This problem requires finding the dimensions $h$ (height of the rectangle) and $w$ (width of the rectangle) that minimize the perimeter of the window, given that:

1. The cross-sectional area is fixed at $A$.
2. The window consists of a rectangle with height $h$ and width $w$, surmounted by a triangle with height $T = 0.4w$ (0.4 times the width of the rectangle).

Step-by-Step Solution:

Step 1: Cross-sectional Area

The total area $A$ consists of the rectangle's area and the triangular area: $A = \text{Area of rectangle} + \text{Area of triangle}$ $A = h \cdot w + \frac{1}{2} \cdot w \cdot T$ Substitute $T = 0.4w$: $A = h \cdot w + \frac{1}{2} \cdot w \cdot (0.4w)$ $A = h \cdot w + 0.2w^2$

Step 2: Perimeter of the Window

The perimeter consists of:

1. Two vertical sides of length $h$,
2. The base of the rectangle $w$,
3. The two slanted sides of the triangle.

The slanted sides of the triangle are calculated using the Pythagorean theorem: $\text{Length of one slanted side} = \sqrt{\left(\frac{w}{2}\right)^2 + T^2}$ Substitute $T = 0.4w$: $\text{Length of one slanted side} = \sqrt{\left(\frac{w}{2}\right)^2 + (0.4w)^2} = \sqrt{\frac{w^2}{4} + 0.16w^2} = \sqrt{0.41w^2} = 0.64w$

Thus, the total perimeter $P$ is: $P = 2h + w + 2(0.64w)$ $P = 2h + 2.28w$

Step 3: Minimize Perimeter

To minimize the perimeter $P$, we substitute $h$ in terms of $w$ and $A$ using the area equation: $A = h \cdot w + 0.2w^2$ Solve for $h$: $h = \frac{A - 0.2w^2}{w}$

Substitute $h$ into $P$: $P = 2\left(\frac{A - 0.2w^2}{w}\right) + 2.28w$ $P = \frac{2(A - 0.2w^2)}{w} + 2.28w$ $P = \frac{2A}{w} - 0.4w + 2.28w$ $P = \frac{2A}{w} + 1.88w$

Step 4: Differentiate and Solve

To find the $w$ that minimizes $P$, take the derivative of $P$ with respect to $w$: $\frac{dP}{dw} = -\frac{2A}{w^2} + 1.88$ Set $\frac{dP}{dw} = 0$ to find critical points: $-\frac{2A}{w^2} + 1.88 = 0$ $\frac{2A}{w^2} = 1.88$ $w^2 = \frac{2A}{1.88}$ $w = \sqrt{\frac{2A}{1.88}}$

Step 5: Calculate $h$

Substitute $w$ into the equation for $h$: $h = \frac{A - 0.2w^2}{w}$

Final Dimensions

1. Compute $w = \sqrt{\frac{2A}{1.88}}$.
2. Compute $h = \frac{A - 0.2w^2}{w}$.

Would you like me to proceed with a specific value for $A$, or should I summarize the process further?