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
⇒ (π + 4)l – 112 = 0
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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