NCERT Solutions · Class 12 Mathematics Application of Derivatives

Let the can have base radius \(\displaystyle r\) cm and height \(\displaystyle h\) cm. The volume is fixed: \[\pi r^{2}h=100\ \Longrightarrow\ h=\frac{100}{\pi r^{2}}. \] The can is closed, so it has two circular ends and the curved surface: \[S=2\pi r^{2}+2\pi rh=2\pi r^{2}+2\pi r\cdot\frac{100}{\pi r^{2}}=2\pi r^{2}+\frac{200}{r},\qquad r>0. \] \[\frac{dS}{dr}=4\pi r-\frac{200}{r^{2}}=0\ \Longrightarrow\ 4\pi r^{3}=200\ \Longrightarrow\ r^{3}=\frac{50}{\pi}\ \Longrightarrow\ r=\left(\frac{50}{\pi}\right)^{1/3}. \] \[\frac{d^{2}S}{dr^{2}}=4\pi+\frac{400}{r^{3}}>0\quad\text{for all }r>0, \] so this \(\displaystyle r\) gives the minimum surface area. For the height, use \(\displaystyle \pi r^{3}=50\): \[h=\frac{100}{\pi r^{2}}=\frac{100r}{\pi r^{3}}=\frac{100r}{50}=2r. \] The can of least surface area has \[r=\left(\frac{50}{\pi}\right)^{1/3}\ \text{cm}\approx 2.52\ \text{cm},\qquad h=2\left(\frac{50}{\pi}\right)^{1/3}\ \text{cm}\approx 5.03\ \text{cm}, \] that is, its height equals the diameter of its base.

Let one piece have length \(\displaystyle x\) m and be bent into the square; the other piece then has length \(\displaystyle (28-x)\) m and is bent into the circle, where \(\displaystyle 0\le x\le 28\). Square: perimeter \(\displaystyle x\Rightarrow\) side \(\displaystyle \frac{x}{4}\Rightarrow\) area \(\displaystyle \frac{x^{2}}{16}\). Circle: circumference \(\displaystyle 28-x=2\pi r\Rightarrow r=\frac{28-x}{2\pi}\Rightarrow\) area \(\displaystyle \pi r^{2}=\pi\cdot\frac{(28-x)^{2}}{4\pi^{2}}=\frac{(28-x)^{2}}{4\pi}\). Combined area: \[A(x)=\frac{x^{2}}{16}+\frac{(28-x)^{2}}{4\pi}. \] \[A'(x)=\frac{x}{8}+\frac{2(28-x)(-1)}{4\pi}=\frac{x}{8}-\frac{28-x}{2\pi}=0\ \Longrightarrow\ 2\pi x=8(28-x)\ \Longrightarrow\ \pi x+4x=112\ \Longrightarrow\ x=\frac{112}{\pi+4}. \] \[A''(x)=\frac{1}{8}+\frac{1}{2\pi}>0, \] so this \(\displaystyle x\) gives the minimum combined area. The second piece is \[28-\frac{112}{\pi+4}=\frac{28(\pi+4)-112}{\pi+4}=\frac{28\pi}{\pi+4}. \] The wire should be cut into a piece of \(\displaystyle \dfrac{112}{\pi+4}\) m \(\displaystyle (\approx 15.68\) m, bent into the square\(\displaystyle )\) and a piece of \(\displaystyle \dfrac{28\pi}{\pi+4}\) m \(\displaystyle (\approx 12.32\) m, bent into the circle\(\displaystyle )\).

