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].
Both answers are wrong... please help...
To determine the dimensions of the window that minimize the perimeter, we need to find the values of h and w that minimize the given expression for the perimeter p.
First, let's revisit the cross-sectional area A that you correctly derived:
A = hw + 0.65w^2
Now, let's differentiate the expression for the perimeter p with respect to w and set it equal to zero to find the critical point(s):
p'(w) = -2A/w^2 + 2.48 = 0
Multiplying through by w^2 gives us:
-2A + 2.48w^2 = 0
Solving for w^2:
w^2 = 2A/2.48
w^2 = 0.8065A
w = sqrt(0.8065A)
Therefore, the width of the window that minimizes the perimeter is given by w = sqrt(0.8065A).
To find the corresponding height h, we substitute this value of w back into the formula for the cross-sectional area A:
A = h * (sqrt(0.8065A)) + 0.65 * (sqrt(0.8065A))^2
Let's simplify this equation:
A = h * sqrt(0.8065A) + 0.528725A
Rearranging the equation:
A - 0.528725A = h * sqrt(0.8065A)
0.471275A = h * sqrt(0.8065A)
Squaring both sides of the equation:
(0.471275A)^2 = h^2 * 0.8065A
0.222063363A^2 = 0.8065A * h^2
Simplifying further:
0.27590551A = h^2
Taking the square root of both sides:
h = sqrt(0.27590551A)
Therefore, the height of the window that minimizes the perimeter is given by h = sqrt(0.27590551A).
To summarize, the dimensions of the window that minimize the perimeter are:
h = sqrt(0.27590551A)
w = sqrt(0.8065A)