85618
The altitude for a right circular cone of minimum volume circumscribed about a sphere of radius \(r\) is
1 \(2 \mathrm{r}\)
2 \(3 \mathrm{r}\)
3 \(5 \mathrm{r}\)
4 \(4 \mathrm{r}\)
Explanation:
(D) : Let the radius Of cone be \(\mathrm{R}\) and Height be \(\mathrm{h}\). In \(\triangle \mathrm{ABC}\), \(\sin \theta=\frac{r}{h-r}\) In \(\triangle \mathrm{ADE}\), \(\sin \theta=\frac{\mathrm{R}}{\sqrt{\mathrm{h}^{2}+\mathrm{R}^{2}}}\) Equating (i) and (ii), we get \(\frac{r}{h-r}=\frac{R}{\sqrt{h^{2}+R^{2}}}\) Squaring both sides, we get \(r^{2}\left(h^{2}+R^{2}\right)=R^{2}(h-r)^{2}\) \(\Rightarrow \mathrm{R}^{2}=\frac{\mathrm{r}^{2} \mathrm{~h}}{\mathrm{~h}-2 \mathrm{r}}\) Volume of cone is given by \(\mathrm{V}=\frac{1}{3} \pi \mathrm{R}^{2} \mathrm{~h}\) \(\Rightarrow \mathrm{V}=\frac{1}{3} \pi\left(\frac{\mathrm{r}^{2} \mathrm{~h}}{\mathrm{~h}-2 \mathrm{r}}\right) \mathrm{h}\) [Using (iii)] So, \(\frac{d V}{d h}=\frac{\pi r^{2}}{3}\left[\frac{(h-2 r) 2 h-h^{2}(1)}{(h-2 r)^{2}}\right]\) \(\Rightarrow h(h-4 r)=0\) Now, for minimum or maximum volume \(\frac{\mathrm{dV}}{\mathrm{dh}}=0 \Rightarrow \mathrm{h}(\mathrm{h}-4 \mathrm{r})=0\) \(\Rightarrow \mathrm{h}=4 \mathrm{r} \quad[\because \mathrm{h} \neq 0]\) Also, \(\left(\frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dh}^{2}}\right)_{\mathrm{h}=4 \mathrm{r}}>0\) \(\therefore\) Altitude \(\mathrm{h}=4 \mathrm{r}\).
Shift-I
Application of Derivatives
85621
The largest value of \(y=2 x^{3}-3 x^{2}-12 x+5\) for \(-2 \leq x \leq 2\) occurs at \(x\) is equal to
1 -2
2 -1
3 2
4 4
Explanation:
(B) : Given equation : \(y=2 x^{3}-3 x^{2}-12 x+5\) ....(i) Differentiate equation (i), \(\frac{d y}{d x}=6 x^{2}-6 x-12\) For max value, put \(\left(\frac{d y}{d x}\right)=0\) \(0=6 x^{2}-6 x-12 \Rightarrow x^{2}-x-2=0\) \((x-2)(x+1)=0 \Rightarrow x=-1,2\) Again differentiating equation (ii), we get \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=12 \mathrm{x}-6\) at \(\mathrm{x}=-1 \Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=12(-1)-6\) \(\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-12-6 \Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-18(-\mathrm{ve})\) Therefore, \(\mathrm{y}\) is \(\max\) at \(\mathrm{x}=-1\).
