a.The sales level needed to maximize the annual profit is the x-value of the vertex of the quadratic function. The highest point is the vertex of the parabola represented by the equation in the form y=ax2+bx+c with a<0 where the x-coordinate is given by x=−b2a.

Substitute a=−0.0001 and b=70:

\(\displaystyle{x}=-{\left(\frac{{70}}{{{2}{\left(-{0.0001}\right)}}}\right)}={350000}\)

So, 350,000 units will maximize the profit.

b.Using the x-value from part (a) and the profit function, the maximum annual profit is:

\(\displaystyle{P}{\left({350000}\right)}=-{0.0001}{\left({350000}\right)}^{{2}}+{70}{\left({350000}\right)}+{125000}\)

\(P(350000)=12262500\)

So, the maximum profit is $12,262,500.