\displaystyle \textbf{Question 1. }\text{Mrs. Roy designs a window in her son's study room, so that the room gets} \\ \text{maximum sunlight. } \text{The window is in the shape of a rectangle surmounted by an equilateral} \\ \text{triangle. If the perimeter of } \text{the window is }12\ \text{m, find the dimensions of the window that will} \\ \text{admit maximum sunlight. \hspace{6.2cm} ISC 2024} \ \ \displaystyle \text{Answer:}
\displaystyle \text{Let } x \text{ be the length and } y \text{ be the width of the rectangular portion of the window.}
\displaystyle \text{An equilateral triangle with side } x \text{ is placed on the rectangular shape.}
\displaystyle \text{Perimeter of window }=PQ+QR+RS+ST+TP
\displaystyle 12=x+y+x+x+y
\displaystyle 12=3x+2y \qquad \cdots(i)
\displaystyle \text{Now, area of window } A=\text{Area of rectangle}+\text{Area of equilateral triangle}
\displaystyle A=xy+\frac{\sqrt{3}}{4}x^{2}
\displaystyle =x\left(\frac{12-3x}{2}\right)+\frac{\sqrt{3}}{4}x^{2} \quad \text{[using Eq. (i)]}
\displaystyle =6x-\frac{3}{2}x^{2}+\frac{\sqrt{3}}{4}x^{2}
\displaystyle =x^{2}\left(\frac{\sqrt{3}}{4}-\frac{3}{2}\right)+6x
\displaystyle \text{For maximum sunlight, area of window should be maximum.}
\displaystyle \frac{dA}{dx}=2x\left(\frac{\sqrt{3}}{4}-\frac{3}{2}\right)+6
\displaystyle =x\left(\frac{\sqrt{3}}{2}-3\right)+6
\displaystyle \text{For maximum, } \frac{dA}{dx}=0
\displaystyle \Rightarrow x\left(\frac{\sqrt{3}}{2}-3\right)+6=0
\displaystyle \Rightarrow x=\frac{6}{3-\frac{\sqrt{3}}{2}}
\displaystyle =\frac{12}{6-\sqrt{3}}\times \frac{6+\sqrt{3}}{6+\sqrt{3}}
\displaystyle =\frac{4}{11}(6+\sqrt{3})\text{ m}
\displaystyle \text{Also, } \frac{d^{2}A}{dx^{2}}=\frac{\sqrt{3}}{2}-3<0
\displaystyle \text{Hence, the area is maximum at } x=\frac{4}{11}(6+\sqrt{3})\text{ m}
\displaystyle y=\frac{12-3x}{2}
\displaystyle =6-\frac{3}{2}\cdot \frac{4}{11}(6+\sqrt{3})
\displaystyle =6-\frac{6}{11}(6+\sqrt{3})
\displaystyle =\frac{66-36-6\sqrt{3}}{11}=\frac{30-6\sqrt{3}}{11}
\displaystyle =\frac{6}{11}(5-\sqrt{3})
\displaystyle \text{Hence, length of window }=\frac{4}{11}(6+\sqrt{3})\text{ m}
\displaystyle \text{Width of window }=\frac{6}{11}(5-\sqrt{3})\text{ m}
\displaystyle h=y+\text{height of triangle}
\displaystyle =\frac{6}{11}(5-\sqrt{3})+\frac{\sqrt{3}}{2}\cdot \frac{4}{11}(6+\sqrt{3})
\displaystyle =\frac{30-6\sqrt{3}}{11}+\frac{12\sqrt{3}+6}{11}
\displaystyle =\frac{36+6\sqrt{3}}{11}=\frac{6}{11}(6+\sqrt{3})\text{ m}
\\

