NCERT Solutions · Class 12 Mathematics Application of Derivatives

The domain is \(\displaystyle x>0\), since \(\displaystyle \log x\) is defined only there.By the quotient rule, \[f'(x)=\frac{x\cdot\dfrac{1}{x}-\log x\cdot 1}{x^{2}}=\frac{1-\log x}{x^{2}}. \]Stationary points: \(\displaystyle x^{2}\neq 0\) on the domain, so \(\displaystyle f'(x)=0\) only when \(\displaystyle 1-\log x=0\), i.e. \(\displaystyle \log x=1\), i.e. \(\displaystyle x=e\).Apply the second derivative test. Differentiating \(\displaystyle f'(x)=(1-\log x)x^{-2}\), \[f''(x)=\frac{\left(-\dfrac{1}{x}\right)x^{2}-(1-\log x)(2x)}{x^{4}}=\frac{-1-2(1-\log x)}{x^{3}}=\frac{2\log x-3}{x^{3}}. \] At \(\displaystyle x=e\), \(\displaystyle \log e=1\), so \[f''(e)=\frac{2-3}{e^{3}}=-\frac{1}{e^{3}}<0. \]Since \(\displaystyle f''(e)<0\), \(\displaystyle f\) has a maximum at \(\displaystyle x=e\).This maximum is in fact the absolute maximum on \(\displaystyle (0,\infty)\): \(\displaystyle f'(x)>0\) for \(\displaystyle 0<x<e\) (there \(\displaystyle \log x<1\)) and \(\displaystyle f'(x)<0\) for \(\displaystyle x>e\).Hence \(\displaystyle f\) has its maximum at \(\displaystyle x=e\), the maximum value being \(\displaystyle f(e)=\dfrac{1}{e}\).

Let each of the equal sides be \(\displaystyle x\) cm at time \(\displaystyle t\) seconds; the base \(\displaystyle b\) is fixed. Given \(\displaystyle \dfrac{dx}{dt}=-3\) cm/s (negative because the sides are decreasing).The perpendicular from the apex to the base bisects the base, so by Pythagoras the height is \(\displaystyle \sqrt{x^{2}-\dfrac{b^{2}}{4}}\) and \[A=\frac{1}{2}\,b\,\sqrt{x^{2}-\frac{b^{2}}{4}}. \]Differentiate with respect to \(\displaystyle t\) by the chain rule: \[\frac{dA}{dt}=\frac{b}{2}\cdot\frac{1}{2\sqrt{x^{2}-\dfrac{b^{2}}{4}}}\cdot 2x\cdot\frac{dx}{dt}=\frac{b\,x}{2\sqrt{x^{2}-\dfrac{b^{2}}{4}}}\cdot\frac{dx}{dt}. \]Now put \(\displaystyle x=b\) (the instant asked for). Then \[\sqrt{b^{2}-\frac{b^{2}}{4}}=\sqrt{\frac{3b^{2}}{4}}=\frac{\sqrt{3}}{2}b, \] so \[\frac{dA}{dt}=\frac{b\cdot b}{2\cdot\dfrac{\sqrt{3}}{2}b}\cdot(-3)=\frac{b}{\sqrt{3}}\cdot(-3)=-\sqrt{3}\,b. \]The negative sign confirms the area is decreasing. The area is decreasing at the rate \(\displaystyle \sqrt{3}\,b\ \mathrm{cm}^{2}/\mathrm{s}\).

Take the vertex at the end \(\displaystyle A(a,0)\) of the major axis. By symmetry of the ellipse in the \(\displaystyle x\)-axis, the other two vertices of an isosceles triangle with this apex are \(\displaystyle (x,y)\) and \(\displaystyle (x,-y)\) with \(\displaystyle -a<x<a\) and \[y=b\sqrt{1-\frac{x^{2}}{a^{2}}}=\frac{b}{a}\sqrt{a^{2}-x^{2}}. \]The base is the vertical chord of length \(\displaystyle 2y\) and the height is \(\displaystyle a-x\), so the area is \[A(x)=\frac{1}{2}(2y)(a-x)=\frac{b}{a}(a-x)\sqrt{a^{2}-x^{2}}=\frac{b}{a}(a-x)^{3/2}(a+x)^{1/2}, \] using \(\displaystyle a^{2}-x^{2}=(a-x)(a+x)\).Differentiate by the product rule: \[A'(x)=\frac{b}{a}\left[-\frac{3}{2}(a-x)^{1/2}(a+x)^{1/2}+\frac{1}{2}(a-x)^{3/2}(a+x)^{-1/2}\right] \] \[=\frac{b}{2a}\cdot\frac{(a-x)^{1/2}}{(a+x)^{1/2}}\Big[-3(a+x)+(a-x)\Big]=\frac{b}{2a}\cdot\frac{(a-x)^{1/2}}{(a+x)^{1/2}}\,(-2a-4x). \]For \(\displaystyle -a<x<a\) the factor \(\displaystyle \dfrac{(a-x)^{1/2}}{(a+x)^{1/2}}\) is positive, so \[A'(x)=0\iff -2a-4x=0\iff x=-\frac{a}{2}. \] Moreover \(\displaystyle -2a-4x>0\) for \(\displaystyle x<-\dfrac{a}{2}\) and \(\displaystyle <0\) for \(\displaystyle x>-\dfrac{a}{2}\), so \(\displaystyle A\) changes from increasing to decreasing: by the first derivative test \(\displaystyle x=-\dfrac{a}{2}\) gives a maximum.At \(\displaystyle x=-\dfrac{a}{2}\): \(\displaystyle a-x=\dfrac{3a}{2}\), \(\displaystyle a+x=\dfrac{a}{2}\), so \[A=\frac{b}{a}\cdot\frac{3a}{2}\cdot\sqrt{\frac{3a}{2}\cdot\frac{a}{2}}=\frac{b}{a}\cdot\frac{3a}{2}\cdot\frac{a\sqrt{3}}{2}=\frac{3\sqrt{3}}{4}ab. \]The maximum area is \(\displaystyle \dfrac{3\sqrt{3}}{4}\,ab\) square units (attained when the base is the chord \(\displaystyle x=-\tfrac{a}{2}\)).

