Q. 22

# A wire of length

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, side of square =

Let r be the radius of the circle,

Then, 2πr = 28 – l

r =

Therefore, the required area (a) is given by

A = (side of the square)2 + πr2

=

Then,

Now, if

Then,

(π + 4)l – 112 = 0

So, when

Then,

Then, by second derivative test, the area (A) is the minimum when .

Therefore, the combined area is the minimum when the length of the wire in making the square is cm while the length of the wire in making the circle is

28 - = cm.

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