85596
If \(x+y=\frac{\pi}{2}\), then the maximum value of \(\sin x\), \(\sin \mathrm{y}\) is
1 \(\frac{-1}{\sqrt{2}}\)
2 \(\frac{1}{\sqrt{2}}\)
3 \(\frac{1}{2}\)
4 \(\frac{-1}{2}\)
Explanation:
(C) : Given, \(\mathrm{x}+\mathrm{y}=\frac{\pi}{2}\) \(\therefore \quad \sin \mathrm{x} \sin \mathrm{y}\) \(=\sin \mathrm{x} \sin \left(\frac{\pi}{2}-\mathrm{x}\right)\) \(=\sin \mathrm{x} \cos \mathrm{x}\) \(=\frac{1}{2} 2 \sin \mathrm{x} \cos \mathrm{x}\) \(=\frac{1}{2} \sin 2 \mathrm{x}\) maximum value obtained when \(\sin 2 x\) is maximum \(\sin 2 x=1\) \(\therefore\) maximum value is \(\frac{1}{2}\)
MHT CET-2020
Application of Derivatives
85597
If rectangles are inscribed in a circle of radius \(r\) units. Then the dimensions of the rectangle which has maximum area are
1 2 runits, \(r\) units,
2 2 runits, \(\sqrt{2} \mathrm{r}\) units,
3 \(\sqrt{2} \mathrm{r}\) units, \(\sqrt{2} \mathrm{r}\) units
4 runits, \(\sqrt{2} \mathrm{r}\) units,
Explanation:
(C) : Let PQRS be the rectangle inscribed in circle of radius \(\mathrm{r}\) Let, \(\quad \mathrm{PQ}=\mathrm{x}\) and \(\mathrm{QR}=\mathrm{y}\) Now, by Pythagoras theorem \(x^{2}+y^{2}=4 r^{2}\) \(y=\sqrt{4 r^{2}-x^{2}}\) Area of rectangle \(A=x y\) \(A=x \sqrt{4 r^{2}-x^{2}}\) Squaring both side we get - \(A^{2}=x^{2}\left(4 r^{2}-x^{2}\right)\) Differentiating w.r.t. \(x\), we get- \(\frac{d\left(A^{2}\right)}{d x}=8 x^{2}-4 x^{3}\) For maxima or minima \(\frac{d\left(A^{2}\right)}{d x}=0\) \(8 x r^{2}-4 x^{3}=0\) \(8 r^{2}-4 x^{2}=0\) \(x=\sqrt{2} r\) Again differentiating w.r.t \(x\), we get- \(\frac{d^{2}\left(A^{2}\right)}{d x^{2}}=8 r^{2}-12 x^{2}\) At, \(\quad \mathrm{x}=\sqrt{2} \mathrm{r}\) then, \(\frac{\mathrm{d}^{2}\left(\mathrm{~A}^{2}\right)}{\mathrm{dx}^{2}}=8 \mathrm{r}^{2}-12(\sqrt{2} \mathrm{r})^{2}\) \(8 \mathrm{r}^{2}-24 \mathrm{r}^{2}\lt 0\) \(-16 \mathrm{r}^{2}\lt 0\) \(A\) is maximum of \(x=\sqrt{2} r\) \(y=\sqrt{2} \mathrm{r}=\mathrm{x}\) Hence the dimensions of the rectangle are \(\sqrt{2} \mathrm{r}\) and \(\sqrt{2} \mathrm{r}\)
MHT CET-2020
Application of Derivatives
85598
20 meters wire is available to fence a flower bed in the form of a circular sector. If the flower bed should have the greatest possible area, then the radius of the circle is
1 \(5 \mathrm{~m}\)
2 \(2 \mathrm{~m}\)
3 \(4 \mathrm{~m}\)
4 \(10 \mathrm{~m}\)
Explanation:
(A) : Given perimeter of circular sector \(20=2 \mathrm{r}+l \Rightarrow l=20-2 \mathrm{r}\) Now area, \(\mathrm{A}=\frac{1}{2} \times l \times \mathrm{r} \Rightarrow \mathrm{A}=\frac{1}{2}(20-2 \mathrm{r}) \mathrm{r}\) Differentiating both sides w.r.t. \(x\), we get - \(\frac{\mathrm{dA}}{\mathrm{dr}} =\frac{1}{2}[20-2 \mathrm{r}+\mathrm{r}(-2)]\) \(\frac{\mathrm{dA}}{\mathrm{dr}} =\frac{1}{2}[20-4 \mathrm{r}]\) For greatest area \(\frac{\mathrm{dA}}{\mathrm{dr}}=0 \Rightarrow \frac{1}{2}(20-4 \mathrm{r})=0\) \(4 \mathrm{r}=20\) \(\mathrm{r}=5\)
MHT CET-2022
Application of Derivatives
85599
A metal wire 108 meters long is bent to form a rectangle. If the area of the rectangle is maximum, then its dimensions are
