37)
keep in mind that the perimeter of a rectangle is length + length + width + width or P = 2l + 2w, or P = 2(l+w).
we know the perimeter of the box's width and length is 36, therefore then
[tex]\bf \stackrel{P}{36}=2(\stackrel{length}{l}+\stackrel{width}{w})\implies 18=l+w\implies \boxed{18-w=\stackrel{length}{l}}
\\\\\\
V(w)=4(w)(18-w)\implies V(w)=-4w^2+72w[/tex]
check the first picture below.
now, that parabolic graph, goes up up up reaches a U-turn and the back down, so it has a "maximum" point, and that is when the volume is the highest, namely V(w).
[tex]\bf \textit{ vertex of a vertical parabola, using coefficients}\\\\
\begin{array}{lccclll}
V(w) = &{{ -4}}w^2&{{ +72}}w&{{ +0}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)
\\\\\\
{{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\implies \stackrel{maximum~volume}{0-\cfrac{72^2}{4(-4)}}[/tex]
43)
is pretty much the same thing, checking the vertex coordinates of the parabola, check the second picture below,
[tex]\bf h=64t-16t^2\implies h=-16t^2+64t+0\\\\\\
\textit{ vertex of a vertical parabola, using coefficients}\\\\
\begin{array}{lccclll}
h = &{{ -16}}t^2&{{ +64}}t&{{ +0}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)
\\\\\\
\stackrel{\textit{it takes this many seconds}}{-\cfrac{64}{2(-16)}}\qquad \qquad \stackrel{\textit{it went up this many feet}}{0-\cfrac{64^2}{4(-16)}}[/tex]
