Let r be the radius, h be the height and V be the volume of the sand cone
Also given that, \(\frac{d V}{d t}=12 \mathrm{~cm}^{3} / \mathrm{s}, h=\frac{1}{6} r\) 
\(\Rightarrow r=6 h \text { and } h=4 \mathrm{~cm}\) 
Volume of sand cone,
\(V=\frac{1}{3} \pi r^{2} h\) 
\( \Rightarrow V=\frac{1}{3} \pi(6 h)^{2} h \)
\(\Rightarrow V=\frac{1}{3} \pi \times 36 h^{2} \times h=12 \pi h^{3} \) 
On differentiating both sides w.r.t. t, we get
\( \frac{d V}{d t}=12 \pi \times 3 h^{2} \frac{d h}{d t}=36 \pi h^{2} \frac{d h}{d t} \)
\(\Rightarrow 12=36 \pi(4)^{2} \frac{d h}{d t} \)
\(\Rightarrow \frac{d h}{d t}=\frac{12}{36 \pi \times 16}=\frac{1}{48 \pi} \mathrm{cm} / \mathrm{s} \) 
Hence, the height of the sand cone is increasing at the rate of \(\frac{1}{48 \pi} \mathrm{cm} / \mathrm{s}\). when the height is 4 cm,

Let x be the length of a side, V be the volume, and s be the surface area of the cube.
Then, V = x³ and S = 6x² where x is a function of time t.
 It is given that dVdt=8 cm3/s . " id="MathJax-Element-558-Frame" role="presentation" style="position: relative;" tabindex="0"> It is given that  \( \frac{d V}{d t}=8 \mathrm{~cm}^{3} / \mathrm{s} \text { . }\)
 Then, by using the chain rule, we have: " id="MathJax-Element-559-Frame" role="presentation" style="position: relative;" tabindex="0">Then, by using the chain rule, we have: 
\(\therefore 8=\frac{d V}{d t}=\frac{d}{d t}\left(x^{3}\right)=\frac{d}{d x}\left(x^{3}\right) \cdot \frac{d x}{d t}=3 x^{2} \cdot \frac{d x}{d t}\)
\(\Rightarrow \frac{d x}{d t}=\frac{8}{3 x^{2}} \ \text { (1) } \quad \text { [By chain rule] } \)
\(\text { Now, } \frac{d \mathrm{~S}}{d t}=\frac{d}{d t}\left(6 x^{2}\right)=\frac{d}{d x}\left(6 x^{2}\right) \cdot \frac{d x}{d t} \ 0 .\)
\(=12 x \cdot \frac{d x}{d t}=12 x .\left(\frac{8}{3 x^{2}}\right)=\frac{32}{x}\)
\(\text {Thus, when } x=12 \mathrm{~cm}, \frac{d S}{d t}=\frac{32}{12} \mathrm{~cm}^{2} / \mathrm{s}=\frac{8}{3} \mathrm{~cm}^{2} / \mathrm{s}\)
Hence, if the length of the edge of the cube is 12 cm, then the surface area is increasing at the rate of \(\frac{8}{3}\) cm²/s

