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 function p(w) = 2h + 3.78w, where h and w are the height and width of the rectangle respectively.

To find the minimum values, we can take the derivative of p(w) with respect to w and set it equal to zero:

p'(w) = 0

Differentiating p(w) = 2h + 3.78w with respect to w:

p'(w) = 0 + 3.78 = 3.78

Now, let's solve for w_min using the equation p'(w_min) = 0:

3.78 = 0

This equation has no solution, which means that w does not have a minimum value.

So, the dimensions you obtained using this method are incorrect.

To find the dimensions of the window that minimize the perimeter, we can use another approach.

From the given information, the area of the window is A = hw + 0.65w^2.

Since we want to minimize the perimeter, we can express the perimeter p in terms of the width w and eliminate h.

The perimeter is given by p = 2h + w + 2s, where s is the length of the sloped side of the triangle.

Using the Pythagorean theorem, we can find s in terms of w:

s^2 = (w/2)^2 + (1.3w)^2
s^2 = 0.25w^2 + 1.69w^2
s^2 = 1.94w^2
s = sqrt(1.94)w

Now, substitute the value of s into the perimeter equation:

p = 2h + w + 2sqrt(1.94)w
p = 2h + (1 + 2sqrt(1.94))w

To eliminate h, we can use the equation for the area of the window:

A = hw + 0.65w^2

Solve this equation for h:

h = (A - 0.65w^2)/w

Substitute this value of h into the perimeter equation:

p = 2[(A - 0.65w^2)/w] + (1 + 2sqrt(1.94))w
p = (2A - 1.3w^2)/w + (1 + 2sqrt(1.94))w
p = 2A/w - 1.3w + (1 + 2sqrt(1.94))w
p = 2A/w + (1 + 2sqrt(1.94) - 1.3)w
p = 2A/w + (1 + 2sqrt(1.94) - 1.3)w
p = 2A/w + (1 + 2sqrt(1.94) - 1.3)w

Now, let's simplify the equation:

p = 2A/w + 2.48w

To find the value of w_min that minimizes the perimeter, we need to take the derivative of p with respect to w and set it equal to zero:

p'(w) = -2A/w^2 + 2.48 = 0

Solving for w_min:

2A/w^2 = 2.48
w^2 = A/1.24
w = sqrt(A/1.24)

Now, to find the value of h, substitute the obtained value of w_min into the equation for h:

h = (A - 0.65w^2)/w
h = (A - 0.65(A/1.24))/sqrt(A/1.24)
h = (A - 0.65A/1.24)/(sqrt(A/1.24))
h = (0.39A)/(sqrt(A/1.24))
h = 0.39(sqrt(A/1.24))

Therefore, the correct dimensions of the window that minimize the perimeter are:
h = 0.39(sqrt(A/1.24))
w = sqrt(A/1.24)

To find the dimensions of the window that minimize the perimeter, we need to find the values of h and w that minimize the function p(w) = 2h + 3.78w.

1. First, let's express h in terms of w using the equation A - 0.65w^2 = hw:

A/w - 0.65w = h

2. Now substitute this expression for h into the perimeter function, p(w):

p(w) = 2(A/w - 0.65w) + 3.78w
= 2A/w + 2.48w

3. To minimize p(w), take the derivative with respect to w and set it equal to zero:

p'(w) = (-2)A/w^2 + 2.48 = 0

4. Solve for w to find the value that minimizes p(w):

(-2)A/w^2 + 2.48 = 0
2A/w^2 = 2.48
w^2 = A/1.24
w = sqrt(A/1.24)

Now, substitute this value of w back into the expression for h in terms of w:

h = A/w - 0.65w
= A/(sqrt(A/1.24)) - 0.65sqrt(A/1.24)
= sqrt(1.24A) - 0.65sqrt(A)

So, the dimensions of the window that minimize the perimeter are:

h = sqrt(1.24A) - 0.65sqrt(A)
w = sqrt(A/1.24)