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Twenty meters of wire is available to fence off a flower bed in the

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New post 22 May 2011, 00:51
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Twenty meters of wire is available to fence off a flower bed in the form of a circular sector. What must the radius of the circle in meters be, if we wish to have a flower bed with the greatest possible surface area?

(A) 2√2
(B) 2√5
(C) 5
(D) 4√2
(E) none of these


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New post 24 May 2011, 20:37
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You don't need derivatives in CAT either! Anyway, we can solve it either by plugging in numbers or by using this concept:
If the sum of given numbers is constant, their product is maximum when they are equal.

Given: \(2r + (\frac{Q}{2pi})*2*pi*r = 20\)
i.e. \(2r + Qr = 20\)

Maximize \((\frac{Q}{2pi})*pi*r^2 = \frac{Qr^2}{2}\)
Now \(Qr = 20 - 2r\) from above. So, Maximize \((20-2r)r/2 = (10 - r)r\)
Since \(10 - r + r = 10\), to maximize product, \(10 - r = r\) i.e. \(r = 5\)
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New post 22 May 2011, 01:01
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hussi9 wrote:
Twenty metres of wire is available to fence off a flower bed in the form of a circular sector. What must the radius of the circle in meters be, if we wish to have a flower bed with the greatest possible surface area?

(1) 2√2 (2) 2√5 (3) 5 (4) 4√2 (5) none of these


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Requires knowledge of Mensuration and Max/min differenciation which not tested on GMAT.
Not sure of any other method

Sector of a circle = 2l+r
2r+l = 20 ; l= 20 -2r

Area = 1/2lr = 1/2*(20-2r)*r = 10r-r^2

Differenciation: dA/dX= 10 -2r
10-2r=0
hence r= 5
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New post 22 May 2011, 00:54
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Area of Sector, A = x/360*pi*r^2

Circumference of the sector = 20
=> x/360*2*pi*r +2r= 20
=> 2A/r+2r=20
=> A= r(10-r)
= r10-r^2

We will now max using derivations

A`=10-2r

Max value of A will found at A`=0
i.e 10-2r=0
r=5

=> Choice (3) is the right answer

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New post 22 May 2011, 01:12
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sudhir18n wrote:
hussi9 wrote:
Twenty metres of wire is available to fence off a flower bed in the form of a circular sector. What must the radius of the circle in meters be, if we wish to have a flower bed with the greatest possible surface area?

(1) 2√2 (2) 2√5 (3) 5 (4) 4√2 (5) none of these


****************************
IIM CAT Level Question
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Requires knowledge of Mensuration and Max/min differenciation which not tested on GMAT.
Not sure of any other method

Sector of a circle = 2l+r
2r+l = 20 ; l= 20 -2r

Area = 1/2lr = 1/2*(20-2r)*r = 10r-r^2

Differenciation: dA/dX= 10 -2r
10-2r=0
hence r= 5



Are you sure Max/min is never tested on GMAT?
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New post 22 May 2011, 01:24
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max , min can be tested. But the gmat wont expect you to use calculus. So by simple methods you should be able to get the max value.

eg A = 10r-r^2 = 2.5.r - r^2 - 25 + 25 = 25 - (r-5)^2 => max only when r = 5
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New post 24 May 2011, 02:49
interesting question indeed.
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New post 24 May 2011, 09:35
I'm wondering why we can't go this route. It's obviously wrong because I got the wrong answer but:

We know the circumference is 20. So we should be able to find the Diameter/radius

C = Pi * D

20 = pi* D

D = 20/Pi

D = ~ 6.37

So I picked E
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New post 21 Aug 2013, 08:16
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I know a concept that for a given perimeter equilateral triangle gives maximum area, so this sector must be similar to an equilateral triangle with sides 20/3 = 6.66, it means radius should be less than 6, in answer choices 5 is given so it is the nearest possible sector, other choices are not even around 6.
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New post 16 Sep 2013, 19:33
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VeritasPrepKarishma wrote:
You don't need derivatives in CAT either! Anyway, we can solve it either by plugging in numbers or by using this concept:
If the sum of given numbers is constant, their product is maximum when they are equal.

Given: \(2r + (\frac{Q}{2pi})*2*pi*r = 20\)
i.e. \(2r + Qr = 20\)

Maximize \((\frac{Q}{2pi})*pi*r^2 = \frac{Qr^2}{2}\)
Now \(Qr = 20 - 2r\) from above. So, Maximize \((20-2r)r/2 = (10 - r)r\)
Since \(10 - r + r = 10\), to maximize product, \(10 - r = r\) i.e. \(r = 5\)

Can anyone tell me how one obtained this eq. \(2r + (\frac{Q}{2pi})*2*pi*r = 20\)?
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New post 17 Sep 2013, 05:41
2
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holidevil wrote:
VeritasPrepKarishma wrote:
You don't need derivatives in CAT either! Anyway, we can solve it either by plugging in numbers or by using this concept:
If the sum of given numbers is constant, their product is maximum when they are equal.

