Consider a window the shape of which is a rectangle of height h surmounted by a triangle having a height T that is 0.5 times the width w of the rectangle (as shown in the figure below).
If the cross-sectional area is A, determine the dimensions of the window which minimize the perimeter.
h =
w =
To minimize the perimeter of the window, we need to find the dimensions that minimize the sum of the lengths of all sides.
Let's break down the window into its individual components:
1. Rectangle:
- Height = h
- Width = w
2. Triangle:
- Height = T (0.5 times the width of the rectangle, so T = 0.5w)
- Base = w
Since the perimeter is the sum of the lengths of all sides, we can denote the perimeter as P.
P = 2h + 2w + w
Now, let's find an expression for the cross-sectional area A of the window:
A = Area of Rectangle + Area of Triangle
= h * w + (0.5w * T)
= hw + 0.5w * 0.5w
Simplifying the expression for A:
A = hw + 0.25w^2
To minimize the perimeter P, we need to find the dimensions (h and w) that minimize the area A. To do that, we can take the derivative of A with respect to w, set it to zero, and solve for w:
dA/dw = h + 0.5w = 0
0.5w = -h
w = -2h
Since we can't have a negative width for the window, we can discard this value. Thus, we are left with the condition h = 0.
Now, since h represents the height of the rectangle, it cannot be zero. Therefore, there is no minimum perimeter for this window configuration.
To determine the dimensions of the window that minimize the perimeter, we need to find the dimensions that minimize the perimeter function by finding the values of h and w that minimize the perimeter.
Let's break down the problem into smaller steps:
Step 1: Express the perimeter function in terms of h and w.
The perimeter of the window consists of the sum of the rectangle's perimeter and the triangle's perimeter.
Perimeter = 2*Width + 2*Height + Base + Hypotenuse
The width of the rectangle is w, the height of the rectangle is h, the base of the triangle is w, and the hypotenuse of the triangle can be determined using the Pythagorean theorem.
Base = w
Hypotenuse = √(w^2 + (0.5w)^2) = √(w^2 + 0.25w^2) = √(1.25w^2) = √(1.25)*w
Perimeter = 2w + 2h + w + √(1.25)*w
Step 2: Simplify the perimeter function.
Perimeter = 3w + 2h + √(1.25)*w
Step 3: Determine the derivative of the perimeter function with respect to h and set it equal to zero to find the minimum.
To find the minimum, we need to find the critical point(s) where the derivative of the perimeter function with respect to h is zero.
dPerimeter/dh = 2 = 0
2 = 0 (There is no equation involving h, so we don't need to solve for h in this step)
Step 4: Determine the derivative of the perimeter function with respect to w and set it equal to zero to find the minimum.
To find the minimum, we need to find the critical point(s) where the derivative of the perimeter function with respect to w is zero.
dPerimeter/dw = 3 + √(1.25) = 0
Solving for w:
3 + √(1.25) = 0
√(1.25) = -3
Since we cannot have a negative width, this means the derivative of the perimeter function does not equal zero at any values of w. Hence, there is no critical point for w.
Step 5: Analyze the results.
Since there are no critical points for w, we need to consider the domain of the problem. Realistically, the dimensions h and w should be positive, so we can exclude any negative solutions.
Therefore, we can conclude that there is no minimum for the perimeter. The dimensions of the window can be determined based on other constraints or considerations, such as aesthetic preferences or material availability.
a little check will show that
h = A/w - w/4
So, the perimeter is (assuming the triangle is isosceles)
p = w+2h+w√2 = w(1+√2) + 2(A/w - w/4)
= (1/2 + √2)w + 2A/w
dp/dw = (1/2 + √2) - 2A/w^2
dp/dw = 0 when
w = 2√(A/(2+√2))
h = (1+√2)/2 √(A/(2+√2))