Step 1

Demand function \(\displaystyle{P}={126}-{0.5}{x}{\left({1}\right)}\)

Total cost function \(\displaystyle{C}={50}{x}+{39.75}\)

a). Revenue R(x) =price x number sold

\(\displaystyle{R}{\left({x}\right)}={P}{x}\)

\(\displaystyle{R}{\left({x}\right)}={\left({126}-{0.5}{x}\right)}{x}\)

\(\displaystyle{R}{\left({x}\right)}={126}{x}-{0.5}{x}^{{{2}}}\)

Step 2

Profit P(x)=revenue-cost

\(\displaystyle={\left({126}{x}-{0.5}{x}^{{{2}}}\right)}-{\left({50}{x}+{39.75}\right)}\)

\(\displaystyle={126}{x}-{0.5}{x}^{{{2}}}-{50}{x}-{39.75}\)

\(\displaystyle=-{0.5}{x}^{{{2}}}+{76}{x}-{39.75}\)

To maximize profit P'(x)=0

\(\displaystyle{P}'{\left({x}\right)}=-{0.5}{\left({2}{x}\right)}+{76}\)

\(\displaystyle{P}'{\left({x}\right)}=-{x}+{76}={0}\)

x=76

and \(\displaystyle{P}{''''}{\left({x}\right)}=-{1}{<}{0}\) (maximum)

first and second derivative shows that at x=76, P(x) is maximum.