Answer
See answers.
Work Step by Step
Assume that the recoil energy of the boron nucleus is negligible.
$$ ^{11}_{6}C \rightarrow \left( ^{11}_{5}B \right)+ ^{0}_{1}\beta^+ +\nu $$
The kinetic energy of the $\beta^+$ is maximum if the neutrino $\nu$ has its minimum kinetic energy, zero. See Problem 36 for the energy released.
$$KE=\left( m(^{11}_{6}C) -m(^{11}_{5}B) -2m(^{0}_{-1}e) \right)c^2$$
$$= (11.011434u-11.009305u-2(0.00054858u)) \left( \frac{931.49MeV/c^2}{u}\right)c^2$$
$$=0.9611 MeV$$
The kinetic energy of the $\beta^+$ is minimum when it is zero. In that case the neutrino has energy of 0.9611 MeV.
In summary, the minimum energy of each is zero.
