22) A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

Answers (1)

Area of the square (A) = a^2
Area of the circle(S) = \pi r^2
Given the length of wire  = 28 m
Let the length of one of the piece is x m 
Then the length of the other piece is (28 - x) m
Now,
 4a = x\Rightarrow a = \frac{x}{4}
and 
2 \pi r = (28-x) \Rightarrow r= \frac{28-x}{2\pi}
Area of the combined circle and square f(x) = A + S
                                                          =a^2 + \pi r^2 = (\frac{x}{4})^2+\pi (\frac{28-x}{2\pi})^2
f^{'}(x) = \frac{2x}{16}+\frac{(28-x)(-1)}{2\pi} \\ f^{'}(x) = \frac{x\pi+4x-112}{8\pi}\\ f^{'}(x) = 0\\ \frac{x\pi+4x-112}{8\pi} = 0\\ x(\pi+4) = 112\\ x = \frac{112}{\pi + 4}
Now,
f^{''}(x) = \frac{1}{8}+ \frac{1}{2\pi}\\ f^{''}(\frac{112}{\pi+4}) = \frac{1}{8}+ \frac{1}{2\pi} > 0
Hence, x = \frac{112}{\pi+4}  is the point of minima
Other length is = 28 - x
                         = 28 - \frac{112}{\pi+4} = \frac{28\pi+112-112}{\pi+4} = \frac{28\pi}{\pi+4}
Hence, two lengths are \frac{28\pi}{\pi+4} and \frac{112}{\pi+4} 
 

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