gradient function) and why she was setting it to zero (stationary points will be when gradient is zero). She also checked that the stationary point she found was actually a maximum, by considering the expected shape of the cubic function and also considering points to left and right of the turning point and checking if they were +ve or -ve gradients.

Is this the sort of thing you mean?

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13 hours ago, kiana said:

The only way that I get 1239 is if I round r^2 and h two decimal places prior to substituting into the final formula. 

Which is WRONG. ARGH. 

Thank you so much for joining in with the fun of maximising the cheese 🙂

I got to tell my daughter that a college maths teacher AND an AoPS teacher agree with her answer. She was truly ever so pleased, as she really really really doesn't like getting wrong answers.

Thank you 🌻

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2 hours ago, chocolate-chip chooky said:

I think she's pretty clear on what she's doing and why. When she was showing me her working and trying to work out that lost unit in the final answer, she talked through all of the steps, including why she was differentiating (needing a gradient function) and why she was setting it to zero (stationary points will be when gradient is zero). She also checked that the stationary point she found was actually a maximum, by considering the expected shape of the cubic function and also considering points to left and right of the turning point and checking if they were +ve or -ve gradients.

Is this the sort of thing you mean?

Great, yeah, that’s what I meant! Also noticing that the function is continuous and differentiable everywhere, so the only maxes and mind are at points with derivative 0. 

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This reminds me of a student that my undergrad adviser told me about. They were using a classic book on mathematical probability/statistics -- the issue was that sometimes for 2 SD they'd use 95% and sometimes they'd use 95.45%, and so he wouldn't get exactly what the back of the book said. 

It drove the student nuts. 

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