BITSAT-2005
Application of Derivatives
85622
The function \(f(x)=2 x^{3}-3 x^{2}-12 x+4\), has
1 two points of local maximum
2 two points of local minimum
3 one maxima and one minima
4 no maxima or minima
Explanation:
(C) : \(\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{3}-3 \mathrm{x}^{2}-12 \mathrm{x}+4\) \(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=6 \mathrm{x}^{2}-6 \mathrm{x}-12=6\left(\mathrm{x}^{2}-\mathrm{x}-2\right)\) \(=6(\mathrm{x}-2)(\mathrm{x}+1)\) For maxima and minima \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\therefore 6(\mathrm{x}-2)(\mathrm{x}+1)=0 \Rightarrow \mathrm{x}=2,-1\) Now, \(\mathrm{f}^{\prime \prime}(\mathrm{x})=12 \mathrm{x}-6\) At \(\mathrm{x}=2 ; \mathrm{f}^{\prime \prime}(\mathrm{x})=24-6=18>0\) \(\therefore \mathrm{x}=2\), local minima point At \(\mathrm{x}=-1 ; \mathrm{f}^{\prime \prime}(\mathrm{x})=12(-1)-6=-18\lt 0\) \(\therefore \mathrm{x}=-1\) local max. point
BITSAT-2011
Application of Derivatives
85623
If \(a^{2} x^{4}+b^{2} y^{4}=c^{4}\), then the maximum value of xy i
1 \(\frac{\mathrm{c}}{\sqrt{\mathrm{ab}}}\)
2 \(\frac{c^{2}}{2 \sqrt{a b}}\)
3 \(\frac{\mathrm{c}}{2 \sqrt{\mathrm{ab}}}\)
4 \(\frac{\mathrm{c}^{2}}{\sqrt{2 \mathrm{ab}}}\)
Explanation:
(D) : If the sum of two positive quantities is a constant, then their product is maximum, when they are equal. \(\therefore \mathrm{a}^{2} \mathrm{x}^{4} \cdot \mathrm{b}^{2} \mathrm{y}^{4}\) is maximum when \(a^{2} x^{4}=b^{2} y^{4}=\frac{1}{2}\left(a^{2} x^{4}+b^{2} y^{4}\right)=\frac{c^{4}}{2}\) \(\therefore\) Maximum value of \(a^{2} x^{4} \cdot b^{2} y^{4}=\frac{c^{4}}{2} \cdot \frac{c^{4}}{2}=\frac{c^{8}}{4}\) Maximum value of \(x y=\left(\frac{c^{8}}{4 a^{2} b^{2}}\right)^{1 / 4}=\frac{c^{2}}{\sqrt{2 a b}}\)
85618
The altitude for a right circular cone of minimum volume circumscribed about a sphere of radius \(r\) is
1 \(2 \mathrm{r}\)
2 \(3 \mathrm{r}\)
3 \(5 \mathrm{r}\)
4 \(4 \mathrm{r}\)
Explanation:
(D) : Let the radius Of cone be \(\mathrm{R}\) and Height be \(\mathrm{h}\). In \(\triangle \mathrm{ABC}\), \(\sin \theta=\frac{r}{h-r}\) In \(\triangle \mathrm{ADE}\), \(\sin \theta=\frac{\mathrm{R}}{\sqrt{\mathrm{h}^{2}+\mathrm{R}^{2}}}\) Equating (i) and (ii), we get \(\frac{r}{h-r}=\frac{R}{\sqrt{h^{2}+R^{2}}}\) Squaring both sides, we get \(r^{2}\left(h^{2}+R^{2}\right)=R^{2}(h-r)^{2}\) \(\Rightarrow \mathrm{R}^{2}=\frac{\mathrm{r}^{2} \mathrm{~h}}{\mathrm{~h}-2 \mathrm{r}}\) Volume of cone is given by \(\mathrm{V}=\frac{1}{3} \pi \mathrm{R}^{2} \mathrm{~h}\) \(\Rightarrow \mathrm{V}=\frac{1}{3} \pi\left(\frac{\mathrm{r}^{2} \mathrm{~h}}{\mathrm{~h}-2 \mathrm{r}}\right) \mathrm{h}\) [Using (iii)] So, \(\frac{d V}{d h}=\frac{\pi r^{2}}{3}\left[\frac{(h-2 r) 2 h-h^{2}(1)}{(h-2 r)^{2}}\right]\) \(\Rightarrow h(h-4 r)=0\) Now, for minimum or maximum volume \(\frac{\mathrm{dV}}{\mathrm{dh}}=0 \Rightarrow \mathrm{h}(\mathrm{h}-4 \mathrm{r})=0\) \(\Rightarrow \mathrm{h}=4 \mathrm{r} \quad[\because \mathrm{h} \neq 0]\) Also, \(\left(\frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dh}^{2}}\right)_{\mathrm{h}=4 \mathrm{r}}>0\) \(\therefore\) Altitude \(\mathrm{h}=4 \mathrm{r}\).
