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
Solution:
Let x and y be the length and breadth of the rectangular window.
The radius of the semicircular opening be x / 2
It is given that the perimeter of the window is 10m .
Therefore,
x + 2y + πx / 2 = 10
x (1 + π / 2) + 2y = 10
2y = 10 - x (1 + π / 2)
y = 5 - x (1/2 + π/4)
Area of the window (A) is given by,
A = xy + π / 2(x / 2)2
= x [5 - x (1 / 2 + π / 4)] + π / 8 x2
= 5x - x2 (1 / 2 + π / 4) + π / 8 x2
Therefore,
On differentiating wrt x, we get
dA/dx = 5 - 2x (1 / 2 + π / 4) + π / 4 x
= 5 - x (1 + π / 2) + π / 4 x
On further differentiating wrt x,
d2A / dx2 = - (1 + π / 2) + π / 4
= 1 - π / 4
Now,
dA/dx = 0
⇒ 5 - x(1 + π / 2) + π / 4 x = 0
⇒ 5 - x - π / 4 x = 0
⇒ x(1 + π / 4) = 5
⇒ x = 5 / (1 + π / 4)
⇒ x = 20 / (π + 4)
When, x = 20 / (π + 4)
The,
d2A / dx2 < 0
Therefore, by second derivative test, the area is the maximum when length is 20/(π + 4) m.
Now,
y = 5 - 20 / (π + 4) ((2 + π) / 4)
= 5 - 5(2 + π) / (π + 4)
= 10 / (π + 4)
Hence, the required dimensions of the window to admit maximum light is given by length 20 / (π + 4) m and breadth 10 / (π + 4) m
NCERT Solutions Class 12 Maths - Chapter 6 Exercise ME Question 11
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
Summary:
Given that a window is in the form of a rectangle surmounted by a semicircular opening. Hence, the required dimensions of the window to admit maximum light is given by length 20 / (π + 4) m and breadth 10 / (π + 4) m
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