Let x, y, z be three non-negative integers su | vedclass

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(C) We want to maximize $f(x, y, z) = xyz + xy + yz + zx$ subject to $x+y+z=10$ where $x, y, z \in \mathbb{Z}_{\geq 0}$.Note that $(x+1)(y+1)(z+1) = x

Vedclass · Accepted Answer

$69$

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(C) We want to maximize $f(x, y, z) = xyz + xy + yz + zx$ subject to $x+y+z=10$ where $x, y, z \in \mathbb{Z}_{\geq 0}$.Note that $(x+1)(y+1)(z+1) = xyz + xy + yz + zx + x + y + z + 1$.Substituting $x+y+z=10$,we get $(x+1)(y+1)(z+1) = xyz + xy + yz + zx + 11$.Let $A = x+1, B = y+1, C = z+1$. Then $A+B+C = x+y+z+3 = 13$.By the $AM$-$GM$ inequality,$(ABC)^{1/3} \leq \frac{A+B+C}{3} = \frac{13}{3}$.Thus,$ABC \leq (\frac{13}{3})^3 = \frac{2197}{27} \approx 81.37$.Since $A, B, C$ are integers,the maximum product $ABC$ for a fixed sum $13$ occurs when the integers are as close as possible: $4, 4, 5$.$4 	imes 4 	imes 5 = 80$.Therefore,$xyz + xy + yz + zx + 11 \leq 80$.$xyz + xy + yz + zx \leq 80 - 11 = 69$.The maximum value is $69$.

Let $x, y, z$ be three non-negative integers such that $x+y+z=10$. The maximum possible value of $xyz+xy+yz+zx$ is

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