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| Calculus help (Optimization Integration) |
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| Kyuande:
This is fake lol |
| Neventii:
what a loving cunt, I'll make sure to ban him from my education system |
| Ipquarx:
--- Quote from: Pastrey Crust on February 13, 2017, 10:49:01 AM ---1) Part the number 100 into two which the sum of their cubes is minimum. --- End quote --- Okay, so we have 1 variable here: 100 = a + (100 - a) And we're trying to minimize: a^3 + (100 - a)^3 In the domain [0, 100] We get the derivative of this: d/dx (a^3 + (100-a)^3) = 3a^2 - 3*(100-a)^2 Simplifies into: 600 (a - 50) Find where this is zero, in this case a = 50. This is where the minimum is. I forget what theorem this comes from. So your minimum is 250,000 for the sum of cubes. |
| Pastrey Crust:
--- Quote from: Ipquarx on February 13, 2017, 05:05:45 PM ---Okay, so we have 1 variable here: 100 = a + (100 - a) And we're trying to minimize: a^3 + (100 - a)^3 In the domain [0, 100] We get the derivative of this: d/dx (a^3 + (100-a)^3) = 3a^2 - 3*(100-a)^2 Simplifies into: 600 (a - 50) Find where this is zero, in this case a = 50. This is where the minimum is. I forget what theorem this comes from. So your minimum is 250,000 for the sum of cubes. --- End quote --- Aha, here's where I messed up --- Quote from: Ipquarx on February 13, 2017, 05:05:45 PM ---d/dx (a^3 + (100-a)^3) = 3a^2 - 3*(100-a)^2 --- End quote --- I wrote --- Quote ---d/dx (a^3 + (100-a)^3) = 3a - 3*(100-a) --- End quote --- |
| Punished Toxicology:
--- Quote from: Insert Name Here² on February 13, 2017, 10:52:17 AM ---I'll be sure to ban Calculus Help (Optimization Integration) from my server, thanks for warning us! --- End quote --- you are a comedy revolutionary |
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