Answer: x = 2
If x is the width of the window, the value for x that results in the greatest area-- presumably admitting the greatest amount of light.
Explanation:
Initial calculations to find a starting value for w.
Determine dimensions that will have a sum of 8 as the perimeter. The length of the arc plus three sides of the rectangle.
The length of the arc is half the circumference of the circle, the width is the diameter. So, modifying the equation: C= πd, a = πw/2
Three sides of a rectangle (presumably a square for the initial calculation) 3w
Initial equation: 3w + πw/2 = 8 Substitute 3.14 for π and solve for w
2(3w) + 2(3.14w/2) = 2(8) multiply all terms by 2 to eliminate denominator
6w + 3.14w = 16 combine like terms
9.14w = 16 divide both sides by 9.14
w ≈ 1.75
Calculate the area using this value:
Area of semicircle: πr²/2 r= 1.75/2
3.14(.875)²/2 ≈ 1.203
Area of rectangle 1.75 × 1.75 = 3.0625
Total Area with width 1.75 : 4.265 ft²
Increasing the width to 2
The length of the arc becomes 3.14 This is twice the area of the semicircle.
3.14/2 = 1.57
The height of the rectangle reduces to 1.43
1.43 × 2 = 2.86
Total area with width of 2 ft 2.86 + 1.57 = 4.43 ft²
So 2 ft seems to be the optimal width.