Minimum and Maximum #3

Let $f(t)=(t-x)(t-y)(t-z)$ .

Then $f(t)=t^3-11t^2+40t-xyz$ .

$f'(t)=3t^2-22t+40=(t-4)(3t-10)$ .

$f'(t)=0$ when $t=\dfrac{10}{3},~4$ .

For the roots of $f(t)=0$ to be all real, $f\left(\dfrac{10}{3}\right)\ge0$ and $f(4)\le0$ .

$f\left(\dfrac{10}{3}\right)=\dfrac{1300}{27}-xyz\ge0$ , therefore

$xyz\le\dfrac{1300}{27}=48.148148\cdots$ .

$f(4)=48-xyz\le0$ , therefore

$xyz\ge48$ .

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From above, $4800\le 100xyz\le 4814$ .

Therefore, $48.00,~48.01,~48.02,~\cdots,~48.14$ are possible for $xyz$ .

$48.00+48.01+~\cdots~+48.14=48\times15+1.05=\boxed{721.05}$ .