Answer :

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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