Given: \(2r + (\frac{Q}{2pi})*2*pi*r = 20\)
i.e. \(2r + Qr = 20\)

Maximize \((\frac{Q}{2pi})*pi*r^2 = \frac{Qr^2}{2}\)
Now \(Qr = 20 - 2r\) from above. So, Maximize \((20-2r)r/2 = (10 - r)r\)
Since \(10 - r + r = 10\), to maximize product, \(10 - r = r\) i.e. \(r = 5\)

Can anyone tell me how one obtained this eq. \(2r + (\frac{Q}{2pi})*2*pi*r = 20\)?


What is a sector of a circle? It looks like a slice of pizza. Its circumference will be given by

Radius + Length of the arc + Radius

Length of the arc =\((\frac{Q}{2pi})*2*pi*r\)
i.e. it is a fraction of the circumference of the complete circle. Q is the angle the arc makes at the center.

Since the rope will be used along the circumference of the sector, the circumference of the sector can be at most equal to the length of the rope i.e. 20

\(r + r + (\frac{Q}{2pi})*2*pi*r = 20\)
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New post 21 Oct 2013, 13:32
VeritasPrepKarishma wrote:
You don't need derivatives in CAT either! Anyway, we can solve it either by plugging in numbers or by using this concept:
If the sum of given numbers is constant, their product is maximum when they are equal.

Given: \(2r + (\frac{Q}{2pi})*2*pi*r = 20\)
i.e. \(2r + Qr = 20\)

Maximize \((\frac{Q}{2pi})*pi*r^2 = \frac{Qr^2}{2}\)
Now \(Qr = 20 - 2r\) from above. So, Maximize \((20-2r)r/2 = (10 - r)r\)
Since \(10 - r + r = 10\), to maximize product, \(10 - r = r\) i.e. \(r = 5\)


Hello Karishma

Can I assume that the max surface area will be reached in a 90 degree sector? since the biggest sector that can be reached is a right angle, if it exceeds 90, would it still be called a sector?
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New post 22 Oct 2013, 21:42
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SaraLotfy wrote:
VeritasPrepKarishma wrote:
You don't need derivatives in CAT either! Anyway, we can solve it either by plugging in numbers or by using this concept:
If the sum of given numbers is constant, their product is maximum when they are equal.

Given: \(2r + (\frac{Q}{2pi})*2*pi*r = 20\)
i.e. \(2r + Qr = 20\)

Maximize \((\frac{Q}{2pi})*pi*r^2 = \frac{Qr^2}{2}\)
Now \(Qr = 20 - 2r\) from above. So, Maximize \((20-2r)r/2 = (10 - r)r\)
Since \(10 - r + r = 10\), to maximize product, \(10 - r = r\) i.e. \(r = 5\)


Hello Karishma

Can I assume that the max surface area will be reached in a 90 degree sector? since the biggest sector that can be reached is a right angle, if it exceeds 90, would it still be called a sector?


Yes, it is still a sector even if the angle exceeds 90. The angle could even go up to almost 360 though we usually deal with minor sectors i.e. when the angle is less than 180. So we cannot assume that the maximum surface area will be reached in a 90 degree sector.
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New post 23 Oct 2013, 01:47
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VeritasPrepKarishma wrote:
You don't need derivatives in CAT either! Anyway, we can solve it either by plugging in numbers or by using this concept:
If the sum of given numbers is constant, their product is maximum when they are equal.

Given: \(2r + (\frac{Q}{2pi})*2*pi*r = 20\)
i.e. \(2r + Qr = 20\)

Maximize \((\frac{Q}{2pi})*pi*r^2 = \frac{Qr^2}{2}\)
Now \(Qr = 20 - 2r\) from above. So, Maximize \((20-2r)r/2 = (10 - r)r\)
Since \(10 - r + r = 10\), to maximize product, \(10 - r = r\) i.e. \(r = 5\)


Thanks Karishma for your explanation.
I need to clarify one more small point. The length of the arc = 2*pi*r * (Q/360) for which Q is the enclosed central angle between the 2 radii. I noticed that you computed (Q/360) as (Q/2pi). I must be missing something behind this logic, but why do we use Q/2pi here?
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New post 23 Oct 2013, 05:06
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SaraLotfy wrote:
VeritasPrepKarishma wrote:
You don't need derivatives in CAT either! Anyway, we can solve it either by plugging in numbers or by using this concept:
If the sum of given numbers is constant, their product is maximum when they are equal.

