`L L '` is the latus rectum of the parabola `y^2`= 4ax and PP' is a double ordinate drawn between the vertex and the latus rectum. Show that the area of the trapezium `P P^(prime)L L '` is maximum when the distance `P P '` from the vertex is `a//9.`

Explanation

Let x and y be the length and breadth of the rectangular window .Therefore
Radius of the semicircular opening =`(x)/(2)`

It is given that the perimeter of the window is 10 m .Therefore
`x+2y+(pix)/(2)=10`
or `Ty=5-(x)/(2)(pi)/(2)+1`
Therefore area of the window is given by
`A=xy+(pio)/(2)(x)/(2)^(2)`
`=x[(5-(x))/(2)(pi)/((2)+1)]+(pi)/(8)x^(2)`
`=5x-x^(2)(pi)/(4)+(1)/(2)+(pi)/(8)x^(2)`
`(da)/(dx)=-5-2x((pi)/(4)+(1)/(2))+(pi)/(4)x`
Now `(dA)/(dx)=0`
or `5-x((1)+(pi)/(2))+(pi)/(4)x=0`
or `x=(20)/(pi+4)`
Also A is a quadratic function having coefficient of `x^(2)` negative Therefore `x=(20)/(pi+4)` is a point of maxima .Thus ,
`y=5-(20)/((pi)+4)((2+pi)/(4))=(10)/(pi+4)m`
Hence the required demensions of the window to admit maximum light is given by length `=(20)/(pi+4)m` and breadeth =`(10)/(pi+4)m`