Shift-I
Application of Derivatives
85621
The largest value of \(y=2 x^{3}-3 x^{2}-12 x+5\) for \(-2 \leq x \leq 2\) occurs at \(x\) is equal to
1 -2
2 -1
3 2
4 4
Explanation:
(B) : Given equation : \(y=2 x^{3}-3 x^{2}-12 x+5\) ....(i) Differentiate equation (i), \(\frac{d y}{d x}=6 x^{2}-6 x-12\) For max value, put \(\left(\frac{d y}{d x}\right)=0\) \(0=6 x^{2}-6 x-12 \Rightarrow x^{2}-x-2=0\) \((x-2)(x+1)=0 \Rightarrow x=-1,2\) Again differentiating equation (ii), we get \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=12 \mathrm{x}-6\) at \(\mathrm{x}=-1 \Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=12(-1)-6\) \(\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-12-6 \Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-18(-\mathrm{ve})\) Therefore, \(\mathrm{y}\) is \(\max\) at \(\mathrm{x}=-1\).
BITSAT-2005
Application of Derivatives
85622
The function \(f(x)=2 x^{3}-3 x^{2}-12 x+4\), has
1 two points of local maximum
2 two points of local minimum
3 one maxima and one minima
4 no maxima or minima
Explanation:
(C) : \(\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{3}-3 \mathrm{x}^{2}-12 \mathrm{x}+4\) \(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=6 \mathrm{x}^{2}-6 \mathrm{x}-12=6\left(\mathrm{x}^{2}-\mathrm{x}-2\right)\) \(=6(\mathrm{x}-2)(\mathrm{x}+1)\) For maxima and minima \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\therefore 6(\mathrm{x}-2)(\mathrm{x}+1)=0 \Rightarrow \mathrm{x}=2,-1\) Now, \(\mathrm{f}^{\prime \prime}(\mathrm{x})=12 \mathrm{x}-6\) At \(\mathrm{x}=2 ; \mathrm{f}^{\prime \prime}(\mathrm{x})=24-6=18>0\) \(\therefore \mathrm{x}=2\), local minima point At \(\mathrm{x}=-1 ; \mathrm{f}^{\prime \prime}(\mathrm{x})=12(-1)-6=-18\lt 0\) \(\therefore \mathrm{x}=-1\) local max. point
BITSAT-2011
Application of Derivatives
85623
If \(a^{2} x^{4}+b^{2} y^{4}=c^{4}\), then the maximum value of xy i
1 \(\frac{\mathrm{c}}{\sqrt{\mathrm{ab}}}\)
2 \(\frac{c^{2}}{2 \sqrt{a b}}\)
3 \(\frac{\mathrm{c}}{2 \sqrt{\mathrm{ab}}}\)
4 \(\frac{\mathrm{c}^{2}}{\sqrt{2 \mathrm{ab}}}\)
Explanation:
(D) : If the sum of two positive quantities is a constant, then their product is maximum, when they are equal. \(\therefore \mathrm{a}^{2} \mathrm{x}^{4} \cdot \mathrm{b}^{2} \mathrm{y}^{4}\) is maximum when \(a^{2} x^{4}=b^{2} y^{4}=\frac{1}{2}\left(a^{2} x^{4}+b^{2} y^{4}\right)=\frac{c^{4}}{2}\) \(\therefore\) Maximum value of \(a^{2} x^{4} \cdot b^{2} y^{4}=\frac{c^{4}}{2} \cdot \frac{c^{4}}{2}=\frac{c^{8}}{4}\) Maximum value of \(x y=\left(\frac{c^{8}}{4 a^{2} b^{2}}\right)^{1 / 4}=\frac{c^{2}}{\sqrt{2 a b}}\)