1 \(28 \mathrm{~m}, 28 \mathrm{~m}\)
2 \(28 \mathrm{~m}, 26 \mathrm{~m}\)
3 \(25 \mathrm{~m}, 25 \mathrm{~m}\)
4 \(27 \mathrm{~m}, 27 \mathrm{~m}\)
Explanation:
(D) : Let the length and breadth of a rectangle be \(x\) and \(y\) Given perimeter of rectangle \(2(x+y)=108\) \(x+y=54\) \(y=54-x\) Now area of the rectangle \(A=x y\) \(A=x(54-x)\) \(A=54 x-x^{2}\) Differentiating w.r.t. \(\mathrm{x}\) we get- \(\frac{\mathrm{dA}}{\mathrm{dx}}=54-2 \mathrm{x}\) For maximum, \(\frac{\mathrm{dA}}{\mathrm{dx}}=0\) \(54-2 x=0\) \(x=27\) \(y=27\)
MHT CET-2020
Application of Derivatives
85600
The perimeter of a triangle is \(10 \mathrm{~cm}\). If one of its side is \(4 \mathrm{~cm}\), then remaining sides of the triangle, when area of triangle is maximum are
1 \(2 \mathrm{~cm}, 4 \mathrm{~cm}\)
2 \(3.6 \mathrm{~cm}, 2.4 \mathrm{~cm}\)
3 \(3 \mathrm{~cm}, 3 \mathrm{~cm}\)
4 \(5 \mathrm{~cm}, 1 \mathrm{~cm}\)
Explanation:
(C) : Given, that perimeter of triangle \(=10\) \(a+b+c=10\) And, \(a=4\) then \(b+c=6\) We know that, area of triangle \(\Delta=\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}\) And, \(s=\frac{a+b+c}{2}\) \(\mathrm{S}=\frac{10}{2}\) \(\mathrm{~S}=5\) \(\Delta=\sqrt{5(5-4)(5-\mathrm{b})(5-\mathrm{c})}=\sqrt{5 \times 1 \times(5-\mathrm{b})(5-\mathrm{c})}\) \(\Delta^{2}=5(5-\mathrm{b})(5-\mathrm{c})\) \(\mathrm{A}^{2}=5(5-\mathrm{b})(5-6+\mathrm{b})\) \(\mathrm{A}^{2}=5(5-\mathrm{b})(\mathrm{b}-1)\) Differentiating both sides w.r.t. b, we get - \(\frac{\mathrm{dA}^{2}}{\mathrm{db}}=(5-\mathrm{b})+(\mathrm{b}-1)(-1)\) For maximum area of triangle \(\frac{\mathrm{dA}^{2}}{\mathrm{db}}=0\) \(5-\mathrm{b}+1-\mathrm{b}=0\) \(-2 \mathrm{~b}+6=0\) \(\mathrm{~b}=3\) Then, \(\mathrm{c}=3\) Then remaining sides are, 3,3
85596
If \(x+y=\frac{\pi}{2}\), then the maximum value of \(\sin x\), \(\sin \mathrm{y}\) is
1 \(\frac{-1}{\sqrt{2}}\)
2 \(\frac{1}{\sqrt{2}}\)
3 \(\frac{1}{2}\)
4 \(\frac{-1}{2}\)
Explanation:
(C) : Given, \(\mathrm{x}+\mathrm{y}=\frac{\pi}{2}\) \(\therefore \quad \sin \mathrm{x} \sin \mathrm{y}\) \(=\sin \mathrm{x} \sin \left(\frac{\pi}{2}-\mathrm{x}\right)\) \(=\sin \mathrm{x} \cos \mathrm{x}\) \(=\frac{1}{2} 2 \sin \mathrm{x} \cos \mathrm{x}\) \(=\frac{1}{2} \sin 2 \mathrm{x}\) maximum value obtained when \(\sin 2 x\) is maximum \(\sin 2 x=1\) \(\therefore\) maximum value is \(\frac{1}{2}\)
MHT CET-2020
Application of Derivatives
85597
If rectangles are inscribed in a circle of radius \(r\) units. Then the dimensions of the rectangle which has maximum area are
1 2 runits, \(r\) units,
2 2 runits, \(\sqrt{2} \mathrm{r}\) units,
3 \(\sqrt{2} \mathrm{r}\) units, \(\sqrt{2} \mathrm{r}\) units
4 runits, \(\sqrt{2} \mathrm{r}\) units,
Explanation:
(C) : Let PQRS be the rectangle inscribed in circle of radius \(\mathrm{r}\) Let, \(\quad \mathrm{PQ}=\mathrm{x}\) and \(\mathrm{QR}=\mathrm{y}\) Now, by Pythagoras theorem \(x^{2}+y^{2}=4 r^{2}\) \(y=\sqrt{4 r^{2}-x^{2}}\) Area of rectangle \(A=x y\) \(A=x \sqrt{4 r^{2}-x^{2}}\) Squaring both side we get - \(A^{2}=x^{2}\left(4 r^{2}-x^{2}\right)\) Differentiating w.r.t. \(x\), we get- \(\frac{d\left(A^{2}\right)}{d x}=8 x^{2}-4 x^{3}\) For maxima or minima \(\frac{d\left(A^{2}\right)}{d x}=0\) \(8 x r^{2}-4 x^{3}=0\) \(8 r^{2}-4 x^{2}=0\) \(x=\sqrt{2} r\) Again differentiating w.r.t \(x\), we get- \(\frac{d^{2}\left(A^{2}\right)}{d x^{2}}=8 r^{2}-12 x^{2}\) At, \(\quad \mathrm{x}=\sqrt{2} \mathrm{r}\) then, \(\frac{\mathrm{d}^{2}\left(\mathrm{~A}^{2}\right)}{\mathrm{dx}^{2}}=8 \mathrm{r}^{2}-12(\sqrt{2} \mathrm{r})^{2}\) \(8 \mathrm{r}^{2}-24 \mathrm{r}^{2}\lt 0\) \(-16 \mathrm{r}^{2}\lt 0\) \(A\) is maximum of \(x=\sqrt{2} r\) \(y=\sqrt{2} \mathrm{r}=\mathrm{x}\) Hence the dimensions of the rectangle are \(\sqrt{2} \mathrm{r}\) and \(\sqrt{2} \mathrm{r}\)
MHT CET-2020
Application of Derivatives
85598
20 meters wire is available to fence a flower bed in the form of a circular sector. If the flower bed should have the greatest possible area, then the radius of the circle is
1 \(5 \mathrm{~m}\)
2 \(2 \mathrm{~m}\)
3 \(4 \mathrm{~m}\)
4 \(10 \mathrm{~m}\)
Explanation:
(A) : Given perimeter of circular sector \(20=2 \mathrm{r}+l \Rightarrow l=20-2 \mathrm{r}\) Now area, \(\mathrm{A}=\frac{1}{2} \times l \times \mathrm{r} \Rightarrow \mathrm{A}=\frac{1}{2}(20-2 \mathrm{r}) \mathrm{r}\) Differentiating both sides w.r.t. \(x\), we get - \(\frac{\mathrm{dA}}{\mathrm{dr}} =\frac{1}{2}[20-2 \mathrm{r}+\mathrm{r}(-2)]\) \(\frac{\mathrm{dA}}{\mathrm{dr}} =\frac{1}{2}[20-4 \mathrm{r}]\) For greatest area \(\frac{\mathrm{dA}}{\mathrm{dr}}=0 \Rightarrow \frac{1}{2}(20-4 \mathrm{r})=0\) \(4 \mathrm{r}=20\) \(\mathrm{r}=5\)
MHT CET-2022
Application of Derivatives
85599
A metal wire 108 meters long is bent to form a rectangle. If the area of the rectangle is maximum, then its dimensions are
1 \(28 \mathrm{~m}, 28 \mathrm{~m}\)
2 \(28 \mathrm{~m}, 26 \mathrm{~m}\)
3 \(25 \mathrm{~m}, 25 \mathrm{~m}\)
4 \(27 \mathrm{~m}, 27 \mathrm{~m}\)
Explanation:
(D) : Let the length and breadth of a rectangle be \(x\) and \(y\) Given perimeter of rectangle \(2(x+y)=108\) \(x+y=54\) \(y=54-x\) Now area of the rectangle \(A=x y\) \(A=x(54-x)\) \(A=54 x-x^{2}\) Differentiating w.r.t. \(\mathrm{x}\) we get- \(\frac{\mathrm{dA}}{\mathrm{dx}}=54-2 \mathrm{x}\) For maximum, \(\frac{\mathrm{dA}}{\mathrm{dx}}=0\) \(54-2 x=0\) \(x=27\) \(y=27\)
MHT CET-2020
Application of Derivatives
85600
The perimeter of a triangle is \(10 \mathrm{~cm}\). If one of its side is \(4 \mathrm{~cm}\), then remaining sides of the triangle, when area of triangle is maximum are
1 \(2 \mathrm{~cm}, 4 \mathrm{~cm}\)
2 \(3.6 \mathrm{~cm}, 2.4 \mathrm{~cm}\)
3 \(3 \mathrm{~cm}, 3 \mathrm{~cm}\)
4 \(5 \mathrm{~cm}, 1 \mathrm{~cm}\)
Explanation:
(C) : Given, that perimeter of triangle \(=10\) \(a+b+c=10\) And, \(a=4\) then \(b+c=6\) We know that, area of triangle \(\Delta=\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}\) And, \(s=\frac{a+b+c}{2}\) \(\mathrm{S}=\frac{10}{2}\) \(\mathrm{~S}=5\) \(\Delta=\sqrt{5(5-4)(5-\mathrm{b})(5-\mathrm{c})}=\sqrt{5 \times 1 \times(5-\mathrm{b})(5-\mathrm{c})}\) \(\Delta^{2}=5(5-\mathrm{b})(5-\mathrm{c})\) \(\mathrm{A}^{2}=5(5-\mathrm{b})(5-6+\mathrm{b})\) \(\mathrm{A}^{2}=5(5-\mathrm{b})(\mathrm{b}-1)\) Differentiating both sides w.r.t. b, we get - \(\frac{\mathrm{dA}^{2}}{\mathrm{db}}=(5-\mathrm{b})+(\mathrm{b}-1)(-1)\) For maximum area of triangle \(\frac{\mathrm{dA}^{2}}{\mathrm{db}}=0\) \(5-\mathrm{b}+1-\mathrm{b}=0\) \(-2 \mathrm{~b}+6=0\) \(\mathrm{~b}=3\) Then, \(\mathrm{c}=3\) Then remaining sides are, 3,3
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Application of Derivatives
85596
If \(x+y=\frac{\pi}{2}\), then the maximum value of \(\sin x\), \(\sin \mathrm{y}\) is
1 \(\frac{-1}{\sqrt{2}}\)
2 \(\frac{1}{\sqrt{2}}\)
3 \(\frac{1}{2}\)
4 \(\frac{-1}{2}\)
Explanation:
(C) : Given, \(\mathrm{x}+\mathrm{y}=\frac{\pi}{2}\) \(\therefore \quad \sin \mathrm{x} \sin \mathrm{y}\) \(=\sin \mathrm{x} \sin \left(\frac{\pi}{2}-\mathrm{x}\right)\) \(=\sin \mathrm{x} \cos \mathrm{x}\) \(=\frac{1}{2} 2 \sin \mathrm{x} \cos \mathrm{x}\) \(=\frac{1}{2} \sin 2 \mathrm{x}\) maximum value obtained when \(\sin 2 x\) is maximum \(\sin 2 x=1\) \(\therefore\) maximum value is \(\frac{1}{2}\)
MHT CET-2020
Application of Derivatives
85597
If rectangles are inscribed in a circle of radius \(r\) units. Then the dimensions of the rectangle which has maximum area are
1 2 runits, \(r\) units,
2 2 runits, \(\sqrt{2} \mathrm{r}\) units,
3 \(\sqrt{2} \mathrm{r}\) units, \(\sqrt{2} \mathrm{r}\) units
4 runits, \(\sqrt{2} \mathrm{r}\) units,
Explanation:
(C) : Let PQRS be the rectangle inscribed in circle of radius \(\mathrm{r}\) Let, \(\quad \mathrm{PQ}=\mathrm{x}\) and \(\mathrm{QR}=\mathrm{y}\) Now, by Pythagoras theorem \(x^{2}+y^{2}=4 r^{2}\) \(y=\sqrt{4 r^{2}-x^{2}}\) Area of rectangle \(A=x y\) \(A=x \sqrt{4 r^{2}-x^{2}}\) Squaring both side we get - \(A^{2}=x^{2}\left(4 r^{2}-x^{2}\right)\) Differentiating w.r.t. \(x\), we get- \(\frac{d\left(A^{2}\right)}{d x}=8 x^{2}-4 x^{3}\) For maxima or minima \(\frac{d\left(A^{2}\right)}{d x}=0\) \(8 x r^{2}-4 x^{3}=0\) \(8 r^{2}-4 x^{2}=0\) \(x=\sqrt{2} r\) Again differentiating w.r.t \(x\), we get- \(\frac{d^{2}\left(A^{2}\right)}{d x^{2}}=8 r^{2}-12 x^{2}\) At, \(\quad \mathrm{x}=\sqrt{2} \mathrm{r}\) then, \(\frac{\mathrm{d}^{2}\left(\mathrm{~A}^{2}\right)}{\mathrm{dx}^{2}}=8 \mathrm{r}^{2}-12(\sqrt{2} \mathrm{r})^{2}\) \(8 \mathrm{r}^{2}-24 \mathrm{r}^{2}\lt 0\) \(-16 \mathrm{r}^{2}\lt 0\) \(A\) is maximum of \(x=\sqrt{2} r\) \(y=\sqrt{2} \mathrm{r}=\mathrm{x}\) Hence the dimensions of the rectangle are \(\sqrt{2} \mathrm{r}\) and \(\sqrt{2} \mathrm{r}\)
MHT CET-2020
Application of Derivatives
85598
20 meters wire is available to fence a flower bed in the form of a circular sector. If the flower bed should have the greatest possible area, then the radius of the circle is
1 \(5 \mathrm{~m}\)
2 \(2 \mathrm{~m}\)
3 \(4 \mathrm{~m}\)
4 \(10 \mathrm{~m}\)
Explanation:
(A) : Given perimeter of circular sector \(20=2 \mathrm{r}+l \Rightarrow l=20-2 \mathrm{r}\) Now area, \(\mathrm{A}=\frac{1}{2} \times l \times \mathrm{r} \Rightarrow \mathrm{A}=\frac{1}{2}(20-2 \mathrm{r}) \mathrm{r}\) Differentiating both sides w.r.t. \(x\), we get - \(\frac{\mathrm{dA}}{\mathrm{dr}} =\frac{1}{2}[20-2 \mathrm{r}+\mathrm{r}(-2)]\) \(\frac{\mathrm{dA}}{\mathrm{dr}} =\frac{1}{2}[20-4 \mathrm{r}]\) For greatest area \(\frac{\mathrm{dA}}{\mathrm{dr}}=0 \Rightarrow \frac{1}{2}(20-4 \mathrm{r})=0\) \(4 \mathrm{r}=20\) \(\mathrm{r}=5\)
MHT CET-2022
Application of Derivatives
85599
A metal wire 108 meters long is bent to form a rectangle. If the area of the rectangle is maximum, then its dimensions are
1 \(28 \mathrm{~m}, 28 \mathrm{~m}\)
2 \(28 \mathrm{~m}, 26 \mathrm{~m}\)
3 \(25 \mathrm{~m}, 25 \mathrm{~m}\)
4 \(27 \mathrm{~m}, 27 \mathrm{~m}\)
Explanation:
(D) : Let the length and breadth of a rectangle be \(x\) and \(y\) Given perimeter of rectangle \(2(x+y)=108\) \(x+y=54\) \(y=54-x\) Now area of the rectangle \(A=x y\) \(A=x(54-x)\) \(A=54 x-x^{2}\) Differentiating w.r.t. \(\mathrm{x}\) we get- \(\frac{\mathrm{dA}}{\mathrm{dx}}=54-2 \mathrm{x}\) For maximum, \(\frac{\mathrm{dA}}{\mathrm{dx}}=0\) \(54-2 x=0\) \(x=27\) \(y=27\)
MHT CET-2020
Application of Derivatives
85600
The perimeter of a triangle is \(10 \mathrm{~cm}\). If one of its side is \(4 \mathrm{~cm}\), then remaining sides of the triangle, when area of triangle is maximum are
1 \(2 \mathrm{~cm}, 4 \mathrm{~cm}\)
2 \(3.6 \mathrm{~cm}, 2.4 \mathrm{~cm}\)
3 \(3 \mathrm{~cm}, 3 \mathrm{~cm}\)
4 \(5 \mathrm{~cm}, 1 \mathrm{~cm}\)
Explanation:
(C) : Given, that perimeter of triangle \(=10\) \(a+b+c=10\) And, \(a=4\) then \(b+c=6\) We know that, area of triangle \(\Delta=\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}\) And, \(s=\frac{a+b+c}{2}\) \(\mathrm{S}=\frac{10}{2}\) \(\mathrm{~S}=5\) \(\Delta=\sqrt{5(5-4)(5-\mathrm{b})(5-\mathrm{c})}=\sqrt{5 \times 1 \times(5-\mathrm{b})(5-\mathrm{c})}\) \(\Delta^{2}=5(5-\mathrm{b})(5-\mathrm{c})\) \(\mathrm{A}^{2}=5(5-\mathrm{b})(5-6+\mathrm{b})\) \(\mathrm{A}^{2}=5(5-\mathrm{b})(\mathrm{b}-1)\) Differentiating both sides w.r.t. b, we get - \(\frac{\mathrm{dA}^{2}}{\mathrm{db}}=(5-\mathrm{b})+(\mathrm{b}-1)(-1)\) For maximum area of triangle \(\frac{\mathrm{dA}^{2}}{\mathrm{db}}=0\) \(5-\mathrm{b}+1-\mathrm{b}=0\) \(-2 \mathrm{~b}+6=0\) \(\mathrm{~b}=3\) Then, \(\mathrm{c}=3\) Then remaining sides are, 3,3
85596
If \(x+y=\frac{\pi}{2}\), then the maximum value of \(\sin x\), \(\sin \mathrm{y}\) is
1 \(\frac{-1}{\sqrt{2}}\)
2 \(\frac{1}{\sqrt{2}}\)
3 \(\frac{1}{2}\)
4 \(\frac{-1}{2}\)
Explanation:
(C) : Given, \(\mathrm{x}+\mathrm{y}=\frac{\pi}{2}\) \(\therefore \quad \sin \mathrm{x} \sin \mathrm{y}\) \(=\sin \mathrm{x} \sin \left(\frac{\pi}{2}-\mathrm{x}\right)\) \(=\sin \mathrm{x} \cos \mathrm{x}\) \(=\frac{1}{2} 2 \sin \mathrm{x} \cos \mathrm{x}\) \(=\frac{1}{2} \sin 2 \mathrm{x}\) maximum value obtained when \(\sin 2 x\) is maximum \(\sin 2 x=1\) \(\therefore\) maximum value is \(\frac{1}{2}\)