Given: \(2r + (\frac{Q}{2pi})*2*pi*r = 20\)
i.e. \(2r + Qr = 20\)

Maximize \((\frac{Q}{2pi})*pi*r^2 = \frac{Qr^2}{2}\)
Now \(Qr = 20 - 2r\) from above. So, Maximize \((20-2r)r/2 = (10 - r)r\)
Since \(10 - r + r = 10\), to maximize product, \(10 - r = r\) i.e. \(r = 5\)


Thanks Karishma for your explanation.
I need to clarify one more small point. The length of the arc = 2*pi*r * (Q/360) for which Q is the enclosed central angle between the 2 radii. I noticed that you computed (Q/360) as (Q/2pi). I must be missing something behind this logic, but why do we use Q/2pi here?


There are 2 different units in which you can measure angles - degrees and radians.

In radians, pi = 180 degrees so 2*pi = 360 degrees.

So we can use the formula as 2*pi*r * (Q/360) if Q is in degrees and as 2*pi*r * (Q/2*pi) if Q is in radians. If we don't have the value of Q, we can choose to use either depending on what will make it easier for us. Here, 2*pi makes more sense because pi gets canceled.
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New post 02 Dec 2013, 00:11
Will the GMAT actually test for area of sector of a circle? Are we supposed to have this formula memorized?
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New post 07 Dec 2013, 04:24
VeritasPrepKarishma wrote:
You don't need derivatives in CAT either! Anyway, we can solve it either by plugging in numbers or by using this concept:
If the sum of given numbers is constant, their product is maximum when they are equal.

Given: \(2r + (\frac{Q}{2pi})*2*pi*r = 20\)
i.e. \(2r + Qr = 20\)

Maximize \((\frac{Q}{2pi})*pi*r^2 = \frac{Qr^2}{2}\)
Now \(Qr = 20 - 2r\) from above. So, Maximize \((20-2r)r/2 = (10 - r)r\)
Since \(10 - r + r = 10\), to maximize product, \(10 - r = r\) i.e. \(r = 5\)


Hi Karishma,
In a question where one of the options is "None of the above" can we still test options?

Also with r=5, Q=360/pi = 2 deg, is it correct? quite unintuitive.

My initial thought was > for a given parameter equilateral triangle has the largest area.
Wouldn't the area of triangle + sector largest when the triangle on which sector is formed is equilateral??
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New post 08 Dec 2013, 21:36
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cumulonimbus wrote:

Hi Karishma,
In a question where one of the options is "None of the above" can we still test options?

Also with r=5, Q=360/pi = 2 deg, is it correct? quite unintuitive.

My initial thought was > for a given parameter equilateral triangle has the largest area.
Wouldn't the area of triangle + sector largest when the triangle on which sector is formed is equilateral??


As the original poster said, this is not a GMAT question. GMAT questions do not have 'None of the above' as an option. So you can test options in GMAT if the question permits.

\(2*\pi\) radians = 360 degrees (just like 1 km = 1000 m) (Recall that \(\pi = 3.1416\) approx)
So about 6.28 radians = 360 degrees

As for our question:
Qr = 20 - 2r (where Q is in radians since we used \(Q/2\pi\) in the formula)
Put r = 5 here, you get Q = 2 radians (not degrees)

2 radians = \(360/\pi\) degrees = 114.6 degrees

Also why would you assume that it must be equilateral? What is the logic behind that? It is a sector we are talking about, not a triangle.
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New post 11 Dec 2013, 19:17
Twenty metres of wire is available to fence off a flower bed in the form of a circular sector. What must the radius of the circle in meters be, if we wish to have a flower bed with the greatest possible surface area?

So, we know that that:

r+r+arc = 20
2r+arc=20

To find an arc length we multiply the circumference by (x/360) where x is the measure of the central angle.
Arc Measurement = (x/360) * pi*D

The problem is, we don't know what the central angle is.

Using formula: 2r + (q/2pi)*(pi*d) = 20
(I am familiar with finding the arc measurement/area with (x/360) Where is (q/2Pi) coming from? Is this for when we don't know what the central angle is?

From that point on, I am lost, specifically regarding "maximizing" the value.





(1) 2√2
(2) 2√5
(3) 5
(4) 4√2
(5) none of these
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