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- - **Fastest structured way to get max(abc) if a+b+c=30**
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Fastest structured way to get max(abc) if a+b+c=30What is the fastest and structured way to get maximum of abc if a+b+c=n, say n=30? a,b,c are non-negative and can be non-integer. |

Choose a = b = c. |

What about for a,b,c natural numbers ? And how to find the minimal value? I searched a bit for Lagrange-Multipliers but how to use it for natural numbers? For this type of problem AM-GM works . $\displaystyle (a+b+c)^3 \geq 3^3 abc \;\;$ , $\displaystyle abc\leq 10^3 =max(abc)$ |

Find the minimum in the reals and test the nearest natural solutions. The minimum product should be fairly obvious. |

Other expressions are: $\displaystyle max(abc)=\lfloor (\frac{n}{3})^3 \rfloor $ . |

10 + 10 + 10 = 30 10 * 10 * 10 = 1000 9.99 + 10 + 10.01 = 30 9.99 * 10 * 10.01 = 999.99899.... Get my drift? |

Quote:
$\nabla \left(a b c - \lambda(a + b + c - 30)\right)$ $b c - \lambda = 0$ $a c - \lambda = 0$ $a b - \lambda = 0$ $a+b+c = 30$ $\dfrac b a = 1$ $\dfrac c a = 1$ $\dfrac c b = 1$ $a = b = c$ (As Skipjack mentioned) $3a = 30$ $a=b=c = 10$ |

If p and q are any two of the variables in the original problem, the solution must maximize pq = ((p + q)² - (p - q)²)/4, where p + q is a constant, so p = q. |

F(x,y,z)=xyz, G(x,y,z)=x+y+z=30 dF=Fxdx+Fydy+Fzdz=0, but dx, dy, dz not arbitrary. They are subject to dG=dx+dy+dz=0 -> (Fx-Fz)dx=0, (Fy-Fz)dy=0, dx and dy arbitrary.-> Fx=Fy -> yz=xz -> x=y But I could also have let dx and dz be arbitrary. Repeating the above I would get x=z. So x=y=z=10. |

How to find the minimal value (a fast way)? |

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