MHT CET-2020
Application of Derivatives
85597
If rectangles are inscribed in a circle of radius \(r\) units. Then the dimensions of the rectangle which has maximum area are
1 2 runits, \(r\) units,
2 2 runits, \(\sqrt{2} \mathrm{r}\) units,
3 \(\sqrt{2} \mathrm{r}\) units, \(\sqrt{2} \mathrm{r}\) units
4 runits, \(\sqrt{2} \mathrm{r}\) units,
Explanation:
(C) : Let PQRS be the rectangle inscribed in circle of radius \(\mathrm{r}\) Let, \(\quad \mathrm{PQ}=\mathrm{x}\) and \(\mathrm{QR}=\mathrm{y}\) Now, by Pythagoras theorem \(x^{2}+y^{2}=4 r^{2}\) \(y=\sqrt{4 r^{2}-x^{2}}\) Area of rectangle \(A=x y\) \(A=x \sqrt{4 r^{2}-x^{2}}\) Squaring both side we get - \(A^{2}=x^{2}\left(4 r^{2}-x^{2}\right)\) Differentiating w.r.t. \(x\), we get- \(\frac{d\left(A^{2}\right)}{d x}=8 x^{2}-4 x^{3}\) For maxima or minima \(\frac{d\left(A^{2}\right)}{d x}=0\) \(8 x r^{2}-4 x^{3}=0\) \(8 r^{2}-4 x^{2}=0\) \(x=\sqrt{2} r\) Again differentiating w.r.t \(x\), we get- \(\frac{d^{2}\left(A^{2}\right)}{d x^{2}}=8 r^{2}-12 x^{2}\) At, \(\quad \mathrm{x}=\sqrt{2} \mathrm{r}\) then, \(\frac{\mathrm{d}^{2}\left(\mathrm{~A}^{2}\right)}{\mathrm{dx}^{2}}=8 \mathrm{r}^{2}-12(\sqrt{2} \mathrm{r})^{2}\) \(8 \mathrm{r}^{2}-24 \mathrm{r}^{2}\lt 0\) \(-16 \mathrm{r}^{2}\lt 0\) \(A\) is maximum of \(x=\sqrt{2} r\) \(y=\sqrt{2} \mathrm{r}=\mathrm{x}\) Hence the dimensions of the rectangle are \(\sqrt{2} \mathrm{r}\) and \(\sqrt{2} \mathrm{r}\)
MHT CET-2020
Application of Derivatives
85598
20 meters wire is available to fence a flower bed in the form of a circular sector. If the flower bed should have the greatest possible area, then the radius of the circle is
1 \(5 \mathrm{~m}\)
2 \(2 \mathrm{~m}\)
3 \(4 \mathrm{~m}\)
4 \(10 \mathrm{~m}\)
Explanation:
(A) : Given perimeter of circular sector \(20=2 \mathrm{r}+l \Rightarrow l=20-2 \mathrm{r}\) Now area, \(\mathrm{A}=\frac{1}{2} \times l \times \mathrm{r} \Rightarrow \mathrm{A}=\frac{1}{2}(20-2 \mathrm{r}) \mathrm{r}\) Differentiating both sides w.r.t. \(x\), we get - \(\frac{\mathrm{dA}}{\mathrm{dr}} =\frac{1}{2}[20-2 \mathrm{r}+\mathrm{r}(-2)]\) \(\frac{\mathrm{dA}}{\mathrm{dr}} =\frac{1}{2}[20-4 \mathrm{r}]\) For greatest area \(\frac{\mathrm{dA}}{\mathrm{dr}}=0 \Rightarrow \frac{1}{2}(20-4 \mathrm{r})=0\) \(4 \mathrm{r}=20\) \(\mathrm{r}=5\)
MHT CET-2022
Application of Derivatives
85599
A metal wire 108 meters long is bent to form a rectangle. If the area of the rectangle is maximum, then its dimensions are
1 \(28 \mathrm{~m}, 28 \mathrm{~m}\)
2 \(28 \mathrm{~m}, 26 \mathrm{~m}\)
3 \(25 \mathrm{~m}, 25 \mathrm{~m}\)
4 \(27 \mathrm{~m}, 27 \mathrm{~m}\)
Explanation:
(D) : Let the length and breadth of a rectangle be \(x\) and \(y\) Given perimeter of rectangle \(2(x+y)=108\) \(x+y=54\) \(y=54-x\) Now area of the rectangle \(A=x y\) \(A=x(54-x)\) \(A=54 x-x^{2}\) Differentiating w.r.t. \(\mathrm{x}\) we get- \(\frac{\mathrm{dA}}{\mathrm{dx}}=54-2 \mathrm{x}\) For maximum, \(\frac{\mathrm{dA}}{\mathrm{dx}}=0\) \(54-2 x=0\) \(x=27\) \(y=27\)
