NCERT Solutions · Class 12 Mathematics Application of Derivatives

Let \(\displaystyle f(x)=x^{4}-62x^{2}+ax+9\) on \(\displaystyle [0,2]\). The maximum is given to occur at \(\displaystyle x=1\), which is an interior point of the interval, and \(\displaystyle f\) is differentiable there; so by the necessary condition for an interior extremum, \[f'(1)=0. \] \[f'(x)=4x^{3}-124x+a\ \Longrightarrow\ f'(1)=4-124+a=0\ \Longrightarrow\ a=120. \] Check that \(\displaystyle a=120\) really makes \(\displaystyle x=1\) the point of maximum on \(\displaystyle [0,2]\): \[f'(x)=4x^{3}-124x+120=4\left(x^{3}-31x+30\right)=4(x-1)(x+6)(x-5). \] For \(\displaystyle x\in[0,2]\) we have \(\displaystyle x+6>0\) and \(\displaystyle x-5<0\), so \(\displaystyle (x+6)(x-5)<0\); hence \(\displaystyle f'(x)>0\) when \(\displaystyle x<1\) and \(\displaystyle f'(x)<0\) when \(\displaystyle x>1\). Thus \(\displaystyle f\) rises up to \(\displaystyle x=1\) and falls after it, confirming a maximum at \(\displaystyle x=1\). \(\displaystyle a=120\).

Let \(\displaystyle f(x)=x+\sin 2x\) on the closed interval \(\displaystyle [0,2\pi]\); \(\displaystyle f\) is continuous, so its absolute extrema occur at critical points or at the endpoints. \[f'(x)=1+2\cos 2x=0\ \Longrightarrow\ \cos 2x=-\frac{1}{2}. \] As \(\displaystyle x\) runs over \(\displaystyle [0,2\pi]\), \(\displaystyle 2x\) runs over \(\displaystyle [0,4\pi]\), so \[2x=\frac{2\pi}{3},\ \frac{4\pi}{3},\ \frac{8\pi}{3},\ \frac{10\pi}{3}\qquad\Longrightarrow\qquad x=\frac{\pi}{3},\ \frac{2\pi}{3},\ \frac{4\pi}{3},\ \frac{5\pi}{3}. \] Now evaluate \(\displaystyle f\) at these four points and at \(\displaystyle x=0,\,2\pi\): \(\displaystyle f(0)=0\); \(\displaystyle f\left(\frac{\pi}{3}\right)=\frac{\pi}{3}+\sin\frac{2\pi}{3}=\frac{\pi}{3}+\frac{\sqrt{3}}{2}\approx 1.91\); \(\displaystyle f\left(\frac{2\pi}{3}\right)=\frac{2\pi}{3}+\sin\frac{4\pi}{3}=\frac{2\pi}{3}-\frac{\sqrt{3}}{2}\approx 1.23\); \(\displaystyle f\left(\frac{4\pi}{3}\right)=\frac{4\pi}{3}+\sin\frac{8\pi}{3}=\frac{4\pi}{3}+\frac{\sqrt{3}}{2}\approx 5.06\); \(\displaystyle f\left(\frac{5\pi}{3}\right)=\frac{5\pi}{3}+\sin\frac{10\pi}{3}=\frac{5\pi}{3}-\frac{\sqrt{3}}{2}\approx 4.37\); \(\displaystyle f(2\pi)=2\pi+\sin 4\pi=2\pi\approx 6.28\). Maximum value \(\displaystyle 2\pi\), attained at \(\displaystyle x=2\pi\); minimum value \(\displaystyle 0\), attained at \(\displaystyle x=0\). (Both occur at endpoints, not at critical points.)

Let the two numbers be \(\displaystyle x\) and \(\displaystyle 24-x\), so that their sum is \(\displaystyle 24\), and let their product be \[P(x)=x(24-x)=24x-x^{2}. \] \[P'(x)=24-2x=0\ \Longrightarrow\ x=12,\qquad P''(x)=-2<0, \] so by the second derivative test \(\displaystyle x=12\) gives a maximum (and being a downward parabola, the absolute maximum). The two numbers are \(\displaystyle 12\) and \(\displaystyle 12\), and the largest possible product is \(\displaystyle 144\).