85618
The altitude for a right circular cone of minimum volume circumscribed about a sphere of radius \(r\) is
1 \(2 \mathrm{r}\)
2 \(3 \mathrm{r}\)
3 \(5 \mathrm{r}\)
4 \(4 \mathrm{r}\)
Explanation:
(D) : Let the radius Of cone be \(\mathrm{R}\) and Height be \(\mathrm{h}\). In \(\triangle \mathrm{ABC}\), \(\sin \theta=\frac{r}{h-r}\) In \(\triangle \mathrm{ADE}\), \(\sin \theta=\frac{\mathrm{R}}{\sqrt{\mathrm{h}^{2}+\mathrm{R}^{2}}}\) Equating (i) and (ii), we get \(\frac{r}{h-r}=\frac{R}{\sqrt{h^{2}+R^{2}}}\) Squaring both sides, we get \(r^{2}\left(h^{2}+R^{2}\right)=R^{2}(h-r)^{2}\) \(\Rightarrow \mathrm{R}^{2}=\frac{\mathrm{r}^{2} \mathrm{~h}}{\mathrm{~h}-2 \mathrm{r}}\) Volume of cone is given by \(\mathrm{V}=\frac{1}{3} \pi \mathrm{R}^{2} \mathrm{~h}\) \(\Rightarrow \mathrm{V}=\frac{1}{3} \pi\left(\frac{\mathrm{r}^{2} \mathrm{~h}}{\mathrm{~h}-2 \mathrm{r}}\right) \mathrm{h}\) [Using (iii)] So, \(\frac{d V}{d h}=\frac{\pi r^{2}}{3}\left[\frac{(h-2 r) 2 h-h^{2}(1)}{(h-2 r)^{2}}\right]\) \(\Rightarrow h(h-4 r)=0\) Now, for minimum or maximum volume \(\frac{\mathrm{dV}}{\mathrm{dh}}=0 \Rightarrow \mathrm{h}(\mathrm{h}-4 \mathrm{r})=0\) \(\Rightarrow \mathrm{h}=4 \mathrm{r} \quad[\because \mathrm{h} \neq 0]\) Also, \(\left(\frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dh}^{2}}\right)_{\mathrm{h}=4 \mathrm{r}}>0\) \(\therefore\) Altitude \(\mathrm{h}=4 \mathrm{r}\).
Shift-I
Application of Derivatives
85621
The largest value of \(y=2 x^{3}-3 x^{2}-12 x+5\) for \(-2 \leq x \leq 2\) occurs at \(x\) is equal to
1 -2
2 -1
3 2
4 4
Explanation:
(B) : Given equation : \(y=2 x^{3}-3 x^{2}-12 x+5\) ....(i) Differentiate equation (i), \(\frac{d y}{d x}=6 x^{2}-6 x-12\) For max value, put \(\left(\frac{d y}{d x}\right)=0\) \(0=6 x^{2}-6 x-12 \Rightarrow x^{2}-x-2=0\) \((x-2)(x+1)=0 \Rightarrow x=-1,2\) Again differentiating equation (ii), we get \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=12 \mathrm{x}-6\) at \(\mathrm{x}=-1 \Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=12(-1)-6\) \(\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-12-6 \Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-18(-\mathrm{ve})\) Therefore, \(\mathrm{y}\) is \(\max\) at \(\mathrm{x}=-1\).
