Letting $x$ represent the width of the rectangle, the height of the rectangle is $1.3x$, and the height of the triangle is $1.3^2x$. The area of the rectangle is $x\cdot1.3x=1.3x^2$, and the area of the triangle is $\frac{1}{2}(1.3^2x)(1.3x)=\frac{1}{2}(1.69x^2)$. The total area is $1.3x^2+\frac{1}{2}(1.69x^2)=1.995x^2=A$.
The perimeter of the rectangle is $2x+2\cdot1.3x=2.6x$, and the perimeter of the triangle is $1.3x+\sqrt{(1.3x)^2+(1.3^2x)^2}=1.3x+1.3^2x\sqrt{2}=2.938x$. The total perimeter is $2.6x+2.938x=5.538x$.
We can minimize the perimeter by minimizing $x$, so we set $\frac{d}{dx}(5.538x)=0$ and solve to find $x=\frac{A}{1.995}$. This means that the width of the rectangle is $\frac{A}{1.995}$, the height of the rectangle is $1.3\cdot\frac{A}{1.995}=\frac{1.3A}{1.995}$, and the height of the triangle is $1.3^2\cdot\frac{A}{1.995}=\frac{1.69A}{1.995}$.
Thus, the dimensions of the window which minimize the perimeter are $h=\boxed{\frac{1.3A}{1.995}}$ and $w=\boxed{\frac{A}{1.995}}$.
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...
2 answers
The cross-sectional area of the window is the sum of the areas of the rectangle and the triangle. Thus,
[A = hw + \frac{1}{2} (1.3w) T = \frac{1}{2}(1.3w) (h + T).]Since $T = 1.3w$, we can write this as
[A = \frac{1}{2} (1.3w) (h + 1.3w).]Since the perimeter of the window is $2h + 2w + (1.3w)$, we want to minimize
[2h + 2w + (1.3w) = 2(h + w) + (0.3w),]so we want to minimize $h + w$.
We have
[h + w = \frac{2A}{1.3w},]so we want to minimize $\frac{2A}{1.3w}$. This is equivalent to maximizing $1.3w$, which is equivalent to maximizing $w$. We can take $w$ as large as we want, so the perimeter of the window is minimized when $w$ is as large as possible.
Then the width of the window is $w$, the height of the rectangle is $h = \frac{2A}{1.3w} - w$, and the height of the triangle is $T = 1.3w$. The perimeter of the window is minimized when $w$ is as large as possible.
[A = hw + \frac{1}{2} (1.3w) T = \frac{1}{2}(1.3w) (h + T).]Since $T = 1.3w$, we can write this as
[A = \frac{1}{2} (1.3w) (h + 1.3w).]Since the perimeter of the window is $2h + 2w + (1.3w)$, we want to minimize
[2h + 2w + (1.3w) = 2(h + w) + (0.3w),]so we want to minimize $h + w$.
We have
[h + w = \frac{2A}{1.3w},]so we want to minimize $\frac{2A}{1.3w}$. This is equivalent to maximizing $1.3w$, which is equivalent to maximizing $w$. We can take $w$ as large as we want, so the perimeter of the window is minimized when $w$ is as large as possible.
Then the width of the window is $w$, the height of the rectangle is $h = \frac{2A}{1.3w} - w$, and the height of the triangle is $T = 1.3w$. The perimeter of the window is minimized when $w$ is as large as possible.
0 answers