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 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?

To minimize the perimeter of the window, we need to find the values of h and w that minimize the function p(w) = 2h + 3.78w, while keeping the cross-sectional area A = hw constant.

Let's go through the steps again:

1. Start with the expression for the cross-sectional area:
A = hw + 0.65w^2

2. Rearrange the equation to solve for h in terms of w:
h = (A - 0.65w^2) / w

3. Substitute this expression for h into the expression for the perimeter:
p(w) = 2h + 3.78w
= 2[(A - 0.65w^2) / w] + 3.78w

4. Simplify the expression for p(w):
p(w) = 2A/w + 3.78w - 1.3w

5. To find the minimum of p(w), we need to find where its derivative equals zero:
p'(w) = (-2A/w^2) + 3.78 - 1.3

6. Set p'(w) = 0 and solve for w:
(-2A/w^2) + 3.78 - 1.3 = 0
(-2A/w^2) + 2.48 = 0
(-2A) + 2.48w^2 = 0
w^2 = 2A/2.48
w^2 = A/1.24
w = sqrt(A/1.24)

7. We now have the value of w that minimizes the perimeter. To find h, substitute w into the expression we derived earlier for h in terms of w:
h = (A - 0.65w^2) / w
= (A - 0.65(A/1.24)) / sqrt(A/1.24)
= (A - 0.65A/1.24) / sqrt(A/1.24)
= (A(1 - 0.65/1.24)) / sqrt(A/1.24)
= (A(1.24 - 0.65)) / sqrt(A/1.24)
= A(0.59) / sqrt(A/1.24)
= 0.59 * sqrt(A * (1.24 / A))
= sqrt(1.24A) * 0.59
= 0.59 * sqrt(A) [sqrt(1.24)]

So the final dimensions of the window that minimize the perimeter are:
h = 0.59 * sqrt(A) [sqrt(1.24)]
w = sqrt(A/1.24)

Make sure to substitute the given value of the cross-sectional area A into these equations to get the specific values of h and w.

To find the dimensions of the window that minimize the perimeter, you need to find the critical points of the perimeter function p(w).

First, let's go through your calculation steps to see if there are any errors.

1. The area of the window can be expressed as:
A = hw + 0.65w^2 (corrected)

2. The perimeter of the window can be expressed as:
p = 2h + w + 2.78w
Simplifying, p = 2h + 3.78w

3. Taking the derivative of p with respect to w:
p'(w) = 2(0) + 3.78 = 3.78

4. Setting p'(w) = 0 to find the critical point:
3.78 = 0

Based on these steps, it seems like there was an error in your calculation for p'(w).

Here's the correct calculation:

1. The area of the window: A = hw + 0.65w^2

2. The perimeter of the window: p = 2h + 3.78w

3. Taking the derivative of p with respect to w:
p'(w) = 3.78

Since the derivative is a constant, there are no critical points for p(w).

Therefore, we cannot find the dimensions of the window that minimize the perimeter using the given information. It is possible that there may be additional constraints or information missing from the problem.