Consider a window the shape of which is a rectangle of height h surmounted by a triangle having a height T that is .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 =
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To determine the dimensions of the window that minimize the perimeter, we need to find the values of h and w that minimize the perimeter based on the given conditions.
Let's start by visualizing the window. We have a rectangle with a height h and a width w, surmounted by a triangle with a height T that is 0.3 times the width w of the rectangle.
To find the perimeter, we need to consider the sum of all the sides of the window. The rectangle contributes to the perimeter with its four sides, and the triangle adds three sides.
The perimeter P is given by:
P = 2h + 2w + T + w + w
Simplifying further:
P = 4w + 2h + T
Now, to minimize the perimeter P, we need to find the values of h and w that minimize the expression above. We can approach this by calculus, specifically, by finding the partial derivatives of P with respect to both h and w, and then setting them equal to zero.
Partial derivative with respect to h:
∂P / ∂h = 2
Partial derivative with respect to w:
∂P / ∂w = 4 + 0 + 0 = 4
Setting both derivatives equal to zero and solving for h and w:
2 = 0
4 = 0
We can see that the partial derivatives do not have any meaningful solutions since they contradict reality. Therefore, we need to reconsider our approach.
Let's focus on the expression for the perimeter:
P = 4w + 2h + T
Since the height of the triangle is specified as 0.3 times the width w, we know that:
T = 0.3w
Substituting this value back into the equation for the perimeter:
P = 4w + 2h + 0.3w
Simplifying further:
P = 4.3w + 2h
Now, we need to minimize P given the constraint provided, which is the cross-sectional area A.
The cross-sectional area A is given by:
A = h * w + (0.5w * T) = h * w + 0.5w * 0.3w = h * w + 0.15w^2
We can rearrange this equation to solve for h:
h = (A - 0.15w^2) / w
Substituting this value back into the expression for the perimeter P:
P = 4.3w + 2((A - 0.15w^2) / w)
Simplifying further:
P = 4.3w + 2A/w - 0.3w
To minimize the perimeter P, we can take the derivative of P with respect to w, set it equal to zero, and solve for w:
dP / dw = 4.3 - 2A/w^2 - 0.3 = 0
4.3 - 2A/w^2 - 0.3 = 0
4.3 - 0.3 = 2A/w^2
4 = 2A/w^2
2 = A/w^2
w^2 = A/2
w = sqrt(A/2)
Now that we have the value of w, we can substitute it back into the equation for h:
h = (A - 0.15w^2) / w
h = (A - 0.15(A/2)) / sqrt(A/2)
h = A(1 - 0.075) / sqrt(A/2)
h = 0.925A / sqrt(A/2)
Therefore, the dimensions of the window that minimize the perimeter are:
h = 0.925A / sqrt(A/2)
w = sqrt(A/2)
Please note that the above calculations assume that the given cross-sectional area A is fixed.