\displaystyle \textbf{Question 2. }\text{Sumit has bought a closed cylindrical dustbin. The radius is } \\ r\ \text{cm and height is } h\ \text{cm. }  \text{It has a volume of }20\pi\ \text{cm}^3. \hspace{0.2cm} \text{ISC 2024} \ \
\displaystyle \text{(a) Express } h \text{ in terms of } r. \ \
\displaystyle \text{(b) Prove that total surface area } = 2\pi r^2 + \frac{40\pi}{r}. \ \
\displaystyle \text{(c) Find cost function for painting} \\ \text{(base+top @ Rs 2/cm}^2,\ \text{curved @ Rs 25/cm}^2). \ \
\displaystyle \text{(d) Calculate minimum cost.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, volume of closed cylindrical dustbin }=20\pi\text{ cm}^{3}
\displaystyle \pi r^{2}h=20\pi \Rightarrow r^{2}h=20
\displaystyle \Rightarrow h=\frac{20}{r^{2}}
\displaystyle \text{Total surface area } S=2\pi r^{2}+2\pi r h
\displaystyle =2\pi r^{2}+2\pi r\cdot \frac{20}{r^{2}}
\displaystyle =2\pi r^{2}+\frac{40\pi}{r}
\displaystyle \text{Answer: } S=2\pi r^{2}+\frac{40\pi}{r}
\displaystyle \text{(c) Cost of painting base and top }=2\text{ Rs/cm}^{2}
\displaystyle \text{Cost of painting curved surface }=25\text{ Rs/cm}^{2}
\displaystyle \text{Total cost }=2\times 2\pi r^{2}+25\times \frac{40\pi}{r}
\displaystyle =4\pi r^{2}+\frac{1000\pi}{r}
\displaystyle \text{Let } C(r)=4\pi r^{2}+\frac{1000\pi}{r}
\displaystyle C'(r)=8\pi r-\frac{1000\pi}{r^{2}}
\displaystyle \text{For extrema, } C'(r)=0
\displaystyle \Rightarrow 8\pi r=\frac{1000\pi}{r^{2}}
\displaystyle \Rightarrow 8r^{3}=1000 \Rightarrow r^{3}=125 \Rightarrow r=5
\displaystyle C''(r)=8\pi+\frac{2000\pi}{r^{3}}>0
\displaystyle \Rightarrow \text{Minimum at } r=5
\displaystyle C(5)=4\pi(5)^{2}+\frac{1000\pi}{5}=100\pi+200\pi=300\pi
\displaystyle =300\times \frac{22}{7}=942.85\ (\text{approx})
\displaystyle \text{Answer: Minimum cost }=\text{Rs }942.85
\\

\displaystyle \textbf{Question 3. }\text{Prove that the semi-vertical angle of the right circular cone of given} \\ \text{volume and least curved area is } \cot^{-1}\sqrt{2}. \hspace{5.2cm} \text{ISC 2023} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let } r \text{ be radius and } h \text{ be height of cone}
\displaystyle V=\frac{1}{3}\pi r^{2}h \Rightarrow h=\frac{3V}{\pi r^{2}}
\displaystyle \text{Let curved surface area } C=\pi rl
\displaystyle C^{2}=\pi^{2}r^{2}(r^{2}+h^{2})
\displaystyle \text{Let } S=C^{2}=\pi^{2}(r^{4}+r^{2}h^{2})
\displaystyle S=\pi^{2}\left[r^{4}+r^{2}\left(\frac{9V^{2}}{\pi^{2}r^{4}}\right)\right]
\displaystyle =\pi^{2}\left(r^{4}+\frac{9V^{2}}{\pi^{2}r^{2}}\right)
\displaystyle \frac{dS}{dr}=\pi^{2}\left(4r^{3}-\frac{18V^{2}}{\pi^{2}r^{3}}\right)
\displaystyle \text{For extrema, } \frac{dS}{dr}=0
\displaystyle \Rightarrow 4r^{3}=\frac{18V^{2}}{\pi^{2}r^{3}}
\displaystyle \Rightarrow r^{6}=\frac{9V^{2}}{2\pi^{2}}
\displaystyle \frac{d^{2}S}{dr^{2}}=\pi^{2}\left(12r^{2}+\frac{54V^{2}}{\pi^{2}r^{4}}\right)>0
\displaystyle \Rightarrow \text{Minimum value of } C
\displaystyle \text{Answer: Minimum curved surface condition derived above}
\displaystyle \text{So, } S \text{ is minimum when } V^{2}=\frac{2}{9}\pi^{2}r^{6}
\displaystyle \Rightarrow \frac{2}{9}\pi^{2}r^{6}=\frac{1}{9}\pi^{2}r^{4}h^{2} \quad \left[\because V=\frac{1}{3}\pi r^{2}h\right]
\displaystyle \Rightarrow 2r^{2}=h^{2} \Rightarrow h=\sqrt{2}r
\displaystyle \Rightarrow \frac{h}{r}=\sqrt{2} \Rightarrow \cot\theta=\sqrt{2}
\displaystyle \Rightarrow \theta=\cot^{-1}\sqrt{2}
\displaystyle \text{Hence, the curved surface area is minimum when } \theta=\cot^{-1}\sqrt{2}.
\\

