NCERT Solutions · Class 12 Mathematics Application of Derivatives

\(\displaystyle f\) is continuous on the closed interval \(\displaystyle [0,\pi]\), so by the extreme value theorem the absolute extrema exist and occur either at an interior critical point or at an endpoint.\[f'(x)=2\cos x(-\sin x)+\cos x=\cos x\,(1-2\sin x). \]Set \(\displaystyle f'(x)=0\) on \(\displaystyle (0,\pi)\): \[\cos x=0\Rightarrow x=\frac{\pi}{2};\qquad \sin x=\frac{1}{2}\Rightarrow x=\frac{\pi}{6}\ \text{or}\ x=\frac{5\pi}{6}. \] (Both solutions of \(\displaystyle \sin x=\tfrac12\) lie in \(\displaystyle [0,\pi]\) — dropping \(\displaystyle \tfrac{5\pi}{6}\) is the usual slip.)Evaluate \(\displaystyle f\) at the critical points and endpoints, using \(\displaystyle \cos^{2}\dfrac{\pi}{6}=\cos^{2}\dfrac{5\pi}{6}=\dfrac{3}{4}\): \[f(0)=1+0=1,\qquad f\!\left(\frac{\pi}{6}\right)=\frac{3}{4}+\frac{1}{2}=\frac{5}{4}, \] \[f\!\left(\frac{\pi}{2}\right)=0+1=1,\qquad f\!\left(\frac{5\pi}{6}\right)=\frac{3}{4}+\frac{1}{2}=\frac{5}{4},\qquad f(\pi)=1+0=1. \]Absolute maximum value \(\displaystyle =\dfrac{5}{4}\), attained at \(\displaystyle x=\dfrac{\pi}{6}\) and \(\displaystyle x=\dfrac{5\pi}{6}\).Absolute minimum value \(\displaystyle =1\), attained at \(\displaystyle x=0,\ \dfrac{\pi}{2},\ \pi\).

Let the cone have altitude \(\displaystyle h\) and base radius \(\displaystyle R\), inscribed in the sphere of radius \(\displaystyle r\) so that its axis passes through the centre \(\displaystyle O\) of the sphere (the apex must lie on the sphere, and the base circle must lie on the sphere).The distance from \(\displaystyle O\) to the plane of the base is \(\displaystyle h-r\), and the radius drawn to a point of the base circle gives, by Pythagoras, \[R^{2}=r^{2}-(h-r)^{2}=r^{2}-h^{2}+2rh-r^{2}=2rh-h^{2}. \] (This formula is also correct when the centre lies outside the cone, since only \(\displaystyle (h-r)^{2}\) enters.)Hence the volume, as a function of \(\displaystyle h\) alone, is \[V(h)=\frac{1}{3}\pi R^{2}h=\frac{\pi}{3}\left(2rh^{2}-h^{3}\right),\qquad 0<h<2r. \]Differentiate: \[V'(h)=\frac{\pi}{3}\left(4rh-3h^{2}\right)=\frac{\pi}{3}h(4r-3h). \] Since \(\displaystyle h>0\), \(\displaystyle V'(h)=0\Rightarrow h=\dfrac{4r}{3}\) (which indeed satisfies \(\displaystyle 0<h<2r\)).Second derivative test: \[V''(h)=\frac{\pi}{3}(4r-6h),\qquad V''\!\left(\frac{4r}{3}\right)=\frac{\pi}{3}\left(4r-8r\right)=-\frac{4\pi r}{3}<0. \]So the volume is maximum when \(\displaystyle h=\dfrac{4r}{3}\); that is, the altitude of the cone of maximum volume inscribed in a sphere of radius \(\displaystyle r\) is \(\displaystyle \dfrac{4r}{3}\).

The tool is Lagrange's Mean Value Theorem: if \(\displaystyle g\) is continuous on \(\displaystyle [p,q]\) and differentiable on \(\displaystyle (p,q)\), then there exists \(\displaystyle c\in(p,q)\) with \(\displaystyle g(q)-g(p)=g'(c)(q-p)\).Recall the definition to be proved: \(\displaystyle f\) is increasing on \(\displaystyle (a,b)\) means that for all \(\displaystyle x_{1},x_{2}\in(a,b)\) with \(\displaystyle x_{1}<x_{2}\) we have \(\displaystyle f(x_{1})<f(x_{2})\).So let \(\displaystyle x_{1},x_{2}\in(a,b)\) be arbitrary with \(\displaystyle x_{1}<x_{2}\).The hypothesis \(\displaystyle f'(x)>0\) for all \(\displaystyle x\in(a,b)\) means in particular that \(\displaystyle f\) is differentiable at every point of \(\displaystyle (a,b)\), hence continuous there. Since \(\displaystyle [x_{1},x_{2}]\subset(a,b)\), the function \(\displaystyle f\) is continuous on \(\displaystyle [x_{1},x_{2}]\) and differentiable on \(\displaystyle (x_{1},x_{2})\). The Mean Value Theorem therefore applies on \(\displaystyle [x_{1},x_{2}]\): there exists \(\displaystyle c\in(x_{1},x_{2})\) such that \[f(x_{2})-f(x_{1})=f'(c)\,(x_{2}-x_{1}). \]Now \(\displaystyle c\in(x_{1},x_{2})\subset(a,b)\), so \(\displaystyle f'(c)>0\) by hypothesis, and \(\displaystyle x_{2}-x_{1}>0\) because \(\displaystyle x_{1}<x_{2}\). A product of two positive numbers is positive, so \[f(x_{2})-f(x_{1})>0,\qquad\text{i.e.}\qquad f(x_{1})<f(x_{2}). \]Since \(\displaystyle x_{1}<x_{2}\) were arbitrary points of \(\displaystyle (a,b)\), \(\displaystyle f\) is an increasing function on \(\displaystyle (a,b)\).

