Asked by Anonymous
A rectangle is topped by an equilateral triangle, the sides of which are equal to the width of the rectangle. Find the maximum area of the window if the perimeter is 500 cm.
Answers
Answered by
oobleck
so let the rectangle have
width = x
height = y
The perimeter is 2y+3x = 500
The area of the figure is
a = xy + √3/4 x^2 = x(500-3x)/2 + √3/4 x^2 = (√3-6)/4 x^2 + 250x
find where da/dx=0, or just use your knowledge of parabolas, which says that the vertex is at x = -b/2a = -250/(2 * (√3-6)/4) = 500/(6-√3)
so y = (500-3(500/(6-√3)))/2 = 250/11 (5-√3)
That makes the maximum area
(500/(6-√3))*(250/11 (5-√3)) + √3/4 (500/(6-√3))^2 = 62500/(6-√3)
width = x
height = y
The perimeter is 2y+3x = 500
The area of the figure is
a = xy + √3/4 x^2 = x(500-3x)/2 + √3/4 x^2 = (√3-6)/4 x^2 + 250x
find where da/dx=0, or just use your knowledge of parabolas, which says that the vertex is at x = -b/2a = -250/(2 * (√3-6)/4) = 500/(6-√3)
so y = (500-3(500/(6-√3)))/2 = 250/11 (5-√3)
That makes the maximum area
(500/(6-√3))*(250/11 (5-√3)) + √3/4 (500/(6-√3))^2 = 62500/(6-√3)
Answered by
Anonymous
Just a question, how does (√3 x^2 - 3x^2)/2 become (√3-6)/4 x^2? I'm confused
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