## Calculus with Applications (10th Edition)

$H(S)=f(S)-S =12S^{0.25} -S$ $H'(S)=12(0.25S^{-0.75})-1=3S^{-0.75}-1$ 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.