GPT答：

7i24.cn

To let in the most light, we want to maximize the area of the window. Let's call the radius of the semicircle "r" and the height of the rectangle "h". 

The perimeter of this window is made up of three parts: 
1. The straight sides of the rectangle (2h) 
2. The curved part of the semicircle (πr) 
3. The bottom straight edge connecting the two ends (2r) 

So, we know that: 

2h + πr + 2r = 14 

Simplifying this equation, we get: 

h = (7 - r(π+2))/2

Now, let's find an expression for the area A in terms of r and h:

A = π(r^2)/2 + hr

Substituting h from above into this equation gives us:

A = π(r^2)/ 4 +(7/4)r - (1/4)r(π)(r+7)

We can simplify this expression by expanding it out and collecting like terms:

A= -(1/4)(π)r^3 + (7/4-π/8)r^ 2 + π / 8 * r

To find where A is maximized, we need to take its derivative with respect to r and set it equal to zero:

dA/dr= -(3/4)(π)r^ 2 +(7/4-π /8)* r+ π /8=0

Solving for r using quadratic formula gives us two values:
    
    r ≈ 1.86 m or ≈0.62 m
    
Since a radius cannot be negative or less than half a height which would make no sense for our problem here so only valid solution is:
    
    Therefore, our optimal value for "r" is approximately equal to 1.86 meters.
    
Now that we have found "r", we can use our original equation above to solve for "h":

    h=(7-r(π+22))/22
        =(7-(1.86*11))/(22)
        ≈0 .96m
    
Therefore, our optimal dimensions are approximately:
Height(h)=0 .96m
Radius(r)=1 .86m