the given relation is
[tex]\begin{gathered} R(p)=1600p-0.5p2 \\ R(p)=-0.5p^2+1600p \end{gathered}[/tex]compare the above expression with the standard parabola quadratic equation
[tex]ax^2+bx+c=0[/tex]so the value of a = -0.5
b = 1600
c = 0
the maximum value will be on
[tex]p=-\frac{b}{2a}[/tex][tex]\begin{gathered} p=\frac{-1600}{2\times-0.5}=\frac{16000}{10} \\ =1600 \\ \end{gathered}[/tex]a) so the unit price of the boat should be p = 1600.
and the maximum revenue will be
put p = 1600
[tex]\begin{gathered} R(p)=1600\times1600-0.5(1600)^2=1600^2(1-0.5) \\ R(p)=1600^2\times0.5=1280000 \\ \\ \end{gathered}[/tex]