Let the sphere have centre \(\displaystyle O\) and radius \(\displaystyle R\), and let the inscribed cone have base radius \(\displaystyle r\) and height \(\displaystyle h\). Work in the axial cross-section: if the centre \(\displaystyle O\) lies at a distance \(\displaystyle x\) from the plane of the base \(\displaystyle (0\le x<R)\), then joining \(\displaystyle O\) to a point of the base circle gives a right triangle with hypotenuse \(\displaystyle R\), so by Pythagoras \[r^{2}=R^{2}-x^{2},\qquad h=R+x. \]\[V(x)=\frac{1}{3}\pi r^{2}h=\frac{\pi}{3}\left(R^{2}-x^{2}\right)(R+x),\qquad 0\le x<R. \] \[\frac{dV}{dx}=\frac{\pi}{3}\left[-2x(R+x)+\left(R^{2}-x^{2}\right)\right]=\frac{\pi}{3}\left(R^{2}-2Rx-3x^{2}\right)=\frac{\pi}{3}(R-3x)(R+x). \] Since \(\displaystyle R+x>0\), the only admissible root is \(\displaystyle x=\frac{R}{3}\). \[\frac{d^{2}V}{dx^{2}}=\frac{\pi}{3}\left(-2R-6x\right)<0, \] so \(\displaystyle x=\frac{R}{3}\) gives the maximum volume. Then \[h=R+\frac{R}{3}=\frac{4R}{3},\qquad r^{2}=R^{2}-\frac{R^{2}}{9}=\frac{8R^{2}}{9}, \] \[V_{\max}=\frac{1}{3}\pi\cdot\frac{8R^{2}}{9}\cdot\frac{4R}{3}=\frac{32\pi R^{3}}{81}. \] Comparing with the volume of the sphere, \(\displaystyle \frac{4}{3}\pi R^{3}\): \[\frac{V_{\max}}{\dfrac{4}{3}\pi R^{3}}=\frac{32\pi R^{3}}{81}\times\frac{3}{4\pi R^{3}}=\frac{8}{27}. \] Hence the largest cone inscribed in a sphere of radius \(\displaystyle R\) has volume \(\displaystyle \frac{8}{27}\) of the volume of the sphere.

Let the cone have base radius \(\displaystyle r\), altitude \(\displaystyle h\) and slant height \(\displaystyle l=\sqrt{r^{2}+h^{2}}\), with the volume \(\displaystyle V\) given (constant): \[V=\frac{1}{3}\pi r^{2}h\ \Longrightarrow\ h=\frac{3V}{\pi r^{2}}. \] The curved surface is \(\displaystyle C=\pi r l=\pi r\sqrt{r^{2}+h^{2}}\). Since \(\displaystyle C>0\), minimising \(\displaystyle C\) is the same as minimising \(\displaystyle C^{2}\), which avoids the square root: \[C^{2}=\pi^{2}r^{2}\left(r^{2}+h^{2}\right)=\pi^{2}r^{2}\left(r^{2}+\frac{9V^{2}}{\pi^{2}r^{4}}\right)=\pi^{2}r^{4}+\frac{9V^{2}}{r^{2}}. \] \[\frac{d\left(C^{2}\right)}{dr}=4\pi^{2}r^{3}-\frac{18V^{2}}{r^{3}}=0\ \Longrightarrow\ 4\pi^{2}r^{6}=18V^{2}\ \Longrightarrow\ V^{2}=\frac{2\pi^{2}r^{6}}{9}\ \Longrightarrow\ V=\frac{\sqrt{2}\,\pi r^{3}}{3}. \] \[\frac{d^{2}\left(C^{2}\right)}{dr^{2}}=12\pi^{2}r^{2}+\frac{54V^{2}}{r^{4}}>0, \] so this radius gives the least curved surface. Equating the two expressions for the same volume, \[\frac{1}{3}\pi r^{2}h=\frac{\sqrt{2}}{3}\pi r^{3}\ \Longrightarrow\ h=\sqrt{2}\,r. \] Hence the right circular cone of least curved surface with a given volume has altitude \(\displaystyle \sqrt{2}\) times the radius of its base.

