Let x be the width and y be the length of the rectangle. x/2 is the radius of the semicircle Perimeter of the Norman window is x+2y+(π x)/2 Let P be the perimeter --- 288 in this problem. P = x+2y+(π x)/2--------(1) Solving for y from equation (1) 2y = P-x-πx/2 y = P/2-x/2-πx/4--------(2) Area = xy + π x^2 / 8 A = x(P/2-x/2-π x/4) + π x^2/8 A= Px/2-x^2 /2 -πx^2/4 +πx^2/8 dA/dx = P/2 -2x/2-2πx /4 +2πx / 8 =0 (4p-8x-2πx)/8=0 4p-2x(π+4)=0 4p=2x(π+4) x= 2P / (4+π) The radius is x/2 = P/(4+PI) Substitute P with 288 radius = 288 / (4+PI) will maximize the area of the window.
d^2A/dx^2 =-1-π/2+π/4 < 0, indicates that the area is maximized. You'll have to simplify x and y if you want them in numeric format.
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