How should the wire of 28 m be cut so that the combined area of the circle and square is as small as possible ?
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
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
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)/2Ï€
We know that Area of Circle A2 = π r2
A2 = π[(28-x)/2π]2
Express Areas in terms of Function
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/(Ï€ + 4)
Hence two pieces of wire should be of length x m and (28-x) m
These pieces should be of length 112/(Ï€+4) and 28Ï€/(Ï€ + 4)
Verification
we can calculate the sum of these pieces , it must be 28 m
1st part
112/(Ï€+4) = 112/{(22/7)+4}=112×7/50 = 784/50
2nd part
28Ï€/(Ï€ + 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
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