(a) \[\left ( \frac{C_{L}}{C_{D}} \right )_{max} = \frac{1}{2\sqrt{k\times C_{D,0}}} = \frac{1}{\sqrt{0.04\times 0.016}} = 19.7642\]

(b)\[\left ( \frac{C_{L}^{1/2}}{C_{D}} \right )_{max} = \frac{\frac{C_{D,0}}{3K}}{\frac{4C_{D,0}}{3}}=\frac{\sqrt{\frac{0.016}{3\left ( 0.04 \right )}}}{\frac{4\left ( 0.016 \right )}{3}} = \frac{0.36515}{0.02133}=17.12\]

(c)For \(\left ( \frac{C_{L}}{C_{D}} \right )_{max}\)\[C_{L} = \sqrt{\frac{C_{D,0}}{K}}=\sqrt{\frac{0.016}{0.04}} = 0.63246\]

\[S = 4.4\,m^{2};\rho = 1.225\,kg/m^{3},W = 3700\,N\]

\[V\left ( \frac{C_{L}}{C_{D}} \right )_{max} = \sqrt{\frac{2W}{SC_{L}\rho }} = \sqrt{\frac{2\left ( 3700 \right )}{4.4\left ( 1.225 \right )\left ( 0.63246 \right )}}= 46.59\,m/s\]

For \(\left ( \frac{C_{L}^{1/2}}{C_{D}} \right )_{max}\)\[C_{L} = 0.36515 V\left ( \frac{C_{L}^{1/2}}{C_{D}} \right )_{max}=V\left ( \frac{C_{L}}{C_{D}} \right )_{max}\sqrt{\frac{0.63246}{0.36515}}= 61.316\,m/s\]

(d) At 10,000 ft, \[\frac{{\rho}’}{\rho} = \sigma = 0.73833\]

\[{V}'{\left(\frac{C_{L}}{C_{D}} \right)}_{max}= V\left ( \frac{C_{L}}{C_{D}} \right )_{max}\times \frac{1}{\sigma} = 46.59\times \frac{1}{0.73833}\]

\[\Rightarrow {V}’\left ( \frac{C_{L}}{C_{D}} \right )_{max} = 54.22\,m/s\]

\[{V}'{\left(\frac{C_{L}^{1/2}}{C_{D}} \right)}_{max}= V\left ( \frac{C_{L}^{1/2}}{C_{D}} \right )_{max}\times \frac{1}{\sigma} = 61.316 \times \frac{1}{\sqrt{0.73833}} = 71.36\,m/s\]