 Hi and welcome to the session. I am Shashin and I am going to help you with the following question. Question says, a wire of length 28 meter is to be cut into two pieces. One of the pieces is to be made into a square and other into a circle. What should be the length of the two pieces so that the combined area of the square and circle is minimum? First of all let us understand that if we are given a function f defined on interval i and c belongs to interval i such that f double dash c exists then x is equal to c is a point of local minima if f dash c is equal to 0 and f double dash c is greater than 0. This is the key idea to solve the given question. Now let us start with the solution. First of all let us assume that length of the two pieces is x meter and 28 minus x meter. We can see total length of the wire is 28 meter and it is cut into two pieces such that one of the piece is made into a square and other into a circle. Now let us assume that the wire of length x meter is made into a square and the wire of length 28 minus x meter is made into a circle. Now wire of length x meter is made into a square implies perimeter of the square is x meter. Now side of the square is equal to x divided by 4. We know perimeter of a square is 4 multiplied by side so we can get side of a square if we divide perimeter by 4. Now the wire of length 28 minus x meter is made into a circle implies circumference of the circle is 28 minus x meter. Now let us assume that radius of the circle is r meters. Now we know circumference of the circle is equal to 2 pi r so we can write 2 pi r is equal to 28 minus x meters. Now this implies r is equal to 28 minus x upon 2 pi meters. We know area of the circle is equal to pi r square. Now substituting this value of r in this formula we get pi multiplied by 28 minus x upon 2 pi whole square which is further equal to pi multiplied by 28 minus x whole square upon 4 pi square. Simplifying we get 28 minus x whole square upon 4 pi. Now let us find out area of square area of square. We know it is equal to side square and side of square we have already shown above is x divided by 4. So we get area equal to x square upon 16 let a be the combined area of circle and the square. It is equal to 28 minus x whole square upon 4 pi plus x square upon 16. Now differentiating both sides with respect to x we get dA upon dx is equal to 1 upon 4 pi multiplied by 2 multiplied by 28 minus x multiplied by minus 1 plus 2x upon 16. We can find derivative of both the terms by quotient rule. Now simplifying we get dA upon dx is equal to 2 will cancel 4 and we know 2 multiplied by 8 is equal to 16. So we get x upon 8 minus 28 minus x upon 2 pi. Now we will find all the values of x at which dA upon dx is equal to 0. So we will put dA upon dx equal to 0. Now this implies x upon 8 minus 28 minus x upon 2 pi is equal to 0. Now this implies pi x minus 4 multiplied by 28 minus x upon 8 pi is equal to 0. Now multiplying both sides by 8 pi we get pi x minus 4 multiplied by 28 minus x is equal to 0. Now this further implies pi x minus 112 plus 4x is equal to 0. We have multiplied 4 by this bracket. Now this further implies pi x plus 4x is equal to 112. Adding 112 on both sides of this equation we get pi x plus 4x is equal to 112. Now taking x common in left hand side we get x multiplied by pi plus 4 is equal to 112. This further implies x is equal to 112 upon pi plus 4. Now we know that x is equal to 112 upon pi plus 4. dA upon dx is equal to 0. And we also know that combined area would be minimum at x equal to 112 upon pi plus 4. If dA upon dx is equal to 0 and d square A upon dx square is greater than 0. So now we will find value of d square A upon dx square. We know dA upon dx is equal to x upon 8 minus 28 minus x upon 2 pi. Now differentiating both sides with respect to x we get d square A upon dx square is equal to 1 upon 8 minus 1 upon 2 pi multiplied by minus 1. Which can be further written as 1 upon 8 plus 1 upon 2 pi. So we get d square A upon dx square is equal to 1 upon 8 plus 1 upon 2 pi which is greater than 0. Now from above discussion we see at x is equal to 112 upon pi plus 4. dA upon dx is equal to 0 and d square A upon dx square is greater than 0. So this implies there exists minimum at x is equal to 112 upon pi plus 4. We know length of the two pieces of wire were x meter and 28 minus x meter. So length of the one piece of wire is equal to 112 upon pi plus 4 meter. Length of the other piece is equal to 28 minus 112 upon pi plus 4. Now this is equal to 28 multiplied by pi plus 4 minus 112 upon pi plus 4. This is further equal to 28 pi plus 112 minus 112 upon pi plus 4. Now plus 112 and minus 112 will cancel each other and we get 28 pi upon pi plus 4 meters. So we get length of two pieces of wire is 112 upon pi plus 4 meters and 28 pi upon pi plus 4 meters. This completes the session. Hope you understood the solution. Take care and have a nice day.