Let the slant height \(\displaystyle l\) be given (constant) and let \(\displaystyle \theta\) be the semi-vertical angle, \(\displaystyle 0<\theta<\frac{\pi}{2}\). Then the radius and height of the cone are \[r=l\sin\theta,\qquad h=l\cos\theta, \] and the volume, expressed in the single variable \(\displaystyle \theta\), is \[V(\theta)=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi l^{3}\sin^{2}\theta\cos\theta. \] \[\frac{dV}{d\theta}=\frac{\pi l^{3}}{3}\left(2\sin\theta\cos^{2}\theta-\sin^{3}\theta\right)=\frac{\pi l^{3}}{3}\sin\theta\left(2\cos^{2}\theta-\sin^{2}\theta\right). \] On \(\displaystyle \left(0,\frac{\pi}{2}\right)\) we have \(\displaystyle \sin\theta\neq 0\), so \[\frac{dV}{d\theta}=0\ \Longrightarrow\ 2\cos^{2}\theta=\sin^{2}\theta\ \Longrightarrow\ \tan^{2}\theta=2\ \Longrightarrow\ \tan\theta=\sqrt{2}. \] This is a maximum: dividing \(\displaystyle 2\cos^{2}\theta-\sin^{2}\theta\) by \(\displaystyle \cos^{2}\theta>0\) gives \(\displaystyle 2-\tan^{2}\theta\), which is positive while \(\displaystyle \tan\theta<\sqrt{2}\) and negative once \(\displaystyle \tan\theta>\sqrt{2}\); so \(\displaystyle V\) increases and then decreases, and \(\displaystyle \frac{dV}{d\theta}\) changes sign from \(\displaystyle +\) to \(\displaystyle -\) at \(\displaystyle \tan\theta=\sqrt{2}\). Hence the cone of maximum volume with a given slant height has semi-vertical angle \(\displaystyle \theta=\tan^{-1}\sqrt{2}\).

Let the cone have base radius \(\displaystyle r\), height \(\displaystyle h\), slant height \(\displaystyle l\) and semi-vertical angle \(\displaystyle \theta\), and let its total surface area \(\displaystyle S=\pi rl+\pi r^{2}\) be given (constant). Solve for \(\displaystyle l\) and then for \(\displaystyle h\): \[l=\frac{S-\pi r^{2}}{\pi r}=\frac{S}{\pi r}-r, \] \[h^{2}=l^{2}-r^{2}=\left(\frac{S}{\pi r}-r\right)^{2}-r^{2}=\frac{S^{2}}{\pi^{2}r^{2}}-\frac{2S}{\pi}+r^{2}-r^{2}=\frac{S^{2}}{\pi^{2}r^{2}}-\frac{2S}{\pi}. \] Since \(\displaystyle V>0\), maximise \(\displaystyle V^{2}\) rather than \(\displaystyle V\) (this clears the square root): \[V^{2}=\frac{1}{9}\pi^{2}r^{4}h^{2}=\frac{1}{9}\pi^{2}r^{4}\left(\frac{S^{2}}{\pi^{2}r^{2}}-\frac{2S}{\pi}\right)=\frac{1}{9}\left(S^{2}r^{2}-2\pi Sr^{4}\right). \] \[\frac{d\left(V^{2}\right)}{dr}=\frac{1}{9}\left(2S^{2}r-8\pi Sr^{3}\right)=\frac{2Sr}{9}\left(S-4\pi r^{2}\right)=0\ \Longrightarrow\ S=4\pi r^{2}\quad(r>0). \] \[\frac{d^{2}\left(V^{2}\right)}{dr^{2}}=\frac{1}{9}\left(2S^{2}-24\pi Sr^{2}\right)\ \overset{S=4\pi r^{2}}{=}\ \frac{1}{9}\left(32\pi^{2}r^{4}-96\pi^{2}r^{4}\right)=-\frac{64\pi^{2}r^{4}}{9}<0, \] so the volume is maximum when \(\displaystyle S=4\pi r^{2}\). Substituting this back, \[l=\frac{S}{\pi r}-r=\frac{4\pi r^{2}}{\pi r}-r=4r-r=3r, \] and therefore \[\sin\theta=\frac{r}{l}=\frac{r}{3r}=\frac{1}{3}\ \Longrightarrow\ \theta=\sin^{-1}\frac{1}{3}. \] Hence the semi-vertical angle of the cone of given surface area and maximum volume is \(\displaystyle \sin^{-1}\frac{1}{3}\).

