The dimensions of the rectangle are:
l = 50 m
b =
m
<h3>What is a perimeter in math?</h3>
The perimeter is the length of the outline of a shape. To find the perimeter of a rectangle or square you have to add the lengths of all the four sides.
<h3>How do we find a perimeter of a rectangle?</h3>
The perimeter of a rectangle,denoted by P is given by the formula, P=2l+2b, where l is the length and b is the breadth of the rectangle.
<h3>
Given:</h3>
As per the question:
Perimeter of the room is given as P = 200 m
The region is rectangular having a semicircle at each end.
Now,
Let 'l' be the length of the rectangle, 'b' be its breadth and 'r' be the radius of the semi-circle at each end.
Then, Area of the given rectangle, A = lb
Perimeter of the room, P is =![\pi r+l+\pi r+l=2\pi r+2l=\pi b+2l](https://tex.z-dn.net/?f=%5Cpi%20r%2Bl%2B%5Cpi%20r%2Bl%3D2%5Cpi%20r%2B2l%3D%5Cpi%20b%2B2l)
Therefore, ![\pi b+2l=200](https://tex.z-dn.net/?f=%5Cpi%20b%2B2l%3D200)
![b=(200-2l)/\pi](https://tex.z-dn.net/?f=b%3D%28200-2l%29%2F%5Cpi)
Now,
Area, A = ![l(200-2l)/\pi=(200l-2l^{2} )/\pi](https://tex.z-dn.net/?f=l%28200-2l%29%2F%5Cpi%3D%28200l-2l%5E%7B2%7D%20%29%2F%5Cpi)
Now, differentiate A w.r.t l:
Again differentiating w.r.t 'l', we get:
< 0
Thus we get maximum are when ![dA/dl=0](https://tex.z-dn.net/?f=dA%2Fdl%3D0)
Therefore,
![(200-4l)/\pi=0](https://tex.z-dn.net/?f=%28200-4l%29%2F%5Cpi%3D0)
l = 50 m
Now, from
![\pi b+2l=200](https://tex.z-dn.net/?f=%5Cpi%20b%2B2l%3D200)
![\pi b=200-2*50](https://tex.z-dn.net/?f=%5Cpi%20b%3D200-2%2A50)
![b=100/\pi](https://tex.z-dn.net/?f=b%3D100%2F%5Cpi)
![r=b/2=50/\pi](https://tex.z-dn.net/?f=r%3Db%2F2%3D50%2F%5Cpi)
To know more about area of a recatangle, visit the link
brainly.com/question/20693059
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