Let the inscribed cylinder have height \(\displaystyle h\) and base radius \(\displaystyle \rho\). Its axis passes through the centre of the sphere, so the centre is at height \(\displaystyle \dfrac{h}{2}\) above the plane of the lower base.The radius drawn from the centre to a point of the rim gives, by Pythagoras, \[R^{2}=\rho^{2}+\left(\frac{h}{2}\right)^{2}\ \Rightarrow\ \rho^{2}=R^{2}-\frac{h^{2}}{4}. \]Hence the volume in terms of \(\displaystyle h\) alone is \[V(h)=\pi\rho^{2}h=\pi\left(R^{2}h-\frac{h^{3}}{4}\right),\qquad 0<h<2R. \]Differentiate: \[V'(h)=\pi\left(R^{2}-\frac{3h^{2}}{4}\right)=0\ \Rightarrow\ h^{2}=\frac{4R^{2}}{3}\ \Rightarrow\ h=\frac{2R}{\sqrt{3}} \] (taking the positive root, since \(\displaystyle h\) is a length).Second derivative test: \[V''(h)=-\frac{3\pi h}{2}<0\quad\text{for }h>0, \] so the volume is maximum at \(\displaystyle h=\dfrac{2R}{\sqrt{3}}\), as required.Maximum volume: with \(\displaystyle h=\dfrac{2R}{\sqrt3}\), \(\displaystyle h^{3}=\dfrac{8R^{3}}{3\sqrt3}\), so \[V=\pi\left(R^{2}\cdot\frac{2R}{\sqrt3}-\frac{1}{4}\cdot\frac{8R^{3}}{3\sqrt3}\right)=\pi\left(\frac{2R^{3}}{\sqrt3}-\frac{2R^{3}}{3\sqrt3}\right)=\pi\cdot\frac{6R^{3}-2R^{3}}{3\sqrt3}=\frac{4\pi R^{3}}{3\sqrt3}. \]So the height is \(\displaystyle \dfrac{2R}{\sqrt3}\) and the maximum volume is \(\displaystyle \dfrac{4\pi R^{3}}{3\sqrt3}=\dfrac{4\sqrt{3}\,\pi R^{3}}{9}\).

The cone has height \(\displaystyle h\) and semi-vertical angle \(\displaystyle \alpha\), so its base radius is \(\displaystyle h\tan\alpha\). Let the inscribed cylinder have radius \(\displaystyle x\) and height \(\displaystyle H\), standing on the base of the cone with its axis along the cone's axis.The top rim of the cylinder lies on the slant surface, at height \(\displaystyle H\) above the base, i.e. at distance \(\displaystyle h-H\) below the apex. The cross-section of the cone at that level is a circle of radius \(\displaystyle (h-H)\tan\alpha\) (similar triangles from the apex), so \[x=(h-H)\tan\alpha\ \Rightarrow\ H=h-x\cot\alpha. \]Hence the volume of the cylinder, in terms of \(\displaystyle x\) alone, is \[V(x)=\pi x^{2}H=\pi x^{2}\left(h-x\cot\alpha\right)=\pi\left(hx^{2}-x^{3}\cot\alpha\right),\qquad 0<x<h\tan\alpha. \]Differentiate: \[V'(x)=\pi\left(2hx-3x^{2}\cot\alpha\right)=\pi x\left(2h-3x\cot\alpha\right). \] Since \(\displaystyle x>0\), \[V'(x)=0\ \Rightarrow\ 3x\cot\alpha=2h\ \Rightarrow\ x=\frac{2h\tan\alpha}{3}. \]Second derivative test: \[V''(x)=\pi\left(2h-6x\cot\alpha\right),\qquad V''\!\left(\frac{2h\tan\alpha}{3}\right)=\pi\left(2h-4h\right)=-2\pi h<0, \] so this \(\displaystyle x\) maximises \(\displaystyle V\).The corresponding height is \[H=h-\frac{2h\tan\alpha}{3}\cot\alpha=h-\frac{2h}{3}=\frac{h}{3}, \] which is one-third the height of the cone, as claimed.The greatest volume is \[V=\pi\left(\frac{2h\tan\alpha}{3}\right)^{2}\cdot\frac{h}{3}=\pi\cdot\frac{4h^{2}\tan^{2}\alpha}{9}\cdot\frac{h}{3}=\frac{4}{27}\pi h^{3}\tan^{2}\alpha. \]

The wheat fills the tank as a cylinder of fixed radius \(\displaystyle r=10\) m and variable depth \(\displaystyle d\), so \[V=\pi r^{2}d=\pi(10)^{2}d=100\pi d. \]The radius is constant, so differentiating with respect to time \(\displaystyle t\) gives \[\frac{dV}{dt}=100\pi\,\frac{dd}{dt}. \]Given \(\displaystyle \dfrac{dV}{dt}=314\ \mathrm{m}^{3}/\mathrm{h}\), \[\frac{dd}{dt}=\frac{314}{100\pi}=\frac{314}{100\times 3.14}=\frac{314}{314}=1. \]The depth of the wheat is increasing at \(\displaystyle 1\ \mathrm{m}/\mathrm{h}\).Hence the correct option is (A) $\displaystyle 1$ m/h.