4) A window is in the form of a rectangle surmounted by a semicircle. The rectangle is of clear glass, whereas the semicircle is of tinted glass that transmits only one-fifth as much light per unit area as clear glass does. The total perimeter is fixed. Find the proportions of the window that will admit the most light. Neglect the thickness of the frame.

Answer:l=0.1401P\\ w =0.2801Pwhere P = perimeterStep-by-step explanation:Given that a window is in the form of a rectangle surmounted by a semicircle. Perimeter of window =2l+\pid/2+w[tex]P= 2l+3.14 (w/2)+w[/tex]Or [tex]P = 2l+2.57w\\l = \frac{P-2.57w}{2}[/tex]To allow maximum light we must have maximum areaArea = area of rectangle + area of semi circle where rectangle width = diameter of semi circle[tex]A=lw +\pi \frac{w^2}{8}[/tex][tex]A=lw +\pi \frac{w^2}{8}\\A=w*\frac{P-2.57w}{2}+0.3925w^2\\2A= Pw-2.57w^2+(0.785w^2)\\2A' = P-5.14w+1.57 w\\2A" =-5.14+1.57<0[/tex]Hence we get maximum area when i derivative is 0i.e. [tex]P-5.14w+1.57 w=0\\3.57w =P\\w = \frac{P}{3.57} =0.2801P[/tex][tex]l = \frac{P-2.57w}{2}\\l = 0.1401P[/tex]Dimensions can be [tex]l=0.1401P\\w =0.2801P[/tex]