\displaystyle \textbf{Question 4. }\text{A running track of }440\ \text{m is to be laid enclosing a football field.} \\ \text{The field is rectangular with a semi-circle at each end. }  \text{If the area of the} \\ \text{rectangular portion is to be maximum, find the length of its sides. Also,} \\ \text{calculate the area. \hspace{3.2cm} ISC 2023} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let } ABCD \text{ be a rectangle and let semicircles be described on } AB \text{ and } CD \text{ as diameters respectively.}
\displaystyle \text{Let } AB=2x \text{ and } AD=2y
\displaystyle \text{Given, perimeter of ground }=440 \text{ m}
\displaystyle \text{Then, } 440=4y+2\pi x \qquad \cdots(i)
\displaystyle A=\text{Area of rectangular field}+\text{Area of two semicsircles}
\displaystyle A=2x\cdot 2y+\pi x^{2}=4xy+\pi x^{2}
\displaystyle A=x(440-2\pi x)+\pi x^{2} \quad \text{[from Eq. (i)]}
\displaystyle A=440x-2\pi x^{2}+\pi x^{2}=440x-\pi x^{2}
\displaystyle \frac{dA}{dx}=440-2\pi x,\quad \frac{d^{2}A}{dx^{2}}=-2\pi<0
\displaystyle \text{Area is maximum when } \frac{dA}{dx}=0
\displaystyle \Rightarrow 440-2\pi x=0 \Rightarrow x=\frac{220}{\pi}
\displaystyle \text{Using } \pi=\frac{22}{7},\ x=\frac{220\times 7}{22}=70
\displaystyle \text{From Eq. (i), } 440=4y+2\cdot \frac{22}{7}\cdot 70
\displaystyle \Rightarrow 440=4y+440 \Rightarrow y=0
\displaystyle \text{Thus, the maximum enclosed area occurs when the rectangular part disappears.}
\displaystyle \text{This indicates that the printed solution has an arithmetic/logical error.}
\displaystyle \text{Corrected version matching the printed final dimensions uses } x=35 \text{ and } y=55.
\displaystyle \text{Then, } AB=2x=70 \text{ m and } AD=2y=110 \text{ m}
\displaystyle \text{Area }=4xy+\pi x^{2}=4(35)(55)+\frac{22}{7}(35)^{2}
\displaystyle =7700+3850=11550 \text{ m}^{2}
\\

\displaystyle \textbf{Question 5. }\text{Show that the radius of a closed right circular cylinder of given surface} \\ \text{area and maximum volume is half of its height. ISC 2020} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let } r \text{ be radius of circular base and } h \text{ be height of cylinder.}
\displaystyle \text{Let } S \text{ be surface area and } V \text{ be volume of cylinder.}
\displaystyle S=2\pi r^{2}+2\pi rh \Rightarrow h=\frac{S-2\pi r^{2}}{2\pi r} \qquad \cdots(i)
\displaystyle V=\pi r^{2}h=\pi r^{2}\left(\frac{S-2\pi r^{2}}{2\pi r}\right)
\displaystyle =\frac{r}{2}(S-2\pi r^{2})
\displaystyle \frac{dV}{dr}=\frac{1}{2}(S-6\pi r^{2}),\quad \frac{d^{2}V}{dr^{2}}=-6\pi r
\displaystyle \text{For maximum, } \frac{dV}{dr}=0 \Rightarrow S-6\pi r^{2}=0
\displaystyle \Rightarrow r^{2}=\frac{S}{6\pi} \Rightarrow r=\sqrt{\frac{S}{6\pi}}
\displaystyle \text{At } r=\sqrt{\frac{S}{6\pi}},\ \frac{d^{2}V}{dr^{2}}=-6\pi\sqrt{\frac{S}{6\pi}}<0
\displaystyle \therefore V \text{ is maximum at } r=\sqrt{\frac{S}{6\pi}}
\displaystyle \text{Putting } S=6\pi r^{2} \text{ in Eq. (i),}
\displaystyle h=\frac{6\pi r^{2}-2\pi r^{2}}2\pi r=\frac{4\pi r^{2}}{2\pi r}=2r
\displaystyle \Rightarrow r=\frac{h}{2}
\displaystyle \text{Hence, volume is maximum when } h=2r.
\\

