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].
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)].
I tried that... it's not it... apparently there must be something wrong with the arythmetic but I can't find what it is...
The cross-sectional area is jus the rectangle...
Does that mean that you've got the right answer?
To determine the dimensions of the window that minimize the perimeter, we need to find the values of h and w that satisfy the condition p'(w_min) = 0, where p'(w) is the derivative of the perimeter function with respect to w and w_min is the value of w that minimizes the perimeter.
Let's start by calculating the derivative of the perimeter function p'(w):
p'(w) = (-1) * (2A / w^2) + 2.48
Now set p'(w_min) equal to zero and solve for w_min:
p'(w_min) = (-1) * (2A / w_min^2) + 2.48 = 0
-2A + 2.48w_min^2 = 0
2.48w_min^2 = 2A
w_min^2 = A / 1.24
w_min = sqrt(A / 1.24)
So the value of w_min is equal to the square root of A divided by 1.24.
Next, we can substitute w_min into the expression for h to find the corresponding value:
h_min = A / w_min - 0.65w_min
h_min = A / (sqrt(A / 1.24)) - 0.65(sqrt(A / 1.24))
h_min = sqrt(1.24A) - 0.65sqrt(A)
Therefore, the dimensions of the window that minimize the perimeter are:
h = sqrt(1.24A) - 0.65sqrt(A)
w = sqrt(A / 1.24)