Let r and h be the radius and height of the cone respectively inscribed in a sphere of radius R.
Let V be the volume of the cone.
Then, \(V=\frac{1}{3} \pi r^{2} h\)
Height of the cone is given by
\(h=R+A B=R+\sqrt{R^{2}-r^{2}} \quad[\mathrm{ABC} \text { is a right triangle }] \)
\(\therefore V=\frac{1}{3} \pi r^{2}\left(R+\sqrt{R^{2}-r^{2}}\right) \)
\(=\frac{1}{3} \pi r^{2} R+\frac{1}{3} \pi r^{2} \sqrt{R^{2}-r^{2}} \)
\(\therefore \frac{d V}{d r}=\frac{2}{3} \pi r R+\frac{2}{3} \pi r \sqrt{R^{2}-r^{2}}+\frac{1}{3} \pi r^{2} \cdot \frac{(-2 r)}{2 \sqrt{R^{2}-r^{2}}} \)
\(=\frac{2}{3} \pi r R+\frac{2}{3} \pi r \sqrt{R^{2}-r^{2}}-\frac{1}{3} \pi \frac{r^{3}}{\sqrt{R^{2}-r^{2}}} \)
\(=\frac{2}{3} \pi r R+\frac{2 \pi r\left(R^{2}-r^{2}\right)-\pi r^{3}}{3 \sqrt{R^{2}-r^{2}}} \)
\(=\frac{2}{3} \pi r R+\frac{2 \pi r R^{2}-3 \pi r^{3}}{3 \sqrt{R^{2}-r^{2}}} \)
\(\frac{d^{2} V}{d r^{2}}=\frac{2 \pi R}{3}+\frac{3 \sqrt{R^{2}-r^{2}}\left(2 \pi R^{2}-9 \pi r^{2}\right)-\left(2 \pi r R^{2}-3 \pi r^{3}\right) \cdot \frac{(-2 r)}{6 \sqrt{R^{2}-r^{2}}}}{9\left(R^{2}-r^{2}\right)} \)
\(=\frac{2}{3} \pi R+\frac{9\left(R^{2}-r^{2}\right)\left(2 \pi R^{2}-9 \pi r^{2}\right)+2 \pi r^{2} R^{2}+3 \pi r^{4}}{27\left(R^{2}-r^{2}\right)^{\frac{3}{2}}} \)
\(\text { Now, } \frac{d V}{d r}=0 \Rightarrow \frac{2}{\pi} r R=\frac{3 \pi r^{3}-2 \pi r R^{2}}{3 \sqrt{R^{2}-r^{2}}} \)
\(\Rightarrow 2 R=\frac{3 r^{2}-2 R^{2}}{\sqrt{R^{2}-r^{2}}} \Rightarrow 2 R \sqrt{R^{2}-r^{2}}=3 r^{2}-2 R^{2} \)
\(\Rightarrow 4 R^{2}\left(R^{2}-r^{2}\right)=\left(3 r^{2}-2 R^{2}\right)^{2} \)
\(\Rightarrow 4 R^{4}-4 R^{2} r^{2}=9 r^{4}+4 R^{4}-12 r^{2} R^{2} \)
\(\Rightarrow 9 r^{4}=8 R^{2} r^{2} \)
\(\Rightarrow r^{2}=\frac{8}{9} R^{2} \)
\(\text { When } r^{2}=\frac{8}{9} R^{2}, \text { then } \frac{d^{2} V}{d r^{2}}<0 \)
∴ By second derivative test, the volume of the cone is the maximum when \(r^{2}=\frac{8}{9} R^{2}\)
\(\text { When } r^{2}=\frac{8}{9} R^{2}, h=R+\sqrt{R^{2}-\frac{8}{9} R^{2}}=R+\sqrt{\frac{1}{9} R^{2}}=R+\frac{R}{3}=\frac{4}{3} R \text { . }\)
\(\text { Therefore, }=\frac{1}{3} \pi\left(\frac{8}{9} R^{2}\right)\left(\frac{4}{3} R\right)\)
\(=\frac{8}{27}\left(\frac{4}{3} \pi R^{3}\right)\)
\(=\frac{8}{27} \times \)(Volume of the sphere)
 Hence, the volume of the largest cone that can be inscribed in the sphere is \( \frac{8}{27} \)

The slant height is given by
\(1=\sqrt{h^2+r^2}\)
Now volume is
\( \mathrm{V}=\frac{\pi}{3}\left(\mathrm{r}^2\right) \mathrm{h}\\ =\frac{\pi}{3}\left(1^2-h^2\right) h\\ \frac{\mathrm{dV}}{\mathrm{dh}}=\frac{\pi}{3}\left[1^2-\mathrm{h}^2-\mathrm{h}(2 \mathrm{~h})\right] \)
Now for critical points
\( \frac{\mathrm{dV}}{\mathrm{dh}}=0 \)
\(Or 1^2-h^2-2 h^2=0 \)
\(1^2=3 h^2 1=\sqrt{3} h\\ \frac{\mathrm{h}}{1}=\frac{1}{\sqrt{3}}=\cos \theta \)
Hence
\( \tan \theta=\frac{\sqrt{1-\cos ^2 \theta}}{\cos \theta} \)
\(=\frac{\frac{\sqrt{2}}{\sqrt{3}}}{\frac{1}{\sqrt{3}}} \)
\(=\sqrt{2} \)
\(Therefore\ \tan \theta=\sqrt{2} \)
\(Or\ \theta=\tan ^{-1}(\sqrt{2}) \)

