SATHEE: Application of Derivatives 4 Question 50

Application of Derivatives 4 Question 50

####53. A window of perimeter (including the base of the arch) is in the form of a rectangle surmounted by a semi-circle. The semi-circular portion is fitted with coloured glass while the rectangular part is fitted with clear glass. The clear glass transmits three times as much light per square meter as the coloured glass does. What is the ratio for the sides of the rectangle so that the window transmits the maximum light? (1991, 4M)

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Answer:

Correct Answer: 53. (2)

Solution:

Let $2 b$ be the diameter of the circular portion and $a$ be the lengths of the other sides of the rectangle.

Total perimeter $= 2 a + 4 b + π b = K$

Now, let the light transmission rate (per square metre) of the coloured glass be $L$ and $Q$ be the total amount of transmitted light.

Then, $Q = 2 a b (3 L) + \frac{1}{2} π b^{2} (L)$

$\Rightarrow Q = \frac{L}{2} (π b^{2} + 12 a b)$

$\Rightarrow Q = \frac{L}{2} [π b^{2} + 6 b (K - 4 b - π b)]$ [from Eq. (i)]

$\Rightarrow Q = \frac{L}{2} (6 K b - 24 b^{2} - 5 π b^{2})$

On differentiating w.r.t. $b$ , we get

$\frac{d Q}{d b} = \frac{L}{2} (6 K - 48 b - 10 π b)$

For maximum, put $\frac{d Q}{d b} = 0$

$\Rightarrow b = \frac{6 K}{48 + 10 π}$

Now, $\frac{d^{2} Q}{d b^{2}} = \frac{L}{2} (- 48 - 10 π) < 0$

Thus, $Q$ is maximum and from Eqs. (i) and (ii), we get $(48 + 10 π) b = 6 2 a + 4 b + π b$

$∴ Ratio = \frac{2 b}{a} = \frac{6}{6 + π} = 6 : 6 + π$

Application of Derivatives 4 Question 50

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Application of Derivatives 4 Question 50

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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