Given the question, the function that represents the cross-sectional area is:
[tex]\begin{gathered} A(h)=h(18-2h) \\ \text{where h is the height and 18-2h is the base.} \end{gathered}[/tex]By expansion:
[tex]A(h)=-2h^2+18h[/tex]By differentiation and equating the result to zero, we have:
[tex]\begin{gathered} A(h)=-4h+18 \\ -4h+18=0 \\ 4h=18 \\ h=\frac{18}{4} \\ h=4.5 \end{gathered}[/tex]Hence, the height of the drain with the maximun cross-sectional area is 4.5 feet which is option B.