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Difficulty:   95% (hard)

Question Stats: 22% (02:39) correct 78% (02:21) wrong based on 993 sessions

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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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IIM CAT Level Question
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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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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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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
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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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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
****************************

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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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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interesting question indeed.
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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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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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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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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

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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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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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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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. $$r = 5$$

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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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. $$r = 5$$

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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Will the GMAT actually test for area of sector of a circle? Are we supposed to have this formula memorized?
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catalysis wrote:
Will the GMAT actually test for area of sector of a circle? Are we supposed to have this formula memorized?

Yes, you should know how it's derived. Check here: math-circles-87957.html

Hope this helps.
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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. $$r = 5$$

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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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)

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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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 Re: Twenty meters of wire is available to fence off a flower bed in the   [#permalink] 11 Dec 2013, 19:17

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