The given function is f(x) = sin x + cos x.
\(\therefore f^{\prime}(x)=\cos x-\sin x\)
\(\text { Now, }f^{\prime}(x)=0 \Rightarrow \sin x=\cos x \Rightarrow \tan x=1 \Rightarrow x=\frac{\pi}{4}\)
Then, we evaluate the value of f at critical point and at the end points\(x=\frac{\pi}{4}\) of the interval [0, π].
\(f\left(\frac{\pi}{4}\right)=\sin \frac{\pi}{4}+\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2} \)
\(f(0)=\sin 0+\cos 0=0+1=1 \)
\(f(\pi)=\sin \pi+\cos \pi=0-1=-1 \)
Hence, we can conclude that the absolute maximum value of f on [0, π] is \(\sqrt{2}\) occurring at \(x=\frac{\pi}{4}\)and the absolute minimum value of f on [0, π] is −1 occurring at x = π.

\(\therefore V=\frac { 4 }{ 3 } \pi r^{ 3 }\)
\(\therefore \frac { dV }{ dr } =\frac { 4 }{ 3 } \pi (3r^{ 3 })=4\pi r^{ 2 }.\)

Hence, \(\left\lfloor \frac { dV }{ dr } \right\rfloor _{ r=10 }=4\pi (10)^{ 2 }\)
\(=400\pi cm^{ 3 }/s\)

\(f(x)=sinx+cosx;x\in [0,2\pi ]\)
\(\therefore f'(x)=cosx-sinx\)
For extreme values \(\therefore f'(x)=0\)
\(\Rightarrow cosx-sinx=0\)
\(\Rightarrow sinx=cosx\Rightarrow tan\quad x=1\)
\(\Rightarrow x=\frac { \pi }{ 4 } ,\frac { 5\pi }{ 4 } \)
Now f(0)=sin0+cos0=0+1=1
\(f\left( \frac { \pi }{ 4 } \right) =sin\frac { \pi }{ 4 } +cos\frac { \pi }{ 4 } =\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 2 } } \)
\( =\frac { 2 }{ \sqrt { 2 } } =\sqrt { 2 } \)
\(f\left( \frac { 5\pi }{ 4 } \right) =sin\frac { 5\pi }{ 4 } +cos\frac { 5\pi }{ 4 } \)
\(=\frac { 1 }{ \sqrt { 2 } } -\frac { 1 }{ \sqrt { 2 } } =-\frac { 2 }{ \sqrt { 2 } } =-\sqrt { 2 } \)
\(f(2\pi )=sin2\pi +cos2\pi =0+1=1\)
Hence, max value of \(f(x)=\sqrt { 2 } \)

We have
f(x) = sin 3x
or f '(x) = 3cos 3x
Therefore, f '(x) = 0 gives cos 3x = 0 which in turn gives \(3 x=\frac{\pi}{2}, \frac{3 \pi}{2} \text { (as } x \in\left[0, \frac{\pi}{2}\right]\)
\(\text {implies } 3 x \in\left[0, \frac{3 \pi}{2}\right] \text { ). So } x=\frac{\pi}{6} \text { and } \frac{\pi}{2} \text { . The point } x=\frac{\pi}{6} \text { divides the interval }\left[0, \frac{\pi}{2}\right]\)
into two disjoint intervals \(\left[0, \frac{\pi}{6}\right) \text { and }\left(\frac{\pi}{6}, \frac{\pi}{2}\right]\)
\(\text {Now, } f^{\prime}(x)>0 \text { for all } x \in\left[0, \frac{\pi}{6}\right) \text { as } 0 \leq x<\frac{\pi}{6} \Rightarrow 0 \leq 3 x<\frac{\pi}{2} \text { and } f^{\prime}(x)<0 \text { for }\)
\(\text { all } x \in\left(\frac{\pi}{6}, \frac{\pi}{2}\right) \text { as } \frac{\pi}{6}
Therefore, f is increasing in \(\left[0, \frac{\pi}{6}\right) \text { and decreasing in }\left(\frac{\pi}{6}, \frac{\pi}{2}\right)\)
Also, the given function is continuous at x = 0 and \(x=\frac{\pi}{6}\) Therefore, by Theorem 1,
f is increasing on \(\left[0, \frac{\pi}{6}\right] \text { and decreasing on }\left[\frac{\pi}{6}, \frac{\pi}{2}\right] \text { . }\)

