Let the length of the rectangle be y and the width of the rectnagle be 2r. Then the radius of the semi circle becomes r.

The perimeter of the Norman window is,P,

`P = y+y+2r+pir`

`P = 2y+2r+pir`

The area of the Norman window is, A,

`A = y*(2r) + (1/2)*pi*r^2`

`A = 2ry + (1/2)*pi*r^2`

P = 29, therefore,

`29 = 2y+2r+pir`

`2y = 29-2r-pir`

`y = 29/2-r-pir/2`

Substitute this in A,

`A = 2r*(29/2-r-pir/2) + (1/2)*pi*r^2`

`A = 29r-2r^2-pir^2) + (1/2)*pi*r^2`

`A = 29r-2r^2- pi/2*r^2`

Differentiate this wrt to r,

`(dA)/(dr) = 29 - 4r - pir`

For maxima and minima, first derivative is zero.

`(dA)/(dr) = 0`

`29-4r-pir =0`

`r = 29/(pi+4)`

Calculate  the second derivative and check for its sign at this r value to check whether this is a maximum or minimum point.

`(d^2A)/(dr^2) = -4-pi`

This is negative at any value, therefore at r = 29/(`pi` +4), the area of the Norman window is maximum. Therefore the maximum area, A is,

`r = 29/(pi+4)` = 4.06 approximately

`A = 29r-2r^2- pi/2*r^2`

`A = 29*4.06-2*4.06^2- pi/2*4.06^2`

A = 58.88

The maximum area is 58.88.