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11 Dec 2013, 20:33
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WholeLottaLove
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
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
11 Dec 2013, 21:20
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VeritasPrepKarishma
WholeLottaLove
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
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?
Paris75
Joined: 26 Aug 2013
Last visit: 22 Jul 2024
Posts: 128
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Status:Student
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Concentration: Finance, General Management
Schools: EMLYON FT'16
GMAT 1: 650 Q47 V32
GPA: 3.44
04 Jan 2014, 17:34
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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 : https://www.mathsisfun.com/calculus/deri ... ction.html
Hope it helps !
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 : https://www.mathsisfun.com/calculus/deri ... ction.html
Hope it helps !
