Q108 / 160 · AP EAMCET 2023 (16 May Shift 1) · Mathematics
Given that the solid obtained by rotating a rectangle about one of its side is a cylinder. If the perimeter of a rectangle is $48 \mathrm{~cm}$ and the volume of the cylinder formed by rotating it is maximum, then the dimensions of that rectangle is
- A14,10
- B20,4
- C18,6
- Correct answerD8,16
Solution
Let $l$ and $b$ be length and width of the rectangle.
$\because$ The perimeter is $48 \mathrm{~cm}$.
$\therefore \quad 2(l+b)=48$
$\Rightarrow l+b=24$ ...(i)
Volume of the cylinder form by rotation is $v=\pi l^2 b$ $\Rightarrow \quad v=\pi l^2(24-l)=24 \pi l^2-\pi l^3$ $\Rightarrow \frac{d v}{d l}=48 \pi l-3 \pi l^2$ ...(ii) For maxima or minima : $\frac{d v}{d l}=0 \Rightarrow 48 \pi l-3 \pi l^2=0$ $\Rightarrow \quad 3 \pi\left(16 l-l^2\right)=0 \Rightarrow l=0,16$ Now, $\frac{d^2 v}{d l^2}=48 \pi-6 \pi l$ At $l=0 \Rightarrow \frac{d^2 v}{d l^2}=48 \pi>0$ So $l=0$ is point of minima : At $l=16 \Rightarrow \frac{d v^2}{d l^2}=48 \pi-96 \pi=-48 \pi < 0$ $\therefore l=16$ is point of manima. So, volume is maximum when $l=16$. From eq. (i) $\Rightarrow 16+b=24 \Rightarrow b=24-16=8$ So, dimension of rectangle is 8,16 .
Volume of the cylinder form by rotation is $v=\pi l^2 b$ $\Rightarrow \quad v=\pi l^2(24-l)=24 \pi l^2-\pi l^3$ $\Rightarrow \frac{d v}{d l}=48 \pi l-3 \pi l^2$ ...(ii) For maxima or minima : $\frac{d v}{d l}=0 \Rightarrow 48 \pi l-3 \pi l^2=0$ $\Rightarrow \quad 3 \pi\left(16 l-l^2\right)=0 \Rightarrow l=0,16$ Now, $\frac{d^2 v}{d l^2}=48 \pi-6 \pi l$ At $l=0 \Rightarrow \frac{d^2 v}{d l^2}=48 \pi>0$ So $l=0$ is point of minima : At $l=16 \Rightarrow \frac{d v^2}{d l^2}=48 \pi-96 \pi=-48 \pi < 0$ $\therefore l=16$ is point of manima. So, volume is maximum when $l=16$. From eq. (i) $\Rightarrow 16+b=24 \Rightarrow b=24-16=8$ So, dimension of rectangle is 8,16 .
Hence, the correct answer is option: 'D'.