Question

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A wire of length 20m is cut into two parys, one of which is bent into a circle and the other into a square. if the sum of the area of the figures is maximum,

a. find the diameter of the circle.

b. find the side of the square.

c. find the sumof the maximum area of the two figures in m^2.

Answer 1

Solution:

Given, length of wire = 20 m.

Let the radius of the circle be $r$ m and side of the square be $s$ m.

Then, circumference of circle + perimeter of square = 20

$2\pi r+4s=20 \\ \Rightarrow \pi r+2s=10 \\ \Rightarrow s=\dfrac{10-\pi r}{2}\ ...(i)$

Let total area of two parts be A. So,

$A=\pi r^2+s^2 \\ \Rightarrow A=\pi r^2 +(\dfrac{10-\pi r}{2})^2$ [Using (i)]

On differentiating w.r.t r,

$\dfrac{dA}{dr}=2\pi r+\dfrac12(10-\pi r)(-\pi) \\=2\pi r-5\pi +\dfrac12\pi^2 r$

(a) Put $\dfrac{dA}{dr}=0$

$\Rightarrow 2\pi r-5\pi +\dfrac12\pi^2 r=0 \\\Rightarrow r(2\pi +\dfrac12\pi^2)=5\pi \\\Rightarrow r=\dfrac{5}{2+\dfrac12\pi} \\\Rightarrow r=\dfrac{10}{4+\pi}\ m \ ...(ii)$

The diameter of the circle is $D=2r=\frac{20}{4+\pi}$

(b) Again differentiating $\dfrac{dA}{dr}$ w.r.t. r,

$\dfrac{dA}{dr}=2\pi r-5\pi +\dfrac12\pi^2 r \\ \dfrac{d^2A}{dr^2}=2\pi+\dfrac12\pi^2>0$

Now put (ii) in (i)

$s=\dfrac{10-\pi (\dfrac{10}{4+\pi})}{2} \\ \Rightarrow s=\dfrac{\dfrac{40+10\pi-10\pi}{4+\pi}}{2} \\ \Rightarrow s=\dfrac{20}{4+\pi}\ m$

(c)

Now, the sum of the maximum area of the two figures in m^2 is

$A=\pi r^2+s^2 \\ \Rightarrow A=\pi (\dfrac{10}{4+\pi})^2+(\dfrac{20}{4+\pi})^2 \\ \Rightarrow A\approx 14\ m^2$