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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES4 Marks
Question
Shreya got a rectangular parallelepiped shaped box and spherical ball inside it as return gift. Sides of the box are x, 2x, and $\frac{\text{x}}{3},$ while radius of the ball is r.

Based on the above information, answer the following questions.
  1. If S represents the sum of volume of parallelepiped and sphere, then Scan be written as.
  1. $\frac{4\text{x}^3}{3}+\frac{2}{2}\pi\text{r}^2$
  2. $\frac{2\text{x}^2}{3}+\frac{4}{3}\pi\text{r}^2$
  3. $\frac{2\text{x}^3}{3}+\frac{4}{3}\pi\text{r}^3$
  4. $\frac{2}{3}\text{x}+\frac{4}{3}\pi\text{r}$
  1. If sum of the surface areas of box and ball are given to be constant $k^2$ then x is equal to.
  1. $\sqrt{\frac{\text{k}^2-4\pi\text{r}^2}{6}}$
  2. $\sqrt{\frac{\text{k}^2-4\pi\text{r}}{6}}$
  3. $\sqrt{\frac{\text{k}^2-4\pi}{6}}$
  4. $\text{None of these}$
  1. The radius of the ball, when Sis minimum, is.
  1. $\sqrt{\frac{\text{k}^2}{54+\pi}}$
  2. $\sqrt{\frac{\text{k}^2}{54+4}}$
  3. $\sqrt{\frac{\text{k}^2}{64+3\pi}}$
  4. $\sqrt{\frac{\text{k}^2}{4\pi+3}}$
  1. Relation between length of the box and radius of the ball can be represented as.
  1. $\text{x} = \frac{2}{\text{r}}$
  2. $\text{x}=\frac{\text{r}}{2}$
  3. $\text{x}=\frac{2}{\text{r}}$
  4. $\text{x}=3\text{r}$
  1. Minimum value of S is.
  1. $\frac{\text{k}^2}{2(3\pi+54)^\frac{2}{3}}$
  2. $\frac{\text{k}}{2(3\pi+54)^\frac{3}{2}}$
  3. $\frac{\text{k}^3}{2(4\pi+54)^\frac{1}{2}}$
  4. $\text{None of these}$
✓

Answer

  1. (c) $\frac{2\text{x}^3}{3}+\frac{4}{3}\pi\text{r}^3$
Solution:
Let S be the sum of volume of parallelepiped and sphere, then
$\text{S}=\text{x}(2\text{x})\Big(\frac{\text{x}}{3}\Big)+\frac{4}{3}\pi\text{r}^3=\frac{2\text{x}^3}{3}+\frac{4}{3}\pi\text{r}^3$
  1. (a) $\sqrt{\frac{\text{k}^2-4\pi\text{r}^2}{6}}$
Solution:
Since, sum of surface area of box and sphere is given to be constant $k^2.$
$\therefore2\Big(\text{x}\times\text{2x}+\text{2x}\times\frac{\text{x}}{3}+\frac{\text{x}}{3}\times\text{x}\Big)+4\pi\text{r}^2=\text{k}^2$
$\Rightarrow6\text{x}^2+4\pi\text{r}^2=\text{k}^2$
$\Rightarrow\text{x}^2=\frac{\text{k}^2-4\pi\text{r}}{6}\Rightarrow\text{x}=\sqrt{\frac{\text{k}^2-4\pi\text{r}}{6}}$
  1. (b) $\sqrt{\frac{\text{k}^2}{54+4}}$
Solution:
From (1) and (2), we get
$\text{s}=\frac{2}{3}\bigg(\frac{\text{k}^2-4\pi\text{r}^2}{6}\bigg)^\frac{3}{2}+\frac{4}{3}\pi\text{r}^3$
$=\frac{2}{3\times6\sqrt{6}}(\text{k}^2-4\pi\text{r}^2)^\frac{3}{2}+\frac{4}{3}\pi\text{r}^3$
$\Rightarrow\frac{\text{ds}}{\text{dr}}=\frac{1}{9\sqrt{6}}\frac{3}{2}(\text{k}^2-4\pi\text{r}^2)^\frac{1}{2}-8\pi\text{r}+4\pi\text{r}^2$
$=4\pi\text{r}\bigg[\text{r}\frac{1}{3\sqrt{6}}\sqrt{\text{k}^2-4\pi\text{r}^2}\bigg]$
For maximum/minimum, $\frac{\text{ds}}{\text{dr}}=0$
$\Rightarrow\frac{4\pi\text{r}}{3\sqrt{6}}\sqrt{\text{k}-4\pi\text{r}^2}=-4\pi\text{r}^2$
$\text{k}^2-4\pi\text{r}^2=54\text{r}^2$
$\Rightarrow\text{r}^2=\frac{\text{k}^2}{54+4\pi}\Rightarrow\text{r}=\sqrt{\frac{\text{k}^2}{54+4\pi}}$
  1. (d) $\text{x}=3\text{r}$
Solution:
$\text{Since},\text{x}^2=\frac{\text{k}-4\pi\text{r}^2}{6}=\frac{1}{6}\bigg[\text{k}^2-4\pi\Big(\frac{\text{k}^2}{54+4\pi}\Big)\bigg]$
[From (2) and (3)
$=\frac{9\text{k}^2}{54+4\pi}=9\Big(\frac{\text{k}^2}{54+4\pi}\Big)=9\text{r}^2=(3\text{r})^2$
⇒ x = 3r
  1. (c) $\frac{\text{k}^3}{2(4\pi+54)^\frac{1}{2}}$
Solution:
Minimum value of S is given by
$\frac{2}{3}(3\text{r})^3+\frac{4}{3}\pi\text{r}^3$
$=18\text{r}^3+\frac{4}{3}\pi\text{r}^3=\bigg(18+\frac{4}{3}\pi\bigg)\text{r}^3$
$\bigg(18+\frac{4}{3}\pi\bigg)\text{r}^3\bigg(\frac{\text{k}^2}{54+4\pi}\bigg)^\frac{3}{2}$
$\frac{1}{3}\frac{\text{k}^3}{(54+4\pi)^\frac{1}{2}}$

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