Let the rectangular base measure \(\displaystyle x\) m by \(\displaystyle y\) m. The depth is \(\displaystyle 2\) m and the volume is \(\displaystyle 8\ \mathrm{m}^{3}\), so \[2xy=8\ \Rightarrow\ xy=4\ \Rightarrow\ y=\frac{4}{x},\qquad x>0. \]Base area \(\displaystyle =xy=4\ \mathrm{m}^{2}\), which is fixed; its cost is \(\displaystyle 70\times 4=\) Rs \(\displaystyle 280\).The tank is open at the top, so the sides are four rectangles: two of size \(\displaystyle x\times 2\) and two of size \(\displaystyle y\times 2\). Their total area is \[2(2x)+2(2y)=4(x+y), \] and their cost is \(\displaystyle 45\times 4(x+y)=180(x+y)\).Hence the total cost is \[C(x)=280+180\left(x+\frac{4}{x}\right). \]Minimise: \(\displaystyle C'(x)=180\left(1-\dfrac{4}{x^{2}}\right)=0\Rightarrow x^{2}=4\Rightarrow x=2\) (reject \(\displaystyle x=-2\), a length). Also \[C''(x)=180\cdot\frac{8}{x^{3}},\qquad C''(2)=180\cdot 1=180>0, \] so \(\displaystyle x=2\) gives a minimum.Then \(\displaystyle y=\dfrac{4}{2}=2\) m, and \[C(2)=280+180(2+2)=280+720=1000. \]The least expensive tank is the one with a \(\displaystyle 2\ \mathrm{m}\times 2\ \mathrm{m}\) base and depth \(\displaystyle 2\) m, costing Rs \(\displaystyle 1000\).

Let the circle have radius \(\displaystyle r\) and the square have side \(\displaystyle s\), with \(\displaystyle r,s>0\). The constraint is \[2\pi r+4s=k\ \Rightarrow\ s=\frac{k-2\pi r}{4}. \]The sum of the areas, as a function of the single variable \(\displaystyle r\), is \[A=\pi r^{2}+s^{2}=\pi r^{2}+\left(\frac{k-2\pi r}{4}\right)^{2}. \]Differentiate, using \(\displaystyle \dfrac{ds}{dr}=-\dfrac{2\pi}{4}=-\dfrac{\pi}{2}\): \[\frac{dA}{dr}=2\pi r+2s\,\frac{ds}{dr}=2\pi r+2s\left(-\frac{\pi}{2}\right)=2\pi r-\pi s=\pi(2r-s). \]So \[\frac{dA}{dr}=0\iff s=2r. \]Check it is a minimum: \[\frac{d^{2}A}{dr^{2}}=2\pi-\pi\frac{ds}{dr}=2\pi-\pi\left(-\frac{\pi}{2}\right)=2\pi+\frac{\pi^{2}}{2}>0 \] for every \(\displaystyle r\), so \(\displaystyle A\) is convex and the stationary point is the absolute minimum.Hence the sum of the areas is least exactly when \(\displaystyle s=2r\), i.e. when the side of the square is double the radius of the circle.

