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| Acmlm's Board - I2 Archive - General Chat - Another homework thread. Help me plz. ;_; |
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Danielle Local Moderator  Level: 76  Posts: 2321/3359 EXP: 3958078 For next: 47982 Since: 09-15-04 From: RATE Since last post: 3 hours Last activity: 3 hours |
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This is a math project of mine.  It has to do with volume of retangular boxes. As the base increases in size, the height decreases. But it isn't proportional, the height decreases a lot and the volume therefore gets smaller. There's probably some geometry related explanation behind this that I can't remember, and I need help being able to understand this. It's part of a large project. The exact question is: Of all rectangular boxes with surface areas of 216 square inches and bases of x inches by 2x inches, which has the maximum volume? Explain. I know that the box with the smallest base has the largest volume, but.. why? |
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ErkDog Fuzz Ball Level: 47  Posts: 960/982 EXP: 752190 For next: 14013 Since: 03-15-04 From: Richmond, VA Since last post: 40 days Last activity: 19 days |
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| you did not specify how the height changes... you simply stated the dimensions of the bottom portion. Perhaps if you could give us the rest of the information we could help you |
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Danielle Local Moderator Level: 76 Posts: 2323/3359 EXP: 3958078 For next: 47982 Since: 09-15-04 From: RATE Since last post: 3 hours Last activity: 3 hours |
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| There's not a pattern.. it just decreases. Here's the assignment exactly.. if you need to know some of my work, tell me and I'll type it out. Page 1 Page 2 (#4 is what I'm stuck on, ignore the other 3 questions.) |
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blackhole89 LOLSEALS Moderator of ROM hacking EmuNET IRC network admin Head GM of TwilightRO Level: 47  Posts: 884/971 EXP: 739208 For next: 26995 Since: 03-15-04 From: Dresden/Germany Since last post: 14 hours Last activity: 12 hours |
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Originally posted by Danielle Gah. I solved it, but it is freakin' complicated... let's try to explain: A=2x²+2 + nx*2 + 2xn*2 = 216 (where n is the height of the box) 4x²+6nx=216 => n=(108-2x²)/(3x) We also know that V=x*2x*n=2x²n and are looking for the x, n where V=max. so we can insert the equation we got for n. With some simplifying, we get -(4/3)x³+72x = max. We know this is a cubic function with a positive factor at x^1, and its constant term is zero, so the maximum of it for positive x is positive as well, while it is zero for x=0. Hence, we look at where its derivation is zero. (-(4/3)x³+72x)' = -4x²+72 = 0 x = 3 * sqrt(2) where sqrt is the square root. Let's insert this into the equation for n to get n=4 * sqrt(2). Hence, the box of this kind which has the largest volume is 6 * sqrt(2) inches wide 3 * sqrt(2) inches deep 4 * sqrt(2) inches high You understand my solution so far? I suck at explaining maths in English.  By the way, what grade are you in? Just curious~ |
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Danielle Local Moderator Level: 76 Posts: 2329/3359 EXP: 3958078 For next: 47982 Since: 09-15-04 From: RATE Since last post: 3 hours Last activity: 3 hours |
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Wow.  I solved to find the volumes a different way.. Ashley helped me quite a bit. If you look on page one that I linked to in my previous post, if gives you a formula: 216=x²+x²+4xh. In problem 4 that I'm stuck at, the base isn't square any longer, so you have to change the formula accordingly: x(2x)+x(2x)+2(xh)+2(xh), which can be simplified to 216=4x²+6xh. That's the formula I used to find the volumes of all the boxes, now rectangular. Then I went through, plugging in all the x bases, to get the height. Enter confusion. ..Wait, I haven't even FOUND the volume yet, only the height. CRAP.  So blackhole89... I need to use V=x*2x*n=2x²n to find the volume, with n being the height? I don't really understand all that you said, but it's just because I suck at math horribly. =[ And to answer you, I'm in 11th grade, Math Analysis AKA Precalculus. (edited by Danielle on 09-10-05 01:01 PM) |
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blackhole89 LOLSEALS Moderator of ROM hacking EmuNET IRC network admin Head GM of TwilightRO Level: 47 Posts: 885/971 EXP: 739208 For next: 26995 Since: 03-15-04 From: Dresden/Germany Since last post: 14 hours Last activity: 12 hours |
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Did you already have differential equations at school? Sorry, I know quite few about the US school system. (And to answer any questions that might come up, no, I haven't had it either. Taught myself  )The volume of the "heaviest" box is 6√2*4√2*3√2 = 144√2. As for the formula you mentioned, I already used it in my calculation (look... right up there, only with n instead of h) to express h (n) through x. What I did then was replacing h with its representation in dependency of x in the volume formula and finding its maximum for positive x. |
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Danielle Local Moderator Level: 76 Posts: 2330/3359 EXP: 3958078 For next: 47982 Since: 09-15-04 From: RATE Since last post: 3 hours Last activity: 3 hours |
