[填空题]
The diagram below shows theaoe n2igj8 rb+e; 6knye :kw:1 cross-sectional area of a mohjtduu3iv-+o) (5dx dund of beach sand created after3oh- du(xu i)5vd+djt a high tide.
The curve of the cross section can be modelled by the following equation
$y=\frac{x^{2}(90-x)}{1800}$
where y represents the vertical height of the mound in $\mathrm{cm}$ and x denotes the horizontal width in $\mathrm{cm}$ , from the start of the mound.
1. At a horizontal width of x=30 , determine
1. The vertical height of the mound at this point; y= cm.
2. The gradient of the mound curve at this point. y'=$\frac{a}{b}$;a= ,b= .
2. 1. Find the value of x which corresponds to the maximum the vertical height of the mound;it is .
2. Hence, find the maximum vertical height of the mound. y= cm
3. Calculate the cross-sectional area of the mound, rounding your answer to one decimal place.
A child uses a toy shovel to remove the top of sand mound, as illustrated by the line segment $\mathrm{MN}$ below. Point M has coordinates at (30,30) .
4. Determine the coordinates of point N .
The cross-sectional area removed by the child can be expressed by the following integral
