Monotonicity, Maxima and Minima
8.0 Second Derivative Test
8.0 Second Derivative Test
Let the given function $y=f(x)$ is twice differentiable. To find the points of local maxima and local minima, following steps should be followed:
1. Differentiate the given function and find $f'(x)$.
2. Put $f'(x)=0$ and find the value of $x$. Assume it to be $x=c$.
3. Again differentiate the function and find $f''(x)$. Put the value of $c$ and find the value of $f''(c)$.
4. If $f''(c)>0$, then $x=c$ is the point of local minima.
5. If $f''(c)<0$, then $x=c$ is the point of local maxima.
6. If $f''(c)=0$, then $x=c$ is neither the point of local maxima or local minima.
Let us solve some questions based on application of maxima and minima
Question 14. Find two numbers whose sum is 12 and the sum of whose squares in minimum.
Solution: Let us assume one number be $x$. Therefore, the other number should be $12-x$ as it is given that the sum of two numbers is $12$. Let $S(x)$ be the sum of the squares of these numbers. We can write $$\begin{equation} \begin{aligned} S(x) = {x^2} + {(12 - x)^2} \\ S(x) = {x^2} + 144 + {x^2} - 24x \\ S(x) = 2{x^2} - 24x + 144 \\\end{aligned} \end{equation} $$ We have to minimize $S(x)$, so we apply the rules to find local minima of the function. Differentiating $S(x)$ with respect to $x$, we get $$\begin{equation} \begin{aligned} S'(x) = 4x - 24 = 4(x - 6) = 0 \\ \Rightarrow x = 6 \\\end{aligned} \end{equation} $$ Therefore, $x=6$ is the critical point. So, Using Second derivative tesr, again differentiate it, we get $$\begin{equation} \begin{aligned} S''(x) = 4 \\ \Rightarrow S''(6) = 4 > 0 \\\end{aligned} \end{equation} $$ Since, $f''(6)>0$, we can say that $x=6$ is the point of local minima. Hence, the sum of squares of numbers is minimum when two numbers are $x=6$ and $12-x-12-6=6$.
Question 15. If a right circular cylinder is inscribed in a given cone. Find the dimensions of the cylinder such that its volume is maximum.
Solution: Let $r$ be the radius of the cylinder and $h$ be its height. Volume $(V)$ of cylinder is $$V = \pi {r^2}h$$ Let us
assume the radius of cone be $R$ and height of cone be $H$. From figure, we can say that $$\Delta ABC \sim \Delta DEC$$ Therefore, using similar triangle property, we can write $$\begin{equation} \begin{aligned} \frac{{AB}}{{DE}} = \frac{{BC}}{{EC}} \\ \frac{H}{h} = \frac{R}{{R - r}} \\ \Rightarrow h = \frac{H}{R}(R - r) \\\end{aligned} \end{equation} $$ Put the value of $h$ in the volume of cylinder, we get $$\begin{equation} \begin{aligned} V(r) = \pi {r^2}\frac{H}{R}(R - r) \\ V(r) = \frac{{\pi H}}{R}(R{r^2} - {r^3}) \\\end{aligned} \end{equation} $$ In the question, it is given that we have to maximize the volume of cylinder. So, the dimensions of cone are constant i.e., $R$ and $H$ are constants. To maximize the volume, we use the method of finding local maxima. Differentiating $V(r)$ with respect to $r$, we get $$\begin{equation} \begin{aligned} V'(r) = \frac{{\pi H}}{R}(2Rr - 3{r^2}) \\ V'(r) = \frac{{\pi Hr}}{R}(2R - 3r) = 0 \\ \Rightarrow r = 0{\text{ and }}r = \frac{{2R}}{3} \\\end{aligned} \end{equation} $$ Since, $r \ne 0$, Using Second derivative test, again differentiating it, and put $r = \frac{{2R}}{3}$, we get $$\begin{equation} \begin{aligned} V'(r) = \frac{{\pi H}}{R}(2Rr - 3{r^2}) \\ V''(r) = \frac{{\pi H}}{R}(2R - 6r) \\ V''(r) = \frac{{2\pi H}}{R}(R - 3r) \\ \therefore V''\left( {\frac{{2R}}{3}} \right) = \frac{{2\pi H}}{R}(R - 3 \times \frac{{2R}}{3}) = - 2\pi H < 0 \\\end{aligned} \end{equation} $$ therefore, we can say that the Volume of cylinder is maximum at $$r = \frac{{2R}}{3}$$ and $$\begin{equation} \begin{aligned} h = \frac{H}{R}(R - \frac{{2R}}{3}) \\ h = \frac{H}{R}\left( {\frac{{3R - 2R}}{3}} \right) = \frac{H}{3} \\\end{aligned} \end{equation} $$
assume the radius of cone be $R$ and height of cone be $H$. From figure, we can say that $$\Delta ABC \sim \Delta DEC$$ Therefore, using similar triangle property, we can write $$\begin{equation} \begin{aligned} \frac{{AB}}{{DE}} = \frac{{BC}}{{EC}} \\ \frac{H}{h} = \frac{R}{{R - r}} \\ \Rightarrow h = \frac{H}{R}(R - r) \\\end{aligned} \end{equation} $$ Put the value of $h$ in the volume of cylinder, we get $$\begin{equation} \begin{aligned} V(r) = \pi {r^2}\frac{H}{R}(R - r) \\ V(r) = \frac{{\pi H}}{R}(R{r^2} - {r^3}) \\\end{aligned} \end{equation} $$ In the question, it is given that we have to maximize the volume of cylinder. So, the dimensions of cone are constant i.e., $R$ and $H$ are constants. To maximize the volume, we use the method of finding local maxima. Differentiating $V(r)$ with respect to $r$, we get $$\begin{equation} \begin{aligned} V'(r) = \frac{{\pi H}}{R}(2Rr - 3{r^2}) \\ V'(r) = \frac{{\pi Hr}}{R}(2R - 3r) = 0 \\ \Rightarrow r = 0{\text{ and }}r = \frac{{2R}}{3} \\\end{aligned} \end{equation} $$ Since, $r \ne 0$, Using Second derivative test, again differentiating it, and put $r = \frac{{2R}}{3}$, we get $$\begin{equation} \begin{aligned} V'(r) = \frac{{\pi H}}{R}(2Rr - 3{r^2}) \\ V''(r) = \frac{{\pi H}}{R}(2R - 6r) \\ V''(r) = \frac{{2\pi H}}{R}(R - 3r) \\ \therefore V''\left( {\frac{{2R}}{3}} \right) = \frac{{2\pi H}}{R}(R - 3 \times \frac{{2R}}{3}) = - 2\pi H < 0 \\\end{aligned} \end{equation} $$ therefore, we can say that the Volume of cylinder is maximum at $$r = \frac{{2R}}{3}$$ and $$\begin{equation} \begin{aligned} h = \frac{H}{R}(R - \frac{{2R}}{3}) \\ h = \frac{H}{R}\left( {\frac{{3R - 2R}}{3}} \right) = \frac{H}{3} \\\end{aligned} \end{equation} $$