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| I don't remember problems like this, but I don't remember much of anything in math soo.. I might have before. Certainly nothing this complicated. So what I'm not understanding.. the formula to find the volume using the height? On the first page I linked to, they give an example for how to find the volume with a square base, I just never realized that. D'oh. It was late last night. So. 6√2*4√2*3√2 = 144√2. I don't understand what that is.. you already solved for the volumes? I'm sure when Ashley gets on she can help explain what you're saying, I'm just too poor at math to understand it. |
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ErkDog Fuzz Ball Level: 47 Posts: 965/982 EXP: 752190 For next: 14013 Since: 03-15-04 From: Richmond, VA Since last post: 40 days Last activity: 19 days |
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| umm okay first of all I don't konw what the hell you did blackhole but all you have to do is plug those numbers in that charge into the formula from page 1... furthemore.... all the ansewrs are slightly under 216 look at the friggin chart RIGHT ABOVE THE PROBLEM... when x is 6 the volume is greatest it goes down on both sides of that, just like it should... so... all you have to do is plug the 5.9 and 6.1 numbers into the friggin equation 54x - 1/2 x^3 PROBLEM SOLVED |
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SamuraiX Paratroopa Level: 19 Posts: 13/140 EXP: 32602 For next: 3175 Since: 10-11-04 From: ^___^ Since last post: 2 hours Last activity: 2 hours |
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blackhole89 LOLSEALS Moderator of ROM hacking EmuNET IRC network admin Head GM of TwilightRO Level: 47 Posts: 886/971 EXP: 739208 For next: 26995 Since: 03-15-04 From: Dresden/Germany Since last post: 14 hours Last activity: 12 hours |
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Originally posted by Danielle so how is "54x - 1/2 x^3" the solution?  Originally posted by ErkDog For x=6, the volume is (6*12*(108-2*36)/(3*6) = 144. My answer, being ca. 203, beats it. In other words, Erk... I guess your only intension is to have your revenge for that money thread. Think a little before pushing the "reply" button. |
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Danielle Local Moderator Level: 76 Posts: 2332/3359 EXP: 3958078 For next: 47982 Since: 09-15-04 From: RATE Since last post: 3 hours Last activity: 3 hours |
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Originally posted by Danielle Okay, now I've solved for volume.  I had a blonde moment and forgot to solve for volumes before, I was looking at the height as volume. I realized it now, and I did that work. As the height increases, the volume increases. What I still need to know is WHY. I have to explain why it does this, and I have no idea. I did the math already, that's not what I'm looking for, but thanks for all that effort blackhole89. You made me realize all I had was the height.  So, when I changed everything to meet the x by 2x base(which is different from the chart on page 2, which represents square bases only), I got new heights. Then by plugging the heights into x(2x)h, I got the volumes. Erk, you're solving for volume of square bases, which isn't what I need for this problem. Read #4, not the chart. Blackhole, you were literally solving for the volumes, which is what I didn't do at the time, which was why I didn't have a clue what you were doing. Now I've done that, I just needan explanation as to why the volume changes, why the maximum volume is different with a rectangular base than it is for a square base. Why does the volume increase as the base increases and height decreases? And at 4 it's the biggest, and after that it decreases again..? The surface area is the same, it never changes, but the volume changes. I have to explain this in words, and I have no clue how. Here's my results for volume: x --------------- h ------------------- volume 1 ------------- 35 1/3 --------------- 70.667 1.5 ------------- 23 ------------------ 103.5 2 -------------16.667 --------------133.334 2.5 --------- 12.233 ------------- 159.163 3 --------------- 10 ------------------- 180 3.5 ---------- 7.952 -------------- 194.824 4 ----------- 6 1/3 --------------- 202.667 4.5 ---------- 5 -------------------- 202.5 5 ----------- 3.866 ----------------- 193.3 5.5 -------- 2.878 ---------------- 174.119 6 -------------- 2 --------------------- 144 6.5 -------- 1.205 --------------- 101.823 7 ----------- 10/21 -------------- 46.648 Anything beyond this has a negative height, so forget those. 4 has the maximum volume, but WHY? |
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blackhole89 LOLSEALS Moderator of ROM hacking EmuNET IRC network admin Head GM of TwilightRO Level: 47 Posts: 887/971 EXP: 739208 For next: 26995 Since: 03-15-04 From: Dresden/Germany Since last post: 14 hours Last activity: 12 hours |
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| 4 has the maximum volume if you only check in 0.5 inch steps. As I already stated above (gosh, why can't anybody read my solution as a whole), the actual maximum volume is at x = 3√2 = ca. 4.243 h = 4√2 V= ca. 203.647 Come on IRC please. I'll try to explain there~ (DA #nobodysworld and Emunet are both ok.) |
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ErkDog Fuzz Ball Level: 47 Posts: 966/982 EXP: 752190 For next: 14013 Since: 03-15-04 From: Richmond, VA Since last post: 40 days Last activity: 19 days |
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| the equation for the volume of the box is clearly 2x^2 + 2x^2 + 2xh + 4xh = 216 (top) + (bottom) + 2 of the sides (x*h) + 2 of the other sides (2xh) don't be a wacko blackhole, I misread the question I swa question four ignore the other three and thought it said question 3 *shakes head* (edited by ErkDog on 09-10-05 03:52 PM) (edited by ErkDog on 09-10-05 04:13 PM) |
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