Posted by Katrina on Sunday, April 10, 2011 at 11:34pm.
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].

Calculus (pleas help I really need help with this)  MathMate, Monday, April 11, 2011 at 12:01am
There's this line:
The complete area: hw + 0.3w^2.
which should read hw+0.65w^2.
But I think your subsequent calculations use the correct expression.
Your approach appears correct, although I did not check the arithmetic.
That is one problem with computerized exercises.
Were there instructions as to how you present the results, such as:
 in decimals to two decimal places, or
 in fractions
 exact expression in fractions, or
any other instructions.
If the question requires an accuracy to 2 decimal places, I would carry all calculations to 4 places until the last, when I enter only two places, probably rounded.
Another possible source of problem is the interpretation of the "cross section" area. I have not heard of a cross section area of a window. Does it refer to the whole window, as you did, or just the rectangular part?
Finally, the last expression should read:
sqrt(A) [sqrt(1.24)0.65/√(1.24)].

Calculus (pleas help I really need help with this)  Katrina, Monday, April 11, 2011 at 12:20am
I tried that... it's not it... apparently there must be something wrong with the arythmetic but I can't find what it is...

Calculus (pleas help I really need help with this)  Katrina, Monday, April 11, 2011 at 12:55am
The crosssectional area is jus the rectangle...

Calculus (pleas help I really need help with this)  MathMate, Monday, April 11, 2011 at 7:33am
Does that mean that you've got the right answer?
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