Answer:
the dimensions of the rectangle are:
l = 50 m
b = [tex]\frac{100}{\pi}\ m[/tex]
Solution:
As per the question:
Perimeter of the room, 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 rectangle, A = lb
Perimeter of the room, P = [tex]\pi r + l + \pi r + l = 2\pi r + 2l = \pi b + 2l[/tex]
Therefore, we can write:
[tex]\pi b + 2l = 200[/tex]
[tex]b = \frac{200 - 2l}{\pi}[/tex]
Now,
Area, A = [tex]l(\frac{200 - 2l}{\pi}) = \frac{200l - 2l^{2}}{\pi}[/tex]
Now, differentiate A w.r.t l:
[tex]\frac{dA}{dl} = \frac{d}{dl}( \frac{200l - 2l^{2}}{\pi} = \frac{200 - 4l}{\pi}) [/tex]
Again differentiating w.r.t 'l', we get:
[tex]\frac{d^{2}A}{dl^{2}} = (\frac{- 4l}{\pi})[/tex] < 0
Thus we get maximum are when [tex]\frac{dA}{dl} = 0[/tex]
Therefore,
[tex]\frac{200 - 4l}{\pi}) = 0[/tex]
l = 50 m
Now, from
[tex]\pi b + 2l = 200[/tex]
[tex]\pi b = 200 - 2\times 50[/tex]
b = [tex]\frac{100}{\pi}[/tex]
r = [tex]\frac{b}{2} = \frac{50}{\pi}[/tex]
