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March 29, 2015

March 29, 2015

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

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 12:01amThere'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 12:20amI 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 12:55amThe cross-sectional area is jus the rectangle...

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

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