Question

A thin rope \[200{\text{ inches}}\] length is going to be cut into three pieces to form one circle and two equal squares. Which of the following represents the \[r\] of the circle, if the rope was finally cut in such a way that the total area of the shapes is minimized?
A) \[\dfrac{{25}}{{2 + 4\pi }}\]
B) \[\dfrac{{50}}{{1 + 4\pi }}\]
C) \[\dfrac{{100}}{{8 + \pi }}\]
D) \[\dfrac{{100}}{{8\pi + 1}}\]

Answer 1

Hint: We will equate the length of rope to the sum of circumference of the circle and the perimeter of two squares. From this we will find the relationship between the length of the square and the radius of the circle. Then we will find the total area of each shape by adding their individual area. Then we will differentiate the total area with respect to radius of circle and by equating the derivative to zero we will find the value of radius of circle. Then we will check whether the obtained value of radius given the minimum area or the maximum area by double differentiating the area and putting the value of radius in that. If we get a positive value then the radius will give a minimum area and if we get a negative value the radius will give the maximum area.

Complete step by step answer:
Let,
Length of rope \[L = 200\]
Radius of circle \[ = r\]
Side of square \[ = a\]
So, we get;
Circumference of circle \[ = 2\pi r\]
Perimeter of each square \[ = 4a\]
Area of circle \[ = \pi {r^2}\]
Area of square \[ = {a^2}\]
Now,
\[L = 2\pi r + 4a + 4a\]
Adding and putting the values we get;
\[ \Rightarrow 2\pi r + 8a = 200\]
Dividing both sides by \[2\], we get;
\[ \Rightarrow \pi r + 4a = 100\]
Sifting we get;
\[ \Rightarrow 4a = 100 - \pi r\]
\[ \Rightarrow a = \dfrac{{100 - \pi r}}{4}\]
Now,
Total area, \[A = {\text{ area of circle + 2}} \times {\text{ area of each square}}\]
Putting the values, we get;
\[ \Rightarrow A = \pi {r^2} + 2 \times {a^2}\]
Now we will put the value of \[a\], obtained in the above equation. So, we get;
\[ \Rightarrow A = \pi {r^2} + 2{\left( {\dfrac{{100 - \pi r}}{4}} \right)^2}\]
Solving we get;
\[ \Rightarrow A = \pi {r^2} + \dfrac{2}{{16}}{\left( {100 - \pi r} \right)^2}\]
On reducing we get;
\[ \Rightarrow A = \pi {r^2} + \dfrac{1}{8}{\left( {100 - \pi r} \right)^2}\]
Now we will differentiate it with respect to \[r\]. So, we have;
\[ \Rightarrow \dfrac{{dA}}{{dr}} = \dfrac{d}{{dr}}\pi {r^2} + \dfrac{d}{{dr}}\left( {\dfrac{1}{8}{{\left( {100 - \pi r} \right)}^2}} \right)\]
Now we will take the constant term out of the differentiation sign. So, we have;
\[ \Rightarrow \dfrac{{dA}}{{dr}} = \pi \dfrac{d}{{dr}}{r^2} + \dfrac{1}{8}\dfrac{d}{{dr}}{\left( {100 - \pi r} \right)^2}\]
Now differentiating using the chain rule we have;
\[ \Rightarrow \dfrac{{dA}}{{dr}} = 2\pi r + \dfrac{1}{8} \times 2\left( {100 - \pi r} \right) \times \dfrac{{d\left( {100 - \pi r} \right)}}{{dr}}\]
On further simplification we get;
\[ \Rightarrow \dfrac{{dA}}{{dr}} = 2\pi r + \dfrac{1}{4}\left( {100 - \pi r} \right) \times \left( { - \pi } \right)\]
Now we will take \[\pi \] common in the RHS. So, we get;
\[ \Rightarrow \dfrac{{dA}}{{dr}} = \pi \left( {2r - \dfrac{{100 - \pi r}}{4}} \right)\]
Now we will equate it to zero to find the value of \[r\]. So, we have;
\[ \Rightarrow \pi \left( {2r - \dfrac{{100 - \pi r}}{4}} \right) = 0\]
\[ \Rightarrow \left( {2r - \dfrac{{100 - \pi r}}{4}} \right) = 0\]
Now shifting to the RHS we get;
\[ \Rightarrow 2r = \dfrac{{100 - \pi r}}{4}\]
On cross-multiplication we get;
\[ \Rightarrow 8r = 100 - \pi r\]
\[ \Rightarrow 8r + \pi r = 100\]
Now taking \[r\] as common. We get;
\[ \Rightarrow \left( {8 + \pi } \right)r = 100\]
On shifting to RHS we get;
\[ \Rightarrow r = \dfrac{{100}}{{\left( {8 + \pi } \right)}}\]
Now we will double differentiate the Area. We have;
 \[ \Rightarrow \dfrac{{dA}}{{dr}} = \pi \left( {2r - \dfrac{{100 - \pi r}}{4}} \right)\]
now differentiating it once more with respect to \[r\], we get;
\[ \Rightarrow \dfrac{{{d^2}A}}{{d{r^2}}} = \dfrac{d}{{dr}}\left( {\pi \left( {2r - \dfrac{{100 - \pi r}}{4}} \right)} \right)\]
Expanding the bracket, we get;
\[ \Rightarrow \dfrac{{{d^2}A}}{{d{r^2}}} = \dfrac{d}{{dr}}\left( {2\pi r - \dfrac{{100\pi }}{4} + \dfrac{{{\pi ^2}r}}{4}} \right)\]
Now we will differentiate each term individually. So, we get;
\[ \Rightarrow \dfrac{{{d^2}A}}{{d{r^2}}} = \dfrac{d}{{dr}}\left( {2\pi r} \right) - \dfrac{d}{{dr}}\left( {\dfrac{{100\pi }}{4}} \right) + \dfrac{d}{{dr}}\left( {\dfrac{{{\pi ^2}r}}{4}} \right)\]
On differentiating we get;
\[ \Rightarrow \dfrac{{{d^2}A}}{{d{r^2}}} = 2\pi - 0 + \dfrac{{{\pi ^2}}}{4}\]
Taking \[\pi \] as common we get;
\[ \Rightarrow \dfrac{{{d^2}A}}{{d{r^2}}} = \pi \left( {2 + \dfrac{\pi }{4}} \right)\]
Now we can see that all terms are positive in the RHS. So, the total term is positive. Now the double differentiation is positive. Also, we have only one value of radius. So, that value of radius will give the minimum area.
Hence,
\[ \Rightarrow r = \dfrac{{100}}{{\left( {8 + \pi } \right)}}\], will give a minimum area.
So, option c is correct.

Note: One thing to note is that if in any other question we have two values for the radius then we will put each value in the double derivative and check for each value whether the derivative comes to be positive or negative. Although in the options unit is not given, the value of radius will be in inches.