RD Sharma Solutions Class 12 Mathematics Chapter 17 MCQ

Class 12 RD Sharma chapter 17 exercise MCQ solution is a sought-after book that is used by almost all students in class 12. The RD Sharma class 12th exercise MCQ covers questions from the entire NCERT maths book and provides intensive knowledge and essential information for all concepts and chapters. RD Sharma Solutions It is designed to aid students to perform well in any school and entrance exam and expand their knowledge on the maths subject.

RD Sharma Class 12 Solutions Chapter 17 MCQ Maxima and Minima - Other Exercise

• Chapter 17 - Maxima and Minima - Ex 17.1
• Chapter 17 - Maxima and Minima - Ex 17.2
• Chapter 17 - Maxima and Minima - Ex 17.3
• Chapter 17 - Maxima and Minima - Ex 17.4
• Chapter 17 - Maxima and Minima - Ex 17.5
• Chapter 17 - Maxima and Minima - Ex FBQ
• Chapter 17 - Maxima and Minima - Ex CSBQ
• Chapter 17 - Maxima and Minima- Ex VSA

Maxima and Minima exercise MCQ question 19

Answer: option(d) $\frac{2}{3}$
Hint: For local maxima or minima, we must have $f^{\prime }(x)=0$.
Given:$h=R+\sqrt{R^2-r^2}$
Solution:
We have,
Solution:

$AB=2r$
$OC=r$
$CD=R+h$
We have,
$h=R+\sqrt{R^2-r^2}$
$h-R=\sqrt{R^2-r^2}$
Squaring on both sides,
$(h-R{)}^{=}(\sqrt{R^2-r^2}{)}^2$
$h^2+R^2-2hR=R^2-r^2$
$r^2=2hR-h^2$ …(i)
Now,
$V=\frac{1}{3}\pi r^{2}h$
$V=\frac{\pi }{3}(2hR-h^2)h$ …using equation(i)
$\begin{array}{rl}& V=\frac{\pi }{3}(2h^{2}R-h^3)\\ & \frac{dV}{dh}=\frac{\pi }{3}(4hR-3h^2)\end{array}$
For maxima and minima $\frac{dV}{dh}=0$
$\begin{array}{rl}& \Rightarrow \frac{\pi }{3}(4hR-3h^2)=0\\ & \Rightarrow 4hR-3h^2=0\\ & \Rightarrow 3h^2=4hR\\ & \Rightarrow h=\frac{4R}{3}\end{array}$
Now,
$\frac{d^{2}V}{d{h}^2}=\frac{\pi }{3}(4R-6h)=\frac{\pi }{3}(4R-6\times \frac{4R}{3})=\frac{\pi }{3}4R(1-\frac{6}{3})=\frac{\pi }{3}4R\left(\frac{-3}{3}\right)=-\frac{4\pi R}{3}<0$
Volume is maximum, when $h=\frac{4R}{3}$
$h=\frac{2(2R)}{3}$
$\frac{h}{2R}=\frac{2}{3}$