Sunday

December 21, 2014

December 21, 2014

Posted by **Katrina** on Sunday, April 10, 2011 at 8:55pm.

If the cross-sectional 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/w-0.65w) + 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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