#### Answer

12,000 is the maximum sustainable harvest
for this population

#### Work Step by Step

$H(S)=f(S)-S =12S^{0.25} -S$
$H'(S)=12(0.25S^{-0.75})-1=3S^{-0.75}-1$
Set this derivative equal to 0
$3S^{-0.75}-1=0$
$S^{-0.75}=\frac{1}{3}$
$S\approx 4.327$
H'(S)=0 when S=4.327 also is a maximum. Using the capability of the calculator, we find that the harvest is $H(4.327) \approx12.9802$
Thus 12,000 is the maximum sustainable harvest
for this population.