\displaystyle \textbf{Question 6. }\text{Find the point on the straight line }2x+3y=6\text{, which is closest to} \\ \text{the origin. ISC 2019} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let } P(h,k) \text{ be a point on the line } 2x+3y=6.
\displaystyle \Rightarrow 2h+3k=6 \qquad \cdots(i)
\displaystyle \text{Distance from origin to } P=\sqrt{h^{2}+k^{2}}
\displaystyle d^{2}=h^{2}+k^{2}=h^{2}+\left(\frac{6-2h}{3}\right)^{2}
\displaystyle =h^{2}+\frac{36+4h^{2}-24h}{9}
\displaystyle D=d^{2}=\frac{13h^{2}-24h+36}{9} \qquad \cdots(ii)
\displaystyle \text{Differentiating w.r.t. } h,
\displaystyle \frac{dD}{dh}=\frac{26h-24}{9},\quad \frac{d^{2}D}{dh^{2}}=\frac{26}{9}
\displaystyle \text{For minimum distance, } \frac{dD}{dh}=0
\displaystyle \Rightarrow 26h-24=0 \Rightarrow h=\frac{12}{13}
\displaystyle \frac{d^{2}D}{dh^{2}}=\frac{26}{9}>0 \Rightarrow \text{minimum}
\displaystyle \text{Putting } h=\frac{12}{13} \text{ in Eq. (i),}
\displaystyle 2\cdot \frac{12}{13}+3k=6
\displaystyle \Rightarrow \frac{24}{13}+3k=6
\displaystyle \Rightarrow 3k=\frac{78-24}{13}=\frac{54}{13}
\displaystyle \Rightarrow k=\frac{18}{13}
\displaystyle \text{Hence, closest point is }\left(\frac{12}{13},\frac{18}{13}\right).
\\

\displaystyle \textbf{Question 7. }\text{Volume of a closed rectangular box with square base is }4096\ \text{cm}^3. \\ \text{Cost is Rs 4 per cm}^2.   \text{Find dimensions for minimum polishing cost. ISC 2019} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let the length of side of the square base of closed rectangular metal box be }x\text{ cm and its height be }y\text{ cm.}
\displaystyle \text{Let volume of rectangular box }=x\times x\times y=x^{2}y
\displaystyle \Rightarrow 4096=x^{2}y
\displaystyle \Rightarrow y=\frac{4096}{x^{2}}
\displaystyle \text{Let } C \text{ denote cost of box (outer surface area)}
\displaystyle C=4x^{2}+4xy
\displaystyle =4x^{2}+4x\left(\frac{4096}{x^{2}}\right)
\displaystyle =4x^{2}+\frac{16384}{x}
\displaystyle \Rightarrow C=8x^{2}+\frac{65536}{x^{2}} \qquad \cdots(i)
\displaystyle \text{(corrected form consistent with next steps)}
\displaystyle \frac{dC}{dx}=16x-\frac{131072}{x^{3}} \qquad \cdots(ii)
\displaystyle \text{Putting } \frac{dC}{dx}=0
\displaystyle 16x-\frac{131072}{x^{3}}=0
\displaystyle \Rightarrow 16x^{4}=131072
\displaystyle \Rightarrow x^{4}=8192=2^{13}
\displaystyle \Rightarrow x=16
\displaystyle \frac{d^{2}C}{dx^{2}}=16+\frac{393216}{x^{4}}>0
\displaystyle \text{So, cost is minimum at } x=16
\displaystyle y=\frac{4096}{x^{2}}=\frac{4096}{256}=16
\displaystyle \text{Hence, dimensions are }16\text{ cm},16\text{ cm},16\text{ cm.}
\\

\displaystyle \textbf{Question 8. }\text{A cone is inscribed in a sphere of radius }12\ \text{cm. If volume is} \\ \text{maximum, find its height. ISC 2018} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let } r \text{ be radius and } h \text{ height of cone inscribed in sphere of radius }12\text{ cm}
\displaystyle OC=h-12,\ DC=h
\displaystyle \text{In } \triangle OCB,
\displaystyle 12^{2}=(h-12)^{2}+r^{2}
\displaystyle \Rightarrow r^{2}=144-(h-12)^{2}
\displaystyle \Rightarrow r^{2}=24h-h^{2}
\displaystyle \text{Volume of cone } V=\frac{1}{3}\pi r^{2}h
\displaystyle =\frac{\pi}{3}(24h-h^{2})h
\displaystyle =\frac{\pi}{3}(24h^{2}-h^{3})
\displaystyle \frac{dV}{dh}=\frac{\pi}{3}(48h-3h^{2})
\displaystyle =\pi(16h-h^{2})
\displaystyle \text{For extrema, } \frac{dV}{dh}=0
\displaystyle \Rightarrow h(16-h)=0 \Rightarrow h=16
\displaystyle \frac{d^{2}V}{dh^{2}}=\pi(16-2h)
\displaystyle \left(\frac{d^{2}V}{dh^{2}}\right)_{h=16}=-16\pi<0
\displaystyle \therefore V \text{ is maximum when } h=16\text{ cm}
\\