MHT CET-2020
Application of Derivatives
85600
The perimeter of a triangle is \(10 \mathrm{~cm}\). If one of its side is \(4 \mathrm{~cm}\), then remaining sides of the triangle, when area of triangle is maximum are
1 \(2 \mathrm{~cm}, 4 \mathrm{~cm}\)
2 \(3.6 \mathrm{~cm}, 2.4 \mathrm{~cm}\)
3 \(3 \mathrm{~cm}, 3 \mathrm{~cm}\)
4 \(5 \mathrm{~cm}, 1 \mathrm{~cm}\)
Explanation:
(C) : Given, that perimeter of triangle \(=10\) \(a+b+c=10\) And, \(a=4\) then \(b+c=6\) We know that, area of triangle \(\Delta=\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}\) And, \(s=\frac{a+b+c}{2}\) \(\mathrm{S}=\frac{10}{2}\) \(\mathrm{~S}=5\) \(\Delta=\sqrt{5(5-4)(5-\mathrm{b})(5-\mathrm{c})}=\sqrt{5 \times 1 \times(5-\mathrm{b})(5-\mathrm{c})}\) \(\Delta^{2}=5(5-\mathrm{b})(5-\mathrm{c})\) \(\mathrm{A}^{2}=5(5-\mathrm{b})(5-6+\mathrm{b})\) \(\mathrm{A}^{2}=5(5-\mathrm{b})(\mathrm{b}-1)\) Differentiating both sides w.r.t. b, we get - \(\frac{\mathrm{dA}^{2}}{\mathrm{db}}=(5-\mathrm{b})+(\mathrm{b}-1)(-1)\) For maximum area of triangle \(\frac{\mathrm{dA}^{2}}{\mathrm{db}}=0\) \(5-\mathrm{b}+1-\mathrm{b}=0\) \(-2 \mathrm{~b}+6=0\) \(\mathrm{~b}=3\) Then, \(\mathrm{c}=3\) Then remaining sides are, 3,3
85596
If \(x+y=\frac{\pi}{2}\), then the maximum value of \(\sin x\), \(\sin \mathrm{y}\) is
1 \(\frac{-1}{\sqrt{2}}\)
2 \(\frac{1}{\sqrt{2}}\)
3 \(\frac{1}{2}\)
4 \(\frac{-1}{2}\)
Explanation:
(C) : Given, \(\mathrm{x}+\mathrm{y}=\frac{\pi}{2}\) \(\therefore \quad \sin \mathrm{x} \sin \mathrm{y}\) \(=\sin \mathrm{x} \sin \left(\frac{\pi}{2}-\mathrm{x}\right)\) \(=\sin \mathrm{x} \cos \mathrm{x}\) \(=\frac{1}{2} 2 \sin \mathrm{x} \cos \mathrm{x}\) \(=\frac{1}{2} \sin 2 \mathrm{x}\) maximum value obtained when \(\sin 2 x\) is maximum \(\sin 2 x=1\) \(\therefore\) maximum value is \(\frac{1}{2}\)
MHT CET-2020
Application of Derivatives
85597
If rectangles are inscribed in a circle of radius \(r\) units. Then the dimensions of the rectangle which has maximum area are
1 2 runits, \(r\) units,
2 2 runits, \(\sqrt{2} \mathrm{r}\) units,
3 \(\sqrt{2} \mathrm{r}\) units, \(\sqrt{2} \mathrm{r}\) units
4 runits, \(\sqrt{2} \mathrm{r}\) units,
Explanation:
(C) : Let PQRS be the rectangle inscribed in circle of radius \(\mathrm{r}\) Let, \(\quad \mathrm{PQ}=\mathrm{x}\) and \(\mathrm{QR}=\mathrm{y}\) Now, by Pythagoras theorem \(x^{2}+y^{2}=4 r^{2}\) \(y=\sqrt{4 r^{2}-x^{2}}\) Area of rectangle \(A=x y\) \(A=x \sqrt{4 r^{2}-x^{2}}\) Squaring both side we get - \(A^{2}=x^{2}\left(4 r^{2}-x^{2}\right)\) Differentiating w.r.t. \(x\), we get- \(\frac{d\left(A^{2}\right)}{d x}=8 x^{2}-4 x^{3}\) For maxima or minima \(\frac{d\left(A^{2}\right)}{d x}=0\) \(8 x r^{2}-4 x^{3}=0\) \(8 r^{2}-4 x^{2}=0\) \(x=\sqrt{2} r\) Again differentiating w.r.t \(x\), we get- \(\frac{d^{2}\left(A^{2}\right)}{d x^{2}}=8 r^{2}-12 x^{2}\) At, \(\quad \mathrm{x}=\sqrt{2} \mathrm{r}\) then, \(\frac{\mathrm{d}^{2}\left(\mathrm{~A}^{2}\right)}{\mathrm{dx}^{2}}=8 \mathrm{r}^{2}-12(\sqrt{2} \mathrm{r})^{2}\) \(8 \mathrm{r}^{2}-24 \mathrm{r}^{2}\lt 0\) \(-16 \mathrm{r}^{2}\lt 0\) \(A\) is maximum of \(x=\sqrt{2} r\) \(y=\sqrt{2} \mathrm{r}=\mathrm{x}\) Hence the dimensions of the rectangle are \(\sqrt{2} \mathrm{r}\) and \(\sqrt{2} \mathrm{r}\)
