Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector.
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20 meters of wire is available for fencing off a flower-bed in the form of a circular sector.
Then what is the maximum area (in sq. m) of the flower-bed?
circle area
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up vote
2
down vote
favorite
20 meters of wire is available for fencing off a flower-bed in the form of a circular sector.
Then what is the maximum area (in sq. m) of the flower-bed?
circle area
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
20 meters of wire is available for fencing off a flower-bed in the form of a circular sector.
Then what is the maximum area (in sq. m) of the flower-bed?
circle area
20 meters of wire is available for fencing off a flower-bed in the form of a circular sector.
Then what is the maximum area (in sq. m) of the flower-bed?
circle area
asked Apr 3 '17 at 5:51
starunique2016
533326
533326
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add a comment |Â
1 Answer
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up vote
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This is a very simple question. Let $M$ be the angle, the arc of the sector $AOB$ makes with the center $O$. Let OA = OB = radius r.
Given
$$ OA + arc AB + OB = 20 $$
$$2r + rM = 20 $$
$$ M = 20-2rr$$
Area $A$ of the sector =$ fracM2r^2 = 10r-r^2$
$$fracdAdr= 10 - 2r = 0$$
$$r=5$$
Second derivative is negative for $r =5$
We get maximum area at $r=5$
$A = 25 $
can you tidy up the maths typesetting? It's quite difficult to follow.
â Harambe
Apr 5 '17 at 9:48
an edit is suggested @Sanjay Bhatnagar
â starunique2016
Apr 10 '17 at 5:15
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
This is a very simple question. Let $M$ be the angle, the arc of the sector $AOB$ makes with the center $O$. Let OA = OB = radius r.
Given
$$ OA + arc AB + OB = 20 $$
$$2r + rM = 20 $$
$$ M = 20-2rr$$
Area $A$ of the sector =$ fracM2r^2 = 10r-r^2$
$$fracdAdr= 10 - 2r = 0$$
$$r=5$$
Second derivative is negative for $r =5$
We get maximum area at $r=5$
$A = 25 $
can you tidy up the maths typesetting? It's quite difficult to follow.
â Harambe
Apr 5 '17 at 9:48
an edit is suggested @Sanjay Bhatnagar
â starunique2016
Apr 10 '17 at 5:15
add a comment |Â
up vote
2
down vote
accepted
This is a very simple question. Let $M$ be the angle, the arc of the sector $AOB$ makes with the center $O$. Let OA = OB = radius r.
Given
$$ OA + arc AB + OB = 20 $$
$$2r + rM = 20 $$
$$ M = 20-2rr$$
Area $A$ of the sector =$ fracM2r^2 = 10r-r^2$
$$fracdAdr= 10 - 2r = 0$$
$$r=5$$
Second derivative is negative for $r =5$
We get maximum area at $r=5$
$A = 25 $
can you tidy up the maths typesetting? It's quite difficult to follow.
â Harambe
Apr 5 '17 at 9:48
an edit is suggested @Sanjay Bhatnagar
â starunique2016
Apr 10 '17 at 5:15
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
This is a very simple question. Let $M$ be the angle, the arc of the sector $AOB$ makes with the center $O$. Let OA = OB = radius r.
Given
$$ OA + arc AB + OB = 20 $$
$$2r + rM = 20 $$
$$ M = 20-2rr$$
Area $A$ of the sector =$ fracM2r^2 = 10r-r^2$
$$fracdAdr= 10 - 2r = 0$$
$$r=5$$
Second derivative is negative for $r =5$
We get maximum area at $r=5$
$A = 25 $
This is a very simple question. Let $M$ be the angle, the arc of the sector $AOB$ makes with the center $O$. Let OA = OB = radius r.
Given
$$ OA + arc AB + OB = 20 $$
$$2r + rM = 20 $$
$$ M = 20-2rr$$
Area $A$ of the sector =$ fracM2r^2 = 10r-r^2$
$$fracdAdr= 10 - 2r = 0$$
$$r=5$$
Second derivative is negative for $r =5$
We get maximum area at $r=5$
$A = 25 $
edited Aug 14 at 12:23
Nehal Samee
1469
1469
answered Apr 5 '17 at 9:18
Sanjay Mohan Bhatnagar
693
693
can you tidy up the maths typesetting? It's quite difficult to follow.
â Harambe
Apr 5 '17 at 9:48
an edit is suggested @Sanjay Bhatnagar
â starunique2016
Apr 10 '17 at 5:15
add a comment |Â
can you tidy up the maths typesetting? It's quite difficult to follow.
â Harambe
Apr 5 '17 at 9:48
an edit is suggested @Sanjay Bhatnagar
â starunique2016
Apr 10 '17 at 5:15
can you tidy up the maths typesetting? It's quite difficult to follow.
â Harambe
Apr 5 '17 at 9:48
can you tidy up the maths typesetting? It's quite difficult to follow.
â Harambe
Apr 5 '17 at 9:48
an edit is suggested @Sanjay Bhatnagar
â starunique2016
Apr 10 '17 at 5:15
an edit is suggested @Sanjay Bhatnagar
â starunique2016
Apr 10 '17 at 5:15
add a comment |Â
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