A general point of the curve \(\displaystyle x^{2}=2y\) is \(\displaystyle \left(x,\frac{x^{2}}{2}\right)\). Its distance from \(\displaystyle (0,5)\) is least exactly when the square of the distance is least (the square root is an increasing function), so minimise \[D=x^{2}+\left(\frac{x^{2}}{2}-5\right)^{2}. \] Substituting \(\displaystyle t=x^{2}\ (t\ge 0)\) makes it a quadratic: \[D(t)=t+\left(\frac{t}{2}-5\right)^{2}=t+\frac{t^{2}}{4}-5t+25=\frac{t^{2}}{4}-4t+25. \] \[\frac{dD}{dt}=\frac{t}{2}-4=0\ \Longrightarrow\ t=8,\qquad \frac{d^{2}D}{dt^{2}}=\frac{1}{2}>0, \] so \(\displaystyle t=8\) gives the minimum. Then \(\displaystyle x^{2}=8\Rightarrow x=\pm2\sqrt{2}\) and \(\displaystyle y=\frac{x^{2}}{2}=4\). The nearest points are \(\displaystyle \left(\pm2\sqrt{2},\,4\right)\), and the one listed is \(\displaystyle \left(2\sqrt{2},4\right)\). Correct answer: (A) \(\displaystyle \left(2\sqrt{2},4\right)\).

Let \(\displaystyle f(x)=\dfrac{1-x+x^{2}}{1+x+x^{2}}\). First note the denominator never vanishes: \(\displaystyle 1+x+x^{2}=\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4}>0\) for all real \(\displaystyle x\), so \(\displaystyle f\) is defined and differentiable on \(\displaystyle \mathbb{R}\). By the quotient rule, \[f'(x)=\frac{(-1+2x)\left(1+x+x^{2}\right)-\left(1-x+x^{2}\right)(1+2x)}{\left(1+x+x^{2}\right)^{2}}. \] Expanding the two products: \(\displaystyle (2x-1)\left(x^{2}+x+1\right)=2x^{3}+x^{2}+x-1\) and \(\displaystyle \left(x^{2}-x+1\right)(2x+1)=2x^{3}-x^{2}+x+1\), so \[f'(x)=\frac{2x^{2}-2}{\left(1+x+x^{2}\right)^{2}}=\frac{2(x-1)(x+1)}{\left(1+x+x^{2}\right)^{2}}. \] Hence \(\displaystyle f'(x)=0\) at \(\displaystyle x=\pm1\). The denominator is positive, so \(\displaystyle f'<0\) on \(\displaystyle (-1,1)\) and \(\displaystyle f'>0\) outside \(\displaystyle [-1,1]\): at \(\displaystyle x=1\) the sign changes from \(\displaystyle -\) to \(\displaystyle +\), a minimum. \[f(1)=\frac{1-1+1}{1+1+1}=\frac{1}{3}. \] At \(\displaystyle x=-1\), \(\displaystyle f(-1)=\frac{3}{1}=3\) is a local maximum, and \(\displaystyle f(x)\to1\) as \(\displaystyle x\to\pm\infty\), so \(\displaystyle \frac{1}{3}\) is the least value over all real \(\displaystyle x\). Correct answer: (D) \(\displaystyle \frac{1}{3}\).

Let \(\displaystyle F(x)=\left[x(x-1)+1\right]^{1/3}=\left(x^{2}-x+1\right)^{1/3}\) on \(\displaystyle [0,1]\). The cube-root function is increasing, so \(\displaystyle F\) is greatest exactly where \(\displaystyle g(x)=x^{2}-x+1\) is greatest; it is enough to maximise \(\displaystyle g\) on \(\displaystyle [0,1]\). \[g'(x)=2x-1=0\ \Longrightarrow\ x=\frac{1}{2}. \] Compare this critical value with the endpoint values: \[g(0)=1,\qquad g\left(\frac{1}{2}\right)=\frac{1}{4}-\frac{1}{2}+1=\frac{3}{4},\qquad g(1)=1-1+1=1. \] (The interior critical point is in fact the minimum, since \(\displaystyle g''(x)=2>0\); the maximum sits at the endpoints.) So the greatest value of \(\displaystyle g\) on \(\displaystyle [0,1]\) is \(\displaystyle 1\), attained at \(\displaystyle x=0\) and \(\displaystyle x=1\), and therefore \[F_{\max}=1^{1/3}=1. \] Correct answer: (C) 1.