MHT CET-2020
Application of Derivatives
85598
20 meters wire is available to fence a flower bed in the form of a circular sector. If the flower bed should have the greatest possible area, then the radius of the circle is
1 \(5 \mathrm{~m}\)
2 \(2 \mathrm{~m}\)
3 \(4 \mathrm{~m}\)
4 \(10 \mathrm{~m}\)
Explanation:
(A) : Given perimeter of circular sector \(20=2 \mathrm{r}+l \Rightarrow l=20-2 \mathrm{r}\) Now area, \(\mathrm{A}=\frac{1}{2} \times l \times \mathrm{r} \Rightarrow \mathrm{A}=\frac{1}{2}(20-2 \mathrm{r}) \mathrm{r}\) Differentiating both sides w.r.t. \(x\), we get - \(\frac{\mathrm{dA}}{\mathrm{dr}} =\frac{1}{2}[20-2 \mathrm{r}+\mathrm{r}(-2)]\) \(\frac{\mathrm{dA}}{\mathrm{dr}} =\frac{1}{2}[20-4 \mathrm{r}]\) For greatest area \(\frac{\mathrm{dA}}{\mathrm{dr}}=0 \Rightarrow \frac{1}{2}(20-4 \mathrm{r})=0\) \(4 \mathrm{r}=20\) \(\mathrm{r}=5\)
MHT CET-2022
Application of Derivatives
85599
A metal wire 108 meters long is bent to form a rectangle. If the area of the rectangle is maximum, then its dimensions are
1 \(28 \mathrm{~m}, 28 \mathrm{~m}\)
2 \(28 \mathrm{~m}, 26 \mathrm{~m}\)
3 \(25 \mathrm{~m}, 25 \mathrm{~m}\)
4 \(27 \mathrm{~m}, 27 \mathrm{~m}\)
Explanation:
(D) : Let the length and breadth of a rectangle be \(x\) and \(y\) Given perimeter of rectangle \(2(x+y)=108\) \(x+y=54\) \(y=54-x\) Now area of the rectangle \(A=x y\) \(A=x(54-x)\) \(A=54 x-x^{2}\) Differentiating w.r.t. \(\mathrm{x}\) we get- \(\frac{\mathrm{dA}}{\mathrm{dx}}=54-2 \mathrm{x}\) For maximum, \(\frac{\mathrm{dA}}{\mathrm{dx}}=0\) \(54-2 x=0\) \(x=27\) \(y=27\)
MHT CET-2020
Application of Derivatives
85600
The perimeter of a triangle is \(10 \mathrm{~cm}\). If one of its side is \(4 \mathrm{~cm}\), then remaining sides of the triangle, when area of triangle is maximum are
1 \(2 \mathrm{~cm}, 4 \mathrm{~cm}\)
2 \(3.6 \mathrm{~cm}, 2.4 \mathrm{~cm}\)
3 \(3 \mathrm{~cm}, 3 \mathrm{~cm}\)
4 \(5 \mathrm{~cm}, 1 \mathrm{~cm}\)
Explanation:
(C) : Given, that perimeter of triangle \(=10\) \(a+b+c=10\) And, \(a=4\) then \(b+c=6\) We know that, area of triangle \(\Delta=\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}\) And, \(s=\frac{a+b+c}{2}\) \(\mathrm{S}=\frac{10}{2}\) \(\mathrm{~S}=5\) \(\Delta=\sqrt{5(5-4)(5-\mathrm{b})(5-\mathrm{c})}=\sqrt{5 \times 1 \times(5-\mathrm{b})(5-\mathrm{c})}\) \(\Delta^{2}=5(5-\mathrm{b})(5-\mathrm{c})\) \(\mathrm{A}^{2}=5(5-\mathrm{b})(5-6+\mathrm{b})\) \(\mathrm{A}^{2}=5(5-\mathrm{b})(\mathrm{b}-1)\) Differentiating both sides w.r.t. b, we get - \(\frac{\mathrm{dA}^{2}}{\mathrm{db}}=(5-\mathrm{b})+(\mathrm{b}-1)(-1)\) For maximum area of triangle \(\frac{\mathrm{dA}^{2}}{\mathrm{db}}=0\) \(5-\mathrm{b}+1-\mathrm{b}=0\) \(-2 \mathrm{~b}+6=0\) \(\mathrm{~b}=3\) Then, \(\mathrm{c}=3\) Then remaining sides are, 3,3