f(x) = sin x − cos x, 0 < x < 2π
∴ f'(x) = cosx + sin x
f'x = cos x + sin x
f'(x) = 0
⇒ cos x = −sin x
⇒ tan x = −1
⇒ \(x =\frac {3\pi}{4}, \frac{7\pi}{4}=(0, 2\pi) f'x = 0\)
⇒ cos x = -sin x ⇒ tanx = -1 ⇒x = \(\frac{3\pi}{4},\frac{7\pi}{4}\) ∈ 0,
2π f''(x) = −sin x + cos x
f''x = -sinx + cosx 
\(f''\frac{3\pi}{4} = sin\frac{3\pi}{4} +cos\frac{3\pi}{4} = -\sqrt 12- \sqrt 12 = -2<0 \)
\(f''\frac{7\pi}{4} = sin\frac{3\pi}{4} +cos\frac{7\pi}{4} = 12+12=2>0 \)
Therefore, by second derivative test, \(x=\frac{3 \pi}{4}\) is a point of local maxima and the local maximum value of f at is \(x=\frac{3 \pi}{4}\) \(f\left(\frac{3 \pi}{4}\right)=\sin \frac{3 \pi}{4}-\cos \frac{3 \pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}\)
However, \(x=\frac{7 \pi}{4}\) is a point of local minima and the local minimum value of f at \(f\left(\frac{7 \pi}{4}\right)=\sin \frac{7 \pi}{4}-\cos \frac{7 \pi}{4}=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\sqrt{2}\)

Let ABCD be a rectangle inscribed in a given circle with centre at 0 and radius a.
Let AB = 2x and BC = 2y

Then, OA² = OM² + AM² 
\(\Rightarrow a^2=y^2+x^2\)
\(\Rightarrow y=\sqrt{a^2-x^2}\)
Let A be the area of the rectangle.
\(\therefore A=4xy=4x\sqrt{x^2-x^2}\)
\(\Rightarrow \ \ \frac{dA}{dx}=4\{\frac{a^2-2x^2}{\sqrt{a^2-x^2}}\}\)
For maximum or minimum value of A,
\(\frac{dA}{dx}=0\)
\(=4\{\frac{a^2-2x^2}{\sqrt{a^2-x^2}}\}=0\Rightarrow\ \ x=\frac{a}{\sqrt{2}}\)
Now, \(\frac{d^2A}{dx^2}=4\frac{d}{dx}\{(a^2-2x^2)(a^2-x^2)^{-1/2}\}\)
\(\Rightarrow \frac{d^2A}{dx^2}=4[-4x(a^2-x^2)^{-1/2}+(a^2-2x^2)\times(-1/2)(a^2-x^2)^(-3/2)(-2x)]\)
\(=[\frac{-4x}{\sqrt{a^2-x^2}}+\frac{x(a^2-2x^2)}{(a^2-x^2)^{3/2}}]\)
\(\therefore\ (\frac{d^2A}{dx^2})_{x=\frac{a}{\sqrt2}}=-16<0\)
Thus A is maximum when \(x=\frac{a}{\sqrt2}\)
putting \(x=\frac{a}{\sqrt2}\) in (i) \(y=\frac{a}{\sqrt2}\)
Therefore \(x=y=\frac{a}{\sqrt2}\)
Hence area is maximum when x = y ⇒ 2x = 2y
i.e., the rectangle is a square.

the required trapezium is as given in figure. Draw perpendiculars DP and CQ on AB. Let AP = x cm. Note that \(\Delta \)APD \(\cong \) \(\Delta \) BQC.
Therefore, QB = x cm. Also, by pythagoras theorem DP = QC = \(\sqrt { 100-{ x }^{ 2 } } \)
Let A be the area of the trapezium.