BITSAT-2005
Application of Derivatives
85622
The function \(f(x)=2 x^{3}-3 x^{2}-12 x+4\), has
1 two points of local maximum
2 two points of local minimum
3 one maxima and one minima
4 no maxima or minima
Explanation:
(C) : \(\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{3}-3 \mathrm{x}^{2}-12 \mathrm{x}+4\) \(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=6 \mathrm{x}^{2}-6 \mathrm{x}-12=6\left(\mathrm{x}^{2}-\mathrm{x}-2\right)\) \(=6(\mathrm{x}-2)(\mathrm{x}+1)\) For maxima and minima \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\therefore 6(\mathrm{x}-2)(\mathrm{x}+1)=0 \Rightarrow \mathrm{x}=2,-1\) Now, \(\mathrm{f}^{\prime \prime}(\mathrm{x})=12 \mathrm{x}-6\) At \(\mathrm{x}=2 ; \mathrm{f}^{\prime \prime}(\mathrm{x})=24-6=18>0\) \(\therefore \mathrm{x}=2\), local minima point At \(\mathrm{x}=-1 ; \mathrm{f}^{\prime \prime}(\mathrm{x})=12(-1)-6=-18\lt 0\) \(\therefore \mathrm{x}=-1\) local max. point
BITSAT-2011
Application of Derivatives
85623
If \(a^{2} x^{4}+b^{2} y^{4}=c^{4}\), then the maximum value of xy i
1 \(\frac{\mathrm{c}}{\sqrt{\mathrm{ab}}}\)
2 \(\frac{c^{2}}{2 \sqrt{a b}}\)
3 \(\frac{\mathrm{c}}{2 \sqrt{\mathrm{ab}}}\)
4 \(\frac{\mathrm{c}^{2}}{\sqrt{2 \mathrm{ab}}}\)
Explanation:
(D) : If the sum of two positive quantities is a constant, then their product is maximum, when they are equal. \(\therefore \mathrm{a}^{2} \mathrm{x}^{4} \cdot \mathrm{b}^{2} \mathrm{y}^{4}\) is maximum when \(a^{2} x^{4}=b^{2} y^{4}=\frac{1}{2}\left(a^{2} x^{4}+b^{2} y^{4}\right)=\frac{c^{4}}{2}\) \(\therefore\) Maximum value of \(a^{2} x^{4} \cdot b^{2} y^{4}=\frac{c^{4}}{2} \cdot \frac{c^{4}}{2}=\frac{c^{8}}{4}\) Maximum value of \(x y=\left(\frac{c^{8}}{4 a^{2} b^{2}}\right)^{1 / 4}=\frac{c^{2}}{\sqrt{2 a b}}\)
85618
The altitude for a right circular cone of minimum volume circumscribed about a sphere of radius \(r\) is
1 \(2 \mathrm{r}\)
2 \(3 \mathrm{r}\)
3 \(5 \mathrm{r}\)
4 \(4 \mathrm{r}\)
Explanation:
(D) : Let the radius Of cone be \(\mathrm{R}\) and Height be \(\mathrm{h}\). In \(\triangle \mathrm{ABC}\), \(\sin \theta=\frac{r}{h-r}\) In \(\triangle \mathrm{ADE}\), \(\sin \theta=\frac{\mathrm{R}}{\sqrt{\mathrm{h}^{2}+\mathrm{R}^{2}}}\) Equating (i) and (ii), we get \(\frac{r}{h-r}=\frac{R}{\sqrt{h^{2}+R^{2}}}\) Squaring both sides, we get \(r^{2}\left(h^{2}+R^{2}\right)=R^{2}(h-r)^{2}\) \(\Rightarrow \mathrm{R}^{2}=\frac{\mathrm{r}^{2} \mathrm{~h}}{\mathrm{~h}-2 \mathrm{r}}\) Volume of cone is given by \(\mathrm{V}=\frac{1}{3} \pi \mathrm{R}^{2} \mathrm{~h}\) \(\Rightarrow \mathrm{V}=\frac{1}{3} \pi\left(\frac{\mathrm{r}^{2} \mathrm{~h}}{\mathrm{~h}-2 \mathrm{r}}\right) \mathrm{h}\) [Using (iii)] So, \(\frac{d V}{d h}=\frac{\pi r^{2}}{3}\left[\frac{(h-2 r) 2 h-h^{2}(1)}{(h-2 r)^{2}}\right]\) \(\Rightarrow h(h-4 r)=0\) Now, for minimum or maximum volume \(\frac{\mathrm{dV}}{\mathrm{dh}}=0 \Rightarrow \mathrm{h}(\mathrm{h}-4 \mathrm{r})=0\) \(\Rightarrow \mathrm{h}=4 \mathrm{r} \quad[\because \mathrm{h} \neq 0]\) Also, \(\left(\frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dh}^{2}}\right)_{\mathrm{h}=4 \mathrm{r}}>0\) \(\therefore\) Altitude \(\mathrm{h}=4 \mathrm{r}\).
