A) For the first question, we will use the first and second derivative criteria. First, we will compute the first and second derivatives of the given function:
![\begin{gathered} (dP(q))/(dq)=2(-0.02)q+3 \\ (d^(2)P(q))/(dq^(2))=2(-0.02)=-0.04 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/2m6djw8wzgupzj9qpj5rjdp8eueh055mlm.png)
Now, we set the first derivative equals to zero and solve for q:
![\begin{gathered} -0.04q+3=0 \\ q=(-3)/(-0.04)=75 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/3058sp3wtl16vje6r6c8nc56ejlo9n7smh.png)
Evaluating q=75 in the second derivative, we get a negative value since it is a constant, therefore there is a maximum for q=75.
B) We know the maximum is reached for q=75 therefore to find the maximum profit we evaluate the function at q=75:
![P(75)=-0.02(75)^(2)+3(75)-44=68.5](https://img.qammunity.org/2023/formulas/mathematics/high-school/nnh6lszqxlb2htje050vjdoxppc5ivkn5i.png)