Then, A \(\equiv \) A(x)
\(=\frac { 1 }{ 2 } (sum\ of\ parallel\ sides)\times (height)\)
\(=\frac { 1 }{ 2 } (2x+10+10)\sqrt { 100-{ x }^{ 2 } } \)
= (x + 10)\(\sqrt { 100-{ x }^{ 2 } } \)
\(or\ A'(x)=(x+10)\frac { (-2x) }{ 2\sqrt { 100-{ x }^{ 2 } } } +(\sqrt { 100-{ x }^{ 2 }) } \)
\(=\frac { -2{ x }^{ 2 }-10x+100 }{ \sqrt { 100-{ x }^{ 2 } } } \)
Now, A'(x) = 0 gives 2x² + 10x - 100 = 0,
i.e., x = 5 and x = -10
So, x = 5.
Now, A''(x) = \(\frac { \sqrt { 100-{ x }^{ 2 } } (-4x-10)-(-2x^{ 2 }-10x+100)\frac { (-2x) }{ 2\sqrt { 100-{ x }^{ 2 } } } }{ 100-{ x }^{ 2 } } \)
\(=\frac { { 2x }^{ 3 }-300x-1000 }{ { (100-{ x }^{ 2 }) }^{ \frac { 1 }{ 2 } } } \)
(on simplification)
or \(A''(5)=\frac { { 2(5) }^{ 3 }-300(5)-1,000 }{ (100-(5{ ) }^{ 2 })^{ \frac { 3 }{ 2 } } } \)
\(=\frac { -2,250 }{ 75\sqrt { 75 } } =\frac { -30 }{ \sqrt { 75 } } <0\)
Thus, area of trapezium is maximum at x = 5 and the maximum area is given by
A(5) = (5 + 10)\(\sqrt { 100-{ (5) }^{ 2 } } \)
= 15\(\sqrt { 75 } =75\sqrt { 3 } { cm }^{ 2 }\)

Let r be the radius, h be the height, α be semi-vertical
Then \(\tan \alpha=\frac{r}{h}\)
Given, \( \alpha =\tan ^{-1}(0.5) \)
\(\frac{r}{h} =0.5 \\ r =\frac{h}{2} \)
 
Let V be the volume of the cone. Then
\(\mathrm{V}=\frac{1}{3} \pi r^2 h=\frac{1}{3} \pi\left(\frac{h}{2}\right)^2 h=\frac{\pi h^3}{12}\)
Therefore \( \frac{d \mathrm{~V}}{d t} =\frac{d}{d h}\left(\frac{\pi h^3}{12}\right) \cdot \frac{d h}{d t} \)
\(=\frac{\pi}{4} h^2 \frac{d h}{d t} \) 
Now rate of change of volume, i.e., \(\frac{d \mathrm{~V}}{d t}=5 \mathrm{~m}^3 / \mathrm{h} \text { and } h=4 \mathrm{~m}\) 
\(5=\frac{\pi}{4}(4)^2 \cdot \frac{d h}{d t}\) 
\(\frac{d h}{d t}=\frac{5}{4 \pi}=\frac{35}{88} \mathrm{~m} / \mathrm{h}\left(\pi=\frac{22}{7}\right)\)
Thus, the rate of change of water level is \(\frac{35}{88} \mathrm{~m} / \mathrm{h}\)
\(V=\frac{1}{3} \pi r^{2} h=\frac{1}{3} \pi\left(\frac{h}{2}\right) h \quad \text { Ans. } \left.\frac{35}{88} \mathrm{~m} / \mathrm{h}\right]\)