Given \(\displaystyle x+y=60\) with \(\displaystyle x>0,\ y>0\). Eliminate \(\displaystyle x\) using \(\displaystyle x=60-y\) and maximise \[P(y)=(60-y)y^{3}=60y^{3}-y^{4},\qquad 0<y<60 \] (the range of \(\displaystyle y\) comes from \(\displaystyle y>0\) and \(\displaystyle x=60-y>0\)). \[P'(y)=180y^{2}-4y^{3}=4y^{2}(45-y)=0\ \Longrightarrow\ y=0\ \text{(rejected, since } y>0)\ \text{or } y=45. \] \[P''(y)=360y-12y^{2},\qquad P''(45)=16200-24300=-8100<0, \] so \(\displaystyle y=45\) gives the maximum. Then \(\displaystyle x=60-45=15\). The numbers are \(\displaystyle x=15\) and \(\displaystyle y=45\) (maximum value \(\displaystyle xy^{3}=15\times 45^{3}=1366875\)).

Given \(\displaystyle x+y=35\) with \(\displaystyle x>0,\ y>0\). Put \(\displaystyle y=35-x\) and maximise \[P(x)=x^{2}(35-x)^{5},\qquad 0<x<35. \] Differentiate as a product and factorise -- this is the step where the common factors must be pulled out carefully: \[P'(x)=2x(35-x)^{5}+x^{2}\cdot 5(35-x)^{4}(-1)=x(35-x)^{4}\left[2(35-x)-5x\right]=x(35-x)^{4}(70-7x)=7x(35-x)^{4}(10-x). \] For \(\displaystyle 0<x<35\) we have \(\displaystyle x>0\) and \(\displaystyle (35-x)^{4}>0\), so the sign of \(\displaystyle P'\) is the sign of \(\displaystyle 10-x\): \(\displaystyle P'(x)>0\) for \(\displaystyle x<10\) and \(\displaystyle P'(x)<0\) for \(\displaystyle x>10\), so \(\displaystyle P'\) changes from \(\displaystyle +\) to \(\displaystyle -\) at \(\displaystyle x=10\) and \(\displaystyle P\) is maximum there. Then \(\displaystyle y=35-10=25\). The numbers are \(\displaystyle x=10\) and \(\displaystyle y=25\).

Let the numbers be \(\displaystyle x\) and \(\displaystyle 16-x\), with \(\displaystyle 0<x<16\), and minimise the sum of their cubes \[S(x)=x^{3}+(16-x)^{3}. \] By the chain rule on the second term, \[S'(x)=3x^{2}-3(16-x)^{2}=3\left[x-(16-x)\right]\left[x+(16-x)\right]=3(2x-16)(16)=96x-768. \] \[S'(x)=0\ \Longrightarrow\ x=8. \] \[S''(x)=6x+6(16-x)=96>0, \] so \(\displaystyle x=8\) gives a minimum. The two numbers are \(\displaystyle 8\) and \(\displaystyle 8\), and the least possible sum of cubes is \(\displaystyle 8^{3}+8^{3}=1024\).

Let \(\displaystyle x\) cm be the side of each square cut from the corners. After folding, the open box has a square base of side \(\displaystyle (18-2x)\) cm and height \(\displaystyle x\) cm, so \[V(x)=x(18-2x)^{2},\qquad 0<x<9 \] (the domain comes from \(\displaystyle x>0\) and \(\displaystyle 18-2x>0\)). \[V'(x)=(18-2x)^{2}+x\cdot 2(18-2x)(-2)=(18-2x)\left[(18-2x)-4x\right]=(18-2x)(18-6x). \] \[V'(x)=0\ \Longrightarrow\ x=9\ \text{(rejected: the base would vanish)}\ \text{or}\ x=3. \] Expanding, \(\displaystyle V(x)=4x^{3}-72x^{2}+324x\), so \(\displaystyle V''(x)=24x-144\) and \[V''(3)=72-144=-72<0, \] confirming a maximum at \(\displaystyle x=3\). The side of the square to be cut off should be \(\displaystyle 3\) cm, giving the greatest volume \(\displaystyle 3\times(18-6)^{2}=3\times144=432\) cm\(\displaystyle ^{3}\).

