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Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
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Re: Twenty meters of wire is available to fence off a flower bed in the  [#permalink]

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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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Karishma
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Senior Manager  Joined: 13 May 2013
Posts: 396
Re: Twenty meters of wire is available to fence off a flower bed in the  [#permalink]

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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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Re: Twenty meters of wire is available to fence off a flower bed in the  [#permalink]

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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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Re: Twenty meters of wire is available to fence off a flower bed in the  [#permalink]

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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

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

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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Posts: 320
Re: Twenty meters of wire is available to fence off a flower bed in the  [#permalink]

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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 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....
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10105
Location: Pune, India
Re: Twenty meters of wire is available to fence off a flower bed in the  [#permalink]

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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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Re: Twenty meters of wire is available to fence off a flower bed in the  [#permalink]

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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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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
****************************

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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Re: Twenty meters of wire is available to fence off a flower bed in the  [#permalink]

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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)

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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Re: Twenty meters of wire is available to fence off a flower bed in the  [#permalink]

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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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1
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).

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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Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10105
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Re: Twenty meters of wire is available to fence off a flower bed in the  [#permalink]

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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. $$r = 5$$

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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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. Re: Twenty meters of wire is available to fence off a flower bed in the   [#permalink] 01 May 2017, 19:31

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