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Q.

A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.

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a

length = \frac{20}{2 + 3 π}

,

breadth = \frac{10 + 5 π}{4 + 3 π}

b

length = \frac{30}{2 + 3 π}

,

breadth = \frac{20 + 5 π}{4 + 3 π}

c

length = \frac{40}{2 + 3 π}

,

breadth = \frac{30 + 5 π}{4 + 3 π}

d

length = \frac{50}{2 + 3 π}

,

breadth = \frac{40 + 5 π}{4 + 3 π}

answer is A.

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Detailed Solution

Given,
Total perimeter of window = 10 m
Suppose that,
A radius of semi-circle = r
The rectangle part of the window length = 2l
The rectangle part of the window breadth = 2b
Length of rectangle = 1
Breadths of rectangle = 2
Now, we have to find length of the semicircular arc,

1 + 2 b + π l = 10

(1 + π) 1 + 2 b = 10

b = \frac{1}{2} [10 - (1 + π) l]

Area of the window,

A = (21) (2 b) + \frac{π l^{2}}{2}

A = 4 lb + \frac{π l^{2}}{2}

A = 21 [10 - (1 + π) l] + \frac{1}{2} π l^{2}

A = 201 - 2 l^{2} - 2 π l^{2} + \frac{1}{2} π l^{2}

A = 201 - {2 l}^{2} - \frac{3}{2} π l^{2}

Differentiate it with respect to l,

\frac{dA}{dl} = 0 - 2 l^{2} - 3 πl

For area of the window to be maximum,

\frac{dA}{dl} = 0

\Rightarrow 20 - 41 - 3 πl = 0

\Rightarrow l = \frac{20}{4 + 3 π}

So, the breadth,

b = \frac{1}{2} [10 - (1 + π) 1]

b = \frac{1}{2} [10 - (1 + π) \frac{20}{4 + 3 π}]

b = \frac{1}{2} [\frac{40 + 30 π - 20 - 20 π}{4 + 3 π}]

b = \frac{10 + 5 π}{4 + 3 π}

Therefore, we calculate
Breadth of the rectangular window 2b,

2 b = \frac{20 + 10 π}{4 + 3 π}

b = \frac{10 + 5 π}{4 + 3 π}

Length of the window

2 l

2 l = \frac{40}{4 + 3 π}

l = \frac{20}{2 + 3 π}

Radius of the semicircular arc is l then,

l = \frac{20}{4 + 3 π}

Hence the required dimension of the window is

l = \frac{20}{2 + 3 π}

,

b = \frac{10 + 5 π}{4 + 3 π}

Correct option is 1.

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