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.

Question

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].

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/�ã(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 cross-sectional 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?

I reposted this because it's so far away now... I didn't get the right answer, can someone please check what I have wrong?

Answer 1

Did you redo your calculations based on A=w*h?
If you did, you can post what you've got.
Also, how many "lives" do you have left?

Answer 2

A=wh
h=A/w

Perimeter,
P(w)=2h+w+2√((w/2)^2+(1.3w)^2)
=2A/w + (1+(√776)/10)w

P'(w)=2A/w^2 + 1+ sqrt(194)/5

Set P'(w)=0 and solve for w to get:
w=±(sqrt(10A))/sqrt(sqrt(194)+5)
=0.727√A (reject negative root)