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

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New post 11 Dec 2013, 19:33
WholeLottaLove 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?

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


The explanation of \(Q/2\pi\) is given here: twenty-metres-of-wire-is-available-to-fence-off-a-flower-bed-114049.html#p1282412

And this post explains how to maximize: twenty-metres-of-wire-is-available-to-fence-off-a-flower-bed-114049.html#p925363
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New post 11 Dec 2013, 20:20
VeritasPrepKarishma wrote:
WholeLottaLove 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?

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


The explanation of \(Q/2\pi\) is given here: twenty-metres-of-wire-is-available-to-fence-off-a-flower-bed-114049.html#p1282412

And this post explains how to maximize: twenty-metres-of-wire-is-available-to-fence-off-a-flower-bed-114049.html#p925363



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 + (Q/2pi)*2*pi*r = 20
i.e. 2r + Qr = 20

Maximize (Q/2pi)*pi*r^2 = 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


Ok, so you have: Qr^2/2 and 2r + Qr = 20 (i.e. Qr=20-2r) however, when you plug Qr=20-2r into Qr^2/2 why do you not square it?
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New post 04 Jan 2014, 16:34
We must agree that the "derivative" is really really easy to use and worth the time spent on studying those!

I knew them by my studies but for those who do not know, I think it is a really usefull tool and in our case, it enables us to resolve the problem really quickly!

You can find some usefull stuff here : http://www.mathsisfun.com/calculus/deri ... ction.html

Hope it helps !
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New post 05 Jan 2014, 03:15
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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Even though I got it correct using standard CAT techniques, I really doubt maxima and minima question will ever come in GMAT.
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New post 21 Sep 2014, 12:31
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 Karisma,
Can you tell me how you got the last line \(10 - r + r = 10\) ? Also, how did you continue from that to the rest of the line?
I don't see where these equations come from....
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New post 22 Sep 2014, 20:18
ronr34 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. \(r = 5\)

Hi Karisma,
Can you tell me how you got the last line \(10 - r + r = 10\) ? Also, how did you continue from that to the rest of the line?
I don't see where these equations come from....



We need to maximize (10 - r)*r.

The concept used is this: When the sum of two numbers is constant, their product is maximum when they are equal.

Say you have two numbers m and n. If you know that m + n = 10, their product m*n will be maximum when m = n = 10/2 = 5
So m*n has maximum value of 5*5 = 25.

Similarly, we need to find the maximum value of (10 - r)*r. Note that the sum of (10-r) and r is a constant i.e. 10 - r + r = 10. The sum is 10. The product (10-r)*r will be maximum when both (10-r) and r are equal and equal to 10/2 = 5 each. That is how you get r = 5.
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New post 27 Sep 2014, 04:01
Only 5 can help in this , So 5 answer.

Just try using all the options with 2piR - you will understand why 5
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New post 17 Jun 2015, 06:46
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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I just took 3 values for radius ( 4, 5, and 6)
The area for 4 is 7.5 pi approx
Thea area for 5 is 7.96 pi approx
The area for 6 is 7.6 pi approx
So the max area converges to 5 approx
I guess it is third option (3)
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New post 02 Nov 2015, 21:47
VeritasPrepKarishma wrote:
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.



Hey I am stuck at

Qr = 20-2r

not sure how you ended up at the answer.




___

Also would this problem not be much faster to solve with calculus?

2Area=2*pi*r*(C/360)*r
2Area=l*r
2Area=(20-r)r
2Area=20r-2r^2
Area=10r-r^2
dA/dR=10-2r=0
10=2r
5=r
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New post 26 Jan 2016, 05:40
A for me .Hello guys, I don't know whether my answer is correct. For a sector to have greater area and circumference,sector should have greater central angle.In options given A gives higher (305).
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New post 27 Jan 2016, 01:23
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manohar0265 wrote:
A for me .Hello guys, I don't know whether my answer is correct. For a sector to have greater area and circumference,sector should have greater central angle.In options given A gives higher (305).


The correct answer is (C).
Maximising the central angle may not give you the greatest surface area because surface area depends on radius too. Check the solutions given on the previous page.
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New post 05 Jun 2016, 21:13
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. \(r = 5\)


Responding to a pm:

Quote:
How you come up with this step:

Maximize (Q/2pi)∗pi∗r^2=Qr^2

You state that the maximum comes from the product of the two numbers but wouldn't that product be 2r *(Q/2pi)*2*pi*r = 2Q*r^2?


The question says this: " ... with the greatest possible surface area?"

You need to maximise the surface area of a circular sector. The area of a circle is \(\pi * r^2\).
The area of a circular sector is
\((Q/2\pi) * AreaOfCircle\)
\(= (Q/2\pi) * \pi * r^2\)

To maximise this, we need to maximise \(Qr^2\) basically since the rest are constants.

Using Qr = 20 - 2r, we get that we need to maximise (20 - 2r)*r. Again, this is same as 2*(10 - r)*r so we need to maximise (10 - r)*r (ignoring the constants).

This has a constant sum (10 - r) + r = 10

So product is maximum when (10 - r) = r

Hope this clarifies.
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New post 01 May 2017, 19:31
PiyushK wrote:
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.


4\sqrt{2}= 5.656 which is closer to 6 than 5. I don't think this logic is correct at all.
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