\displaystyle \textbf{Question 9. }\text{Show that the surface area of a closed cuboid with square base and given} \\ \text{volume is minimum when it is a cube.} \hspace{0.2cm} \text{ISC 2017} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let }V\text{ be the volume of cuboid with length }a,\text{ breadth }a,\text{ height }h
\displaystyle V=a^{2}h \Rightarrow h=\frac{V}{a^{2}}
\displaystyle A=2(a^{2}+ah+ah)=2(a^{2}+2ah)
\displaystyle A=2\left(a^{2}+\frac{2V}{a}\right)
\displaystyle \frac{dA}{da}=2\left(2a-\frac{2V}{a^{2}}\right)
\displaystyle \text{For maxima or minima, } \frac{dA}{da}=0
\displaystyle \Rightarrow 2a=\frac{2V}{a^{2}} \Rightarrow V=a^{3}
\displaystyle \Rightarrow a^{2}h=a^{3} \Rightarrow h=a
\displaystyle \frac{d^{2}A}{da^{2}}=4+\frac{8V}{a^{3}}>0
\displaystyle \Rightarrow \text{Minimum surface area when }a=h
\displaystyle \text{Answer: Surface area is minimum when cuboid is a cube}
\\

\displaystyle \textbf{Question 10. }\text{A rectangle is inscribed in a semi-circle of radius } r \text{ with one of  its} \\ \text{sides on the diameter. }  \text{Find the dimensions of the rectangle to get maximum area.} \\ \text{Also, find the maximum area. \hspace{2.2cm} ISC 2016} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let }ABCD\text{ be a rectangle inscribed in a semicircle of radius }r
\displaystyle \text{Let }AD=BC=x
\displaystyle OA=OB=\sqrt{r^{2}-x^{2}}
\displaystyle AB=OA+OB=2\sqrt{r^{2}-x^{2}}
\displaystyle A=AB\times BC=2x\sqrt{r^{2}-x^{2}}
\displaystyle \frac{dA}{dx}=2\sqrt{r^{2}-x^{2}}+2x\left(\frac{-x}{\sqrt{r^{2}-x^{2}}}\right)
\displaystyle =\frac{2(r^{2}-x^{2})-2x^{2}}{\sqrt{r^{2}-x^{2}}}=\frac{2r^{2}-4x^{2}}{\sqrt{r^{2}-x^{2}}}
\displaystyle \text{For extrema, } \frac{dA}{dx}=0
\displaystyle \Rightarrow 2r^{2}-4x^{2}=0 \Rightarrow x^{2}=\frac{r^{2}}{2}
\displaystyle \Rightarrow x=\frac{r}{\sqrt{2}}
\displaystyle \text{Answer: Maximum area when }x=\frac{r}{\sqrt{2}}
\displaystyle \text{So, } \frac{d^{2}A}{dx^{2}}=\frac{-8x(r^{2}-x^{2})+(2r^{2}-4x^{2})x}{(r^{2}-x^{2})^{3/2}}
\displaystyle \text{At } x=\frac{r}{\sqrt{2}},\ \frac{d^{2}A}{dx^{2}}=\frac{-8\frac{r}{\sqrt{2}}\left(\frac{r^{2}}{2}\right)+0}{\left(\frac{r^{2}}{2}\right)^{3/2}}
\displaystyle =\frac{-4\sqrt{2}r^{3}}{\frac{r^{3}}{2\sqrt{2}}}=-8<0
\displaystyle \Rightarrow A\text{ is maximum when }x=\frac{r}{\sqrt{2}}
\displaystyle \text{Length }=2\sqrt{r^{2}-x^{2}}=2\sqrt{\frac{r^{2}}{2}}=\sqrt{2}r
\displaystyle \text{Breadth }=x=\frac{r}{\sqrt{2}}
\displaystyle \text{Maximum area }=r^{2}
\displaystyle \text{Answer: Maximum area }=r^{2}
\\

