# 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 x^{2}

= 5x - x^{2} (1 / 2 + π / 4) + π / 8 x^{2}

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,

d^{2}A / dx^{2} = - (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,

d^{2}A / dx^{2} < 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

visual curriculum