\(p_1 ^*=49.5\) since \(\) \(M':\) \(c_1=m c_0 +b\)             

 \(m=\frac{66-0}{0-W_0 ^*}=\frac{66-0}{0-60}=-\frac{11}{10}\)   

 \(c_1=-\frac{11}{10} c_0 +b\)          

Letting \(c_1=66\) and \(c_0 =0\), \((66)=-\frac{11}{10} (0) +b\)    

\(b=66\)         thus \(M'\) : \(c_1=-\frac{11}{10} c_0 +66\)

When \(p_0 ^* =15=c_0\), \(c_1=-\frac{11}{10} (15) +66=49.5=p_1 ^*\)

Similarly, letting \(c_0=c_0 ^*=24\),   \(c_1=\frac{11}{10} (24) +66=39.6=c_1 ^*\)

\(M\) and \(M'\) are parallel which means that their slopes are equivalent.

 This means that \(M\): \(c_1=-\frac{11}{10} W_0 ^Y +b\)           

Using the fact that \(W_0 ^Y =42\) and \(c_1 =0\),        \((0) =- \frac{11}{10} (42) +b\)

\(b=\frac{231}{5}\)                    \(M\) : \(c_1=-\frac{11}{10} c_0 +\frac{231}{5}\)

This means that the \(c_1\) intercept is  \(-\frac{11}{10} (0) +\frac{231}{5}=46.2\)

When \(y_0=c_0=30\) , \(c_1=-\frac{11}{10} (30) +\frac{231}{5}=13.2=y_1\)

We can conclude that from \(Y\) to \(P^*\), we invest \(15\) \(c_0\) and reap \(36.3\) \(c_1\) beceause

\(y_0 -p_0 ^*=30-15=15\) during time \(c_0\)

\(p_1 ^* -y_1=\)\(49.5-13.2=36.3\) during time \(c_1\)

Additionally, from \(P^*\) to \(C^*\), we borrow \(9 c_0\) and repay \(9.9 c_1\) since

\(c_0 ^* -p_0 ^*=\)\(24-15\)\(=9\) in time \(c_0\)

\(p_1 ^*-c_1 ^*=\)\(49.5-39.6=9.9\) in time \(c_1\)