so for each unit they get $100, so if they sell "x" units, they intake 100x, that's just the gross income, and therefore the Revenue.
Now, they have a cost of 20x + 0.25x² if they "x" units, now, bearing in mind that if you pluck out the costs out of the incoming revenue, what's leftover, is a surplus amount, namely the Profit. Therefore, profit p(x) = revenue - costs, or 100x - (20x + 0.25x²), which is p(x) = -0.25x² +80x.
a)
now, what's the highest profit they can make? check the picture below.
[tex]\bf p(x)=-0.25x^2+80x\\\\
-------------------------------\\\\
\textit{ vertex of a vertical parabola, using coefficients}\\\\
\begin{array}{lcclll}
p(x) = &{{ -0.25}}x^2&{{ +80}}x&{{ +0}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)
\\\\\\
\left( -\cfrac{80}{-0.50}~,~0-\cfrac{80^2}{-1} \right)\implies \left( \stackrel{\textit{quantity}}{160}~,~\stackrel{\textit{profit value}}{6400} \right)[/tex]
b)
well, if a tax is imposed on them, and say, they take it out of the revenue, then their new revenue is not 100x, is 90x, thus [tex]\bf p(x)=-0.25x^2+70x[/tex]
and you can use the same formula as above to get the x-coordinate for that vertex.
[tex]\bf p(x)=-0.25x^2+70x\\\\
-------------------------------\\\\
\textit{ vertex of a vertical parabola, using coefficients}\\\\
\begin{array}{lcclll}
p(x) = &{{ -0.25}}x^2&{{ +70}}x&{{ +0}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)
\\\\\\
\left( -\cfrac{70}{-0.50}~,~0-\cfrac{70^2}{-1} \right)[/tex]
