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# A wire of length 28m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

**Solution:**

Maxima and minima are known as the extrema of a function.

Maxima and minima are the maximum or the minimum value of a function within the given set of ranges.

Let a piece of length l be cut from the given wire to make a square.

Then, the other piece of wire to be made into a circle is of length (28 - l) m.

Now, the side of the square is l / 4

Let r be the radius of the circle.

Then,

2πr = 28 - l

⇒ r = l/2π (28 - l)

The combined areas of the square and the circle, (A) is given by,

A = (side of the square)^{2} + π r^{2}

= l^{2}/16 + π [l/2π (28 - l)]^{2}

= l^{2}/16 + 1/4π (28 - l)^{2}

Hence,

dA/dl = 2l/16 + 2/4π (28 - l)(- 1)

= 1/8 - 1/2π (28 - l)

d^{2}A/dl^{2} = 1/8 + 1/2π > 0

Now,

d^{2}A/dl^{2} = 0

⇒ 1/8 - 1/2π (28 - l) = 0

⇒ [πl - 4(28 - l )]/8π = 0

⇒ (π + 4)l - 112 = 0

⇒ l = 112/(π + 4)

When, l = 112/(π + 4)

Then,

d^{2}A/dl^{2} > 0

By the second derivative test, area (A) is the minimum when l = 112/(π + 4).

Hence, the combined area is the minimum when the length of the wire in making the square is 112/(π + 4) m while the length of the wire in making the circle is (28 - 112/(π + 4))

= 28π/(π + 4) m

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.5 Question 22

## A wire of length 28m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

**Summary:**

Given that wire of length 28m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. The length of the two pieces so that the combined area of the square and the circle is minimum is 28π/(π + 4) m and 112/(π + 4)

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