From part c),
A ( t ) = 4 − 1 2 t + 1 48 t 2 {\displaystyle A(t)=4-{\frac {1}{2}}t+{\frac {1}{48}}t^{2}}
To find the critical points of A ′ ( t ) {\displaystyle A'(t)} , calculate
A ′ ( t ) = − 1 2 + 2 48 t = − 1 2 + 1 24 t {\displaystyle A'(t)={\frac {-1}{2}}+{\frac {2}{48}}t={\frac {-1}{2}}+{\frac {1}{24}}t}
and solve A ′ ( t ) = 0 {\displaystyle A'(t)=0} to obtain
− 1 2 + 1 24 t = 0 {\displaystyle {\frac {-1}{2}}+{\frac {1}{24}}t=0}
t = 24 2 = 12 {\displaystyle t={\frac {24}{2}}=12}
Moreover,
A ″ ( t ) = 1 24 {\displaystyle A''(t)={\frac {1}{24}}}
So as A ″ ( t ) > 0 {\displaystyle A''(t)>0} for all t {\displaystyle t} , A ( t ) {\displaystyle A(t)} is concave upward. So the critical point t = 12 {\displaystyle t=12} of A ( t ) {\displaystyle A(t)} is where A ( t ) {\displaystyle A(t)} is minimized. Hence, the solution is t = 12 h o u r s {\displaystyle \color {blue}t=12hours}
, calculate
to obtain
for all 
, 
is concave upward. So the critical point 
of is where is minimized. Hence, the solution is 