Let the rectangle have width \(\displaystyle 2x\) m and height \(\displaystyle y\) m; the semicircle surmounting it then has radius \(\displaystyle x\) m. The perimeter of the window consists of the two vertical sides, the bottom, and the semicircular arc (the common chord is not part of the boundary): \[2y+2x+\pi x=10\ \Rightarrow\ y=\frac{10-2x-\pi x}{2}. \]The light admitted is proportional to the area of the opening, so maximise \[A=\underbrace{2xy}_{\text{rectangle}}+\underbrace{\frac{1}{2}\pi x^{2}}_{\text{semicircle}}=2x\cdot\frac{10-2x-\pi x}{2}+\frac{\pi x^{2}}{2}=10x-2x^{2}-\pi x^{2}+\frac{\pi x^{2}}{2}, \] \[A(x)=10x-2x^{2}-\frac{\pi}{2}x^{2}. \]Then \[A'(x)=10-4x-\pi x=0\ \Rightarrow\ x=\frac{10}{4+\pi}, \] \[A''(x)=-4-\pi<0, \] so this \(\displaystyle x\) gives the maximum.The corresponding height is \[y=\frac{10-(2+\pi)x}{2}=\frac{1}{2}\left(10-\frac{10(2+\pi)}{4+\pi}\right)=\frac{1}{2}\cdot\frac{10\big[(4+\pi)-(2+\pi)\big]}{4+\pi}=\frac{1}{2}\cdot\frac{20}{4+\pi}=\frac{10}{4+\pi}. \]So the window admits most light when \[\text{width of rectangle}=2x=\frac{20}{4+\pi}\ \text{m},\qquad \text{height of rectangle}=y=\frac{10}{4+\pi}\ \text{m}, \] the radius of the semicircle being \(\displaystyle \dfrac{10}{4+\pi}\) m — that is, the rectangle's height equals the radius, so its width is twice its height.

Place the right angle at the origin with the two legs along the axes, and let \(\displaystyle P\) be the given point on the hypotenuse, at distance \(\displaystyle a\) from one leg and \(\displaystyle b\) from the other. Let \(\displaystyle \theta\) be the acute angle the hypotenuse makes with the leg from which \(\displaystyle P\) is at distance \(\displaystyle a\), so \(\displaystyle \theta\in\left(0,\dfrac{\pi}{2}\right)\).\(\displaystyle P\) divides the hypotenuse into two pieces. In the right triangle formed by \(\displaystyle P\), its foot on the horizontal leg and the vertex \(\displaystyle A\), the side opposite \(\displaystyle \theta\) is \(\displaystyle a\), so that piece has length \(\displaystyle a/\sin\theta\). In the right triangle on the other side, the horizontal projection of the piece \(\displaystyle PB\) is \(\displaystyle b\) and it makes angle \(\displaystyle \theta\) with the horizontal, so that piece has length \(\displaystyle b/\cos\theta\). Hence the length of the hypotenuse is \[L(\theta)=\frac{a}{\sin\theta}+\frac{b}{\cos\theta}=a\,\mathrm{cosec}\,\theta+b\sec\theta. \]Differentiate: \[L'(\theta)=-a\,\mathrm{cosec}\,\theta\cot\theta+b\sec\theta\tan\theta=-\frac{a\cos\theta}{\sin^{2}\theta}+\frac{b\sin\theta}{\cos^{2}\theta}. \] Setting \(\displaystyle L'(\theta)=0\) and multiplying by \(\displaystyle \dfrac{\sin^{2}\theta\cos^{2}\theta}{1}\): \[b\sin^{3}\theta=a\cos^{3}\theta\ \Rightarrow\ \tan^{3}\theta=\frac{a}{b}\ \Rightarrow\ \tan\theta=\left(\frac{a}{b}\right)^{1/3}. \]This is the only stationary point in \(\displaystyle \left(0,\dfrac{\pi}{2}\right)\), and \(\displaystyle L(\theta)\to\infty\) as \(\displaystyle \theta\to 0^{+}\) and as \(\displaystyle \theta\to\dfrac{\pi}{2}^{-}\); also \(\displaystyle L'<0\) before it and \(\displaystyle L'>0\) after it (the factor \(\displaystyle b\sin^{3}\theta-a\cos^{3}\theta\) is increasing in \(\displaystyle \theta\)). So this value gives the minimum.From \(\displaystyle \tan\theta=a^{1/3}/b^{1/3}\), the right triangle with legs \(\displaystyle a^{1/3}\) and \(\displaystyle b^{1/3}\) has hypotenuse \(\displaystyle \sqrt{a^{2/3}+b^{2/3}}\), so \[\sin\theta=\frac{a^{1/3}}{\sqrt{a^{2/3}+b^{2/3}}},\qquad \cos\theta=\frac{b^{1/3}}{\sqrt{a^{2/3}+b^{2/3}}}. \] Therefore \[L=\frac{a\sqrt{a^{2/3}+b^{2/3}}}{a^{1/3}}+\frac{b\sqrt{a^{2/3}+b^{2/3}}}{b^{1/3}}=\sqrt{a^{2/3}+b^{2/3}}\left(a^{2/3}+b^{2/3}\right)=\left(a^{2/3}+b^{2/3}\right)^{3/2}. \]Hence the minimum length of the hypotenuse is \(\displaystyle \left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}\).