Let \(\displaystyle x\) cm be the side of each corner square. The open box then has base \(\displaystyle (45-2x)\) cm by \(\displaystyle (24-2x)\) cm and height \(\displaystyle x\) cm, so \[V(x)=x(45-2x)(24-2x)=4x^{3}-138x^{2}+1080x,\qquad 0<x<12 \] (the binding restriction is \(\displaystyle 24-2x>0\), i.e. \(\displaystyle x<12\)). \[V'(x)=12x^{2}-276x+1080=12\left(x^{2}-23x+90\right)=12(x-5)(x-18). \] \[V'(x)=0\ \Longrightarrow\ x=5\ \text{or}\ x=18;\quad x=18\ \text{is rejected as it lies outside}\ 0<x<12. \] \[V''(x)=24x-276,\qquad V''(5)=120-276=-156<0, \] so \(\displaystyle x=5\) gives the maximum volume. The side of the square to be cut off should be \(\displaystyle 5\) cm, giving the greatest volume \(\displaystyle 5\times 35\times 14=2450\) cm\(\displaystyle ^{3}\).

Let the fixed circle have radius \(\displaystyle a\). If a rectangle is inscribed in a circle, its diagonal is a diameter (the angle in a semicircle is a right angle), so with sides \(\displaystyle x\) and \(\displaystyle y\), \[x^{2}+y^{2}=(2a)^{2}=4a^{2}\ \Longrightarrow\ y=\sqrt{4a^{2}-x^{2}},\qquad 0<x<2a. \] Area \(\displaystyle A=xy=x\sqrt{4a^{2}-x^{2}}\). Since \(\displaystyle A>0\) and \(\displaystyle t\mapsto t^{2}\) is increasing for \(\displaystyle t>0\), it is equivalent (and easier) to maximise \[S=A^{2}=x^{2}\left(4a^{2}-x^{2}\right)=4a^{2}x^{2}-x^{4}. \] \[\frac{dS}{dx}=8a^{2}x-4x^{3}=4x\left(2a^{2}-x^{2}\right)=0\ \Longrightarrow\ x=a\sqrt{2}\quad(\text{taking }x>0). \] \[\frac{d^{2}S}{dx^{2}}=8a^{2}-12x^{2};\qquad \text{at }x=a\sqrt{2}:\ 8a^{2}-24a^{2}=-16a^{2}<0, \] so the area is maximum at \(\displaystyle x=a\sqrt{2}\). The other side is then \[y=\sqrt{4a^{2}-2a^{2}}=a\sqrt{2}=x. \] The two adjacent sides are equal, so the rectangle of largest area inscribed in the circle is a square (of area \(\displaystyle 2a^{2}\)), as required.

Let the cylinder have base radius \(\displaystyle r\) and height \(\displaystyle h\). Its total surface area \(\displaystyle S\) is given, i.e. constant: \[S=2\pi r^{2}+2\pi rh\ \Longrightarrow\ h=\frac{S-2\pi r^{2}}{2\pi r}. \] Substitute into the volume so that it depends on the single variable \(\displaystyle r\): \[V=\pi r^{2}h=\pi r^{2}\cdot\frac{S-2\pi r^{2}}{2\pi r}=\frac{1}{2}\left(Sr-2\pi r^{3}\right). \] \[\frac{dV}{dr}=\frac{1}{2}\left(S-6\pi r^{2}\right)=0\ \Longrightarrow\ S=6\pi r^{2}. \] \[\frac{d^{2}V}{dr^{2}}=-6\pi r<0\quad(r>0), \] so this value of \(\displaystyle r\) gives the maximum volume. Putting \(\displaystyle S=6\pi r^{2}\) back into the expression for \(\displaystyle h\): \[h=\frac{6\pi r^{2}-2\pi r^{2}}{2\pi r}=\frac{4\pi r^{2}}{2\pi r}=2r. \] Hence the height equals \(\displaystyle 2r\), which is the diameter of the base.