Part C The yield of component B $($derived in problem $5)$ is given by $($C$\_($B$))/($C$\_($A$0))=($k$\_(1))/($k$\_(2)-$k$\_(1))($e$^(-$k$\_(1)$t$)-$e$^(-$k$\_(2)$t$))($a$)$ Calculate the optimum time $($t$\_($opt $))$ that maximizes the yield of B$.$ t$\_($opt$)=(1)/($k$\_(2)-$k$\_(1))$ln$($k$\_(2))/($k$\_(1))($b$)$ Find the maximum yield of B and optimal time when $($i$)$ k$\_(1)=2$k$\_(2)($ii$)$ k$\_(1)=$k$\_(2).($Hint: When taking the limit k$\_(1)->$k$\_(2)$ use Hospital's rule$)$Part C The yield of component B $($derived in problem $5)$ is given by

$\frac{C_B}{C_{A0}}=\frac{k_1}{k_2-k_1}(e^{-k_{1}t}-e^{-k_{2}t})$

$($a$)$ Calculate the optimum time $(t_{\text{o}\text{p}\text{t}})$ that maximizes the yield of B$.$

$t_{\text{o}\text{p}\text{t}}=\frac{1}{k_2-k_1}ln\frac{k_2}{k_1}$

$($b$)$ Find the maximum yield of B and optimal time when $($i$)k_1=2k_2($ii$)k_1=k_2.$

$($Hint: When taking the limit $k_1{k}_2$ use Hospital's rule$)$