Question 10.SP.7: Find the maximum and minimum values of the inductor current ......

Since D = 0.6 \gt D_{\mathrm{min}} = 0.4286, continuous inductor current is assured. Also, by (10.5),
G_{V} = {\frac{V_{2}}{V_{1}}} = D                (10.5)
V_{2} = D V_{1} = 0.6(24)  =14.4\,\mathrm{V}

By (5) and (6) of Problem 10.6,

I_{\mathrm{max}} = {\frac{V_{2}}{R_{L}}} + {\frac{(V_{1}  –  V_{2})D}{2f_{s}L}} = {\frac{14.4}{7}} + {\frac{(24  –  14.4)(0.6)}{2(30  \times  10^{3})(50  \times  10^{-6})}} = 3.977\,\mathrm{A}
I_{\mathrm{min}} = {\frac{V_{2}}{R_{L}}}  –  {\frac{(V_{1}  –  V_{2})D}{2f_{s}L}} = {\frac{14.4}{7}}  –  {\frac{(24  –  14.4)(0.6)}{2(30  \times  10^{3})(50  \times  10^{-6})}} = 0.137\,\mathrm{A}