Shift-I
Application of Derivatives
85621
The largest value of \(y=2 x^{3}-3 x^{2}-12 x+5\) for \(-2 \leq x \leq 2\) occurs at \(x\) is equal to
1 -2
2 -1
3 2
4 4
Explanation:
(B) : Given equation : \(y=2 x^{3}-3 x^{2}-12 x+5\) ....(i) Differentiate equation (i), \(\frac{d y}{d x}=6 x^{2}-6 x-12\) For max value, put \(\left(\frac{d y}{d x}\right)=0\) \(0=6 x^{2}-6 x-12 \Rightarrow x^{2}-x-2=0\) \((x-2)(x+1)=0 \Rightarrow x=-1,2\) Again differentiating equation (ii), we get \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=12 \mathrm{x}-6\) at \(\mathrm{x}=-1 \Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=12(-1)-6\) \(\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-12-6 \Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-18(-\mathrm{ve})\) Therefore, \(\mathrm{y}\) is \(\max\) at \(\mathrm{x}=-1\).
BITSAT-2005
Application of Derivatives
85622
The function \(f(x)=2 x^{3}-3 x^{2}-12 x+4\), has
1 two points of local maximum
2 two points of local minimum
3 one maxima and one minima
4 no maxima or minima
Explanation:
(C) : \(\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{3}-3 \mathrm{x}^{2}-12 \mathrm{x}+4\) \(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=6 \mathrm{x}^{2}-6 \mathrm{x}-12=6\left(\mathrm{x}^{2}-\mathrm{x}-2\right)\) \(=6(\mathrm{x}-2)(\mathrm{x}+1)\) For maxima and minima \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\therefore 6(\mathrm{x}-2)(\mathrm{x}+1)=0 \Rightarrow \mathrm{x}=2,-1\) Now, \(\mathrm{f}^{\prime \prime}(\mathrm{x})=12 \mathrm{x}-6\) At \(\mathrm{x}=2 ; \mathrm{f}^{\prime \prime}(\mathrm{x})=24-6=18>0\) \(\therefore \mathrm{x}=2\), local minima point At \(\mathrm{x}=-1 ; \mathrm{f}^{\prime \prime}(\mathrm{x})=12(-1)-6=-18\lt 0\) \(\therefore \mathrm{x}=-1\) local max. point
BITSAT-2011
Application of Derivatives
85623
If \(a^{2} x^{4}+b^{2} y^{4}=c^{4}\), then the maximum value of xy i
1 \(\frac{\mathrm{c}}{\sqrt{\mathrm{ab}}}\)
2 \(\frac{c^{2}}{2 \sqrt{a b}}\)
3 \(\frac{\mathrm{c}}{2 \sqrt{\mathrm{ab}}}\)
4 \(\frac{\mathrm{c}^{2}}{\sqrt{2 \mathrm{ab}}}\)
Explanation:
(D) : If the sum of two positive quantities is a constant, then their product is maximum, when they are equal. \(\therefore \mathrm{a}^{2} \mathrm{x}^{4} \cdot \mathrm{b}^{2} \mathrm{y}^{4}\) is maximum when \(a^{2} x^{4}=b^{2} y^{4}=\frac{1}{2}\left(a^{2} x^{4}+b^{2} y^{4}\right)=\frac{c^{4}}{2}\) \(\therefore\) Maximum value of \(a^{2} x^{4} \cdot b^{2} y^{4}=\frac{c^{4}}{2} \cdot \frac{c^{4}}{2}=\frac{c^{8}}{4}\) Maximum value of \(x y=\left(\frac{c^{8}}{4 a^{2} b^{2}}\right)^{1 / 4}=\frac{c^{2}}{\sqrt{2 a b}}\)
