jmc

We complete the square: $$\begin{array}{rl}\frac{1}{2}{b}^2+5b-3& =(\frac{1}{2}{b}^2+5b)-3\\ & =\frac{1}{2}(b^2+10b+25)-3-25\cdot \frac{1}{2}\\ & =\frac{1}{2}(b+5{)}^2-\frac{31}{2}.\end{array}$$ The minimum value of $\frac{1}{2}(b+5{)}^2$ is $0$, since the square of a real number is never negative. Thus, the minimum value of the expression occurs at $b=\boxed{-5}$.