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A wire of length 2 units is cut into two parts which are bent respectively to form a square of side = x units and a circle of radius = r units. If the sum of the areas of the square and the circle so formed is minimum, then: 
  • a)
    x = 2r
  • b)
    2x = r
  • c)
    2x = (π + 4)r
  • d)
    (4 – π) x = πr
Correct answer is option 'A'. Can you explain this answer?
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A wire of length 2 units is cut into two parts which are bent respecti...

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A wire of length 2 units is cut into two parts which are bent respecti...
Let the length of the wire used for the square be L1 and the length used for the circle be L2. We know that L1 + L2 = 2.

The perimeter of a square is given by 4x, so L1 = 4x.
The circumference of a circle is given by 2πr, so L2 = 2πr.

We can rewrite L2 in terms of r: L2 = 2πr = 2(πr).

Now we can substitute these values back into the equation L1 + L2 = 2:

4x + 2(πr) = 2.

Simplifying this equation, we get:

4x + 2πr = 2,
4x = 2 - 2πr,
2x = 1 - πr.

To find the minimum sum of the areas of the square and the circle, we need to minimize the sum of their areas. The area of a square is given by x^2, and the area of a circle is given by πr^2.

So, the sum of the areas is given by x^2 + πr^2.

We want to minimize this sum, so we can take the derivative with respect to x and r and set it equal to zero:

d/dx (x^2 + πr^2) = 2x,
d/dr (x^2 + πr^2) = 2πr.

Setting both derivatives equal to zero, we get:

2x = 0,
2πr = 0.

Solving for x and r, we find that x = 0 and r = 0. However, these values are not physically meaningful.

Therefore, there is no minimum sum of the areas of the square and the circle.
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A wire of length 2 units is cut into two parts which are bent respectively to form a square of side = x units and a circle of radius = r units. If the sum of the areas of the square and the circle so formed is minimum, then:a)x = 2rb)2x = rc)2x = (π + 4)rd)(4 – π) x = πrCorrect answer is option 'A'. Can you explain this answer?
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