## Application of Derivative

A piece of wire 28 cm long is to be cut into two pieces. One piece is to be made into a circle and another into a square. How should the wire be cut so that the combined area of the two figures is as small as possible?

Let the wire be cut at a distance of  x meter  from one end. Therefore then two pieces of wire be x m and (28-x) m.

## Calculate Dimension of Circle and Square

Now 1st part be turned into a square and  the 2nd part be be made into a circle.

Since 1st part of the wire is turned into square. then its perimeter will be x m.
So using formula of perimeter of square , we can calculate side of the square = x/4 m

## Calculate Areas of Circle and Square

Therefore Area of square = (x/4)(x/4) sq m

A1 = x2/16

And  when 2nd part of the wire is turned to circle, then its perimeter ( circumference ) will be 28 - x m. So using formula of perimeter of square , And if  "r" be  radius of the circle , Then
Circumference of circle =  2 π r =  (28-x)
∴  r = (28-x)/2π

We know that Area of Circle A2   = π r2

A2  π[(28-x)/2π]2

## To find value/s of x

Now to find the value of x for which this function A(x) is maximum or minimum ,put A(x) = 0

## To Test the Minimum Value of  Function

Now we have the value of "x" on which either A(x) have maximum or minimum value . To check the maximum or minimum value we have to find A''(x) as follows

So A''(x) has positive value Therefore A(x) shall have maximum value at x = 112/(π + 4)

Hence two pieces of wire should be of length x m and (28-x) m

These pieces should be of length 112/(π+4) and 28π/(π + 4)

## Verification

we can calculate the sum of these pieces , it must be 28 m

#### 1st part

112/(π+4) = 112/{(22/7)+4}=112×7/50 = 784/50

#### 2nd part

28π/(π + 4) = {28×22/7}/{(22/7)+4} = 88×7/50 = 616/50

#### Sum of Two Parts

112×7/50 + 28×7/50 = (784+616)/50

= 1400/50= 28 m

## Conclusion

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