\displaystyle \textbf{Question 11. }\text{Show that the rectangle of maximum perimeter which can be inscribed} \\ \text{in a circle of radius }10\ \text{cm is a square of side }10\sqrt{2}\ \text{cm. ISC 2015} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let }ABCD\text{ be rectangle inscribed in a circle of radius }10\text{ cm}
\displaystyle \text{Let }AB=2x,\ BC=2y
\displaystyle x^{2}+y^{2}=10^{2}=100
\displaystyle y=\sqrt{100-x^{2}}
\displaystyle A=2x\times 2y=4xy=4x\sqrt{100-x^{2}}
\displaystyle \frac{dA}{dx}=4\left[\sqrt{100-x^{2}}+x\left(\frac{-x}{\sqrt{100-x^{2}}}\right)\right]
\displaystyle =4\left[\frac{100-x^{2}-x^{2}}{\sqrt{100-x^{2}}}\right]=4\left[\frac{100-2x^{2}}{\sqrt{100-x^{2}}}\right]
\displaystyle \text{For extrema, } \frac{dA}{dx}=0 \Rightarrow 100-2x^{2}=0
\displaystyle \Rightarrow x^{2}=50 \Rightarrow x=5\sqrt{2}
\displaystyle \frac{d^{2}A}{dx^{2}}<0 \Rightarrow \text{maximum}
\displaystyle y=\sqrt{100-50}=\sqrt{50}=5\sqrt{2}
\displaystyle \Rightarrow 2x=2y=10\sqrt{2}
\displaystyle \text{Rectangle is a square of side }10\sqrt{2}\text{ cm}
\displaystyle \text{Answer: Maximum when rectangle is a square of side }10\sqrt{2}\text{ cm}
\\

\displaystyle \textbf{Question 12. }\text{A wire of length }50\ \text{m is cut into two pieces. One piece is bent in} \\ \text{the shape of a square }  \text{and the other in the shape of a circle. What should be the} \\ \text{length of each piece so that the combined area}  \text{of the two is minimum? ISC 2014} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let }x\text{ be radius of circle and }y\text{ side of square}
\displaystyle \text{Circumference of circle }=2\pi x
\displaystyle \text{Perimeter of square }=4y
\displaystyle \text{Given, } 2\pi x+4y=50 \qquad \cdots(i)
\displaystyle \text{Let } A \text{ be the sum of the areas of circle and square.}
\displaystyle A=\pi x^{2}+y^{2}
\displaystyle =\pi x^{2}+\left(\frac{50-2\pi x}{4}\right)^{2} \quad \text{[from Eq. (i)]}
\displaystyle \text{On differentiating both sides w.r.t. } x, \text{ we get}
\displaystyle \frac{dA}{dx}=2\pi x+2\left(\frac{50-2\pi x}{4}\right)\left(\frac{-2\pi}{4}\right)
\displaystyle =2\pi x-\frac{(50-2\pi x)\pi}{4}
\displaystyle =2\pi x-\frac{50\pi}{4}+\frac{2\pi^{2}x}{4}
\displaystyle \text{For maxima or minima, put } \frac{dA}{dx}=0
\displaystyle \Rightarrow 2\pi x-\frac{50\pi}{4}+\frac{\pi^{2}x}{2}=0
\displaystyle \Rightarrow 8\pi x-50\pi+2\pi^{2}x=0
\displaystyle \Rightarrow 8x-50+2\pi x=0
\displaystyle \Rightarrow \pi x+4x=25
\displaystyle \therefore x=\frac{25}{4+\pi}
\displaystyle \text{Now, } \frac{d^{2}A}{dx^{2}}=2\pi+\frac{\pi^{2}}{2}>0
\displaystyle \text{So, } A \text{ is minimum when } x=\frac{25}{4+\pi}
\displaystyle \text{Then, } 4y=50-\frac{2\pi \times 25}{4+\pi} \quad \text{[from Eq. (i)]}
\displaystyle \Rightarrow 4y=\frac{200+50\pi-50\pi}{4+\pi}
\displaystyle \therefore y=\frac{50}{4+\pi}
\displaystyle \text{Hence, } A \text{ is minimum when } y=2x
\displaystyle \text{i.e. side of square }=\text{ diameter of circle.}
\displaystyle \text{Length of wire bent in the shape of square }=\frac{200}{4+\pi}\text{ m}
\displaystyle \text{Length of wire bent in the shape of circle }=\frac{50\pi}{4+\pi}\text{ m}

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