Calculus (please help!!!!)
posted by Katrina .
Consider a window the shape of which is a rectangle of height h surmounted a triangle having a height T that is 1.3 times the width w of the rectangle.
If the crosssectional area is A, determine the dimensions of the window which minimize the perimeter.
h=______
w=______
What I did was:
Area:
Rectangle: h*w
Triangle: w * 1.3w / 2 = 0.65w^2
The complete area: hw + 0.3w^2.
Perimeter:
3 sides of the rectangle: 2h+w
Twice the sloped side of the triangle: s^2 = (w/2)^2 + (1.3w)^2 = w^2(0.25+1.69)=1.94w^2, so s = 1.39w.
The complete perimeter p = 2h+w+2.78w = 2h+3.78w
A = hw + 0.65w^2
A  0.65w^2 = hw
A/w  0.65w = h
p = p(w) = 2h + 3.78w
= 2(A/w0.65w) + 3.78w
= 2A/w  1.3w + 3.78w
= 2A/w + 2.48w
p'(w) = (1)2A/w^2 + 2.48
p'(w_min) = (1)2A/w_min^2 + 2.48 = 0
2A + 2.48w_min^2 = 0
w_min^2 = A/1.24
w_min = sqrt(A/1.24)
So the dimensions I got are:
w_min = sqrt(A/1.24)
h_min = A/w_min  0.65w_min = A/sqrt(A/1.24)  0.6sqrt(A/1.24) =
= sqrt(1.24A)0.65sqrt(A) = sqrt(A) [sqrt(1.24)0.65].
Both answers are wrong... please help...
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