Question

A wire of length $20 m$ is to be cut into two pieces A piece of length $l_1$ is bent to make a square of area $A_1$ and the other piece of length $l_2$ is made into a circle of area $A_2$ If $2 A_1+3 A_2$ is minimum then $\left(\pi l_1\right): l_2$ is equal to:

• A. $6: 1$
• B. $3: 1$
• C. $4: 1$
• D. $1: 6$

Hint:

For minimizing the total area, always use the method of derivatives and substitution to find optimal values.

Answer 1

The Correct Option is A

Step 1:

Let the total length of the wire be distributed between a square and a circle, such that:

ℓ₁ + ℓ₂ = 20

We are required to minimize the expression:

2A₁ + 3A₂

Where:

• A₁ is the area of a square formed using wire length ℓ₁
• A₂ is the area of a circle formed using wire length ℓ₂

Since the perimeter of the square is ℓ₁, each side of the square is ℓ₁ / 4, so:

A₁ = (ℓ₁ / 4)² = ℓ₁² / 16

Similarly, the circumference of the circle is ℓ₂, and the radius is:

r = ℓ₂ / (2π)

Thus the area of the circle becomes:

A₂ = π (ℓ₂ / (2π))² = ℓ₂² / (4π)

Step 2:

Substitute the expressions for A₁ and A₂ into the objective function:

S = 2A₁ + 3A₂ = 2(ℓ₁² / 16) + 3(ℓ₂² / (4π)) = (ℓ₁² / 8) + (3ℓ₂² / 4π)

Now we express the function S in terms of a single variable using the constraint ℓ₁ + ℓ₂ = 20:

ℓ₂ = 20 - ℓ₁

Substitute into S:

S(ℓ₁) = (ℓ₁² / 8) + (3(20 - ℓ₁)² / 4π)

Step 3:

Differentiate S with respect to ℓ₁ and set the derivative to zero to minimize:

dS/dℓ₁ = (1/4)ℓ₁ - (3(20 - ℓ₁)) / (2π)

Set derivative to zero:

(1/4)ℓ₁ = (3(20 - ℓ₁)) / (2π)

Multiply both sides by 4π to eliminate denominators:

πℓ₁ = 6(20 - ℓ₁)

Expanding and simplifying:

πℓ₁ = 120 - 6ℓ₁
ℓ₁(π + 6) = 120
ℓ₁ = 120 / (π + 6)

Then:

ℓ₂ = 20 - ℓ₁ = 20 - (120 / (π + 6))

Final Ratio:

Now compute the ratio ℓ₁ : ℓ₂:

ℓ₁ / ℓ₂ = [120 / (π + 6)] / [20 - 120 / (π + 6)] = 6

Conclusion:

To minimize the combined weighted area of the square and the circle, the wire should be divided in the ratio:

ℓ₁ : ℓ₂ = 6 : 1