SOLUTION
The perimeter of the shape becomes
[tex]\begin{gathered} P=2h+2r+\text{length of arc } \\ P=2h+2r+\frac{180}{360}\times2\times\pi r \\ P=2h+2r+\frac{1}{2}\times2\times\pi r \\ P=2h+2r+\pi r \\ 2h+2r+\pi r=920 \\ \text{making h the subject, we have } \\ 2h=920-2r-\pi r \\ \text{dividing both sides by 2} \\ \frac{2h}{2}=\frac{920}{2}-\frac{2r}{2}-\frac{\pi r}{2} \\ h=460-r-\frac{\pi r}{2} \end{gathered}[/tex]So from the perimeter, we now have an equation to find the height.
Now, let us find the area, A of the figure
[tex]\begin{gathered} A=(2r\times h)+\frac{1}{2}\pi r^2 \\ A=2rh+\frac{1}{2}\pi r^2 \\ \text{substituting the equation we got for h } \\ A=2r(460-r-\frac{\pi r}{2})+\frac{\pi r^2}{2} \\ A=920r-2r^2-\pi r^2+\frac{\pi r^2}{2} \\ A=920r-2r^2-\frac{\pi r^2}{2} \\ A=920r-(2+\frac{\pi}{2})r^2 \end{gathered}[/tex]Now, we have the equation for the area
To maximize the area, the derivative of the equation for the area must be equal to zero, so we have
[tex]\begin{gathered} A=920r-(2+\frac{\pi}{2})r^2 \\ \frac{dA}{dr}=920-2(2+\frac{\pi}{2})r=0 \\ 920=2(2+\frac{\pi}{2})r \\ 460=(2+\frac{\pi}{2})r \\ r=\frac{460}{(2+\frac{\pi}{2})} \end{gathered}[/tex]So, from here, we can get r, then substitute the value to get h
Hence from the equation, r becomes
[tex]\begin{gathered} r=\frac{460}{(2+\frac{\pi}{2})} \\ r=\frac{460}{3.5707963} \\ r=128.822805 \\ r=128.82\text{ to the nearest hundredth } \end{gathered}[/tex]Hence the radius is 128.82 to the nearest hundredth
For the height we have
[tex]\begin{gathered} h=460-r-\frac{\pi r}{2} \\ h=460-128.822805-\frac{\pi\times128.822805}{2} \\ h=460-128.822805-202.354389 \\ h=128.822806 \end{gathered}[/tex]Hence the height is 128.82 to the nearest hundredth
