A sector of a circle with radius r cm contains an angle of θ radians between the bounding radii. Given that the perimeter of the sector is 7cm, express θ in terms of r and show that the area is r/2(7-2r)cm square. Hence, find the maximum area of this sector as r varies.
I have already done the first part of the question, i only don't know how to do the "hence...". Please help, Thanks!
To find the maximum area of the sector as r varies, we need to maximize the expression r/2(7-2r).
First, let's find the derivative of the expression with respect to r:
dA/dr = d(r/2(7-2r))/dr
= (d(r/2(7-2r))/d(7-2r)) * (d(7-2r)/dr)
Using the chain rule, the derivative of r/2(7-2r) with respect to (7-2r) is 1/2.
The derivative of (7-2r) with respect to r is -2.
So, dA/dr = (1/2) * (-2) = -1.
To maximize the area, we need to find where the derivative is equal to zero, i.e., when dA/dr = 0.
Setting -1 = 0, we can see that there is no solution.
This means that the expression does not have any maximum or minimum values, but it is always decreasing as r increases.
Therefore, the maximum area of the sector is obtained when r is infinitely large, or as r approaches infinity.
To find the area of the sector, we first need to determine the length of the arc, which can be found using the formula for the circumference of a circle.
The circumference of a circle is given by the formula C = 2πr, where r is the radius of the circle.
In this case, the arc length is a fraction of the full circumference, based on the angle θ. Since there are 2π radians in a full circle, the fraction of the circumference that the arc subtends is θ / (2π). Therefore, the length of the arc, denoted by s, is given by s = (θ / (2π)) * 2πr = θr.
Now, we know that the perimeter of the sector is 7cm, which includes the length of the arc and the lengths of the two radii that bound the sector. So we have:
7 = s + 2r
7 = θr + 2r
7 = r(θ + 2)
Rearranging this equation, we get:
r = 7 / (θ + 2)
To find the area of the sector, we can use the formula for the area of a sector, which is given by A = (θ / (2π)) * πr². Simplifying, we have:
A = (θ / 2) * r²
A = (θ / 2) * (7 / (θ + 2))²
A = (θ / 2) * ((49 / ((θ + 2)²))
Now simplify this equation further:
A = (θ * 49) / (2(θ + 2)²)
A = θ(49 / (2(θ + 2)²))
Now, we can find the maximum area of the sector by finding the value of θ that maximizes the area. To do this, we can take the derivative of the area with respect to θ and set it equal to zero, and solve for θ. However, this will lead to a complicated equation.
Alternatively, we can try to simplify the expression for the area further by multiplying out the denominator:
A = θ(49 / (2(θ² + 4θ + 4)))
A = θ(49 / (2θ² + 8θ + 8))
Now, we notice that θ is a common factor in both the numerator and denominator. Canceling θ from both sides:
A = 49 / (2θ + 8)
To maximize the area, we want to find the maximum value of A, so we can set the denominator equal to zero:
2θ + 8 = 0
θ = -4
However, since θ represents an angle, it cannot be negative. Therefore, there is no maximum area for the sector as r varies.
arc length = a = rѲ
Then P = 2r + rѲ
7 = 2r + rѲ
Ѳ = (7-2r)/r
Let the area of the sector be A
by ratios
A/(pir^2) = Ѳ/(2pi)
A = (r^2)Ѳ/2
or
Ѳ = 2A/r^2
then 2A/r^2 = (7-2r)/r
solving for A after cross-multiplying and simplifying
A = (7r-2r^2)/2 or r/2(7-r) as required
second part:
A = (7/2)r - r^2 from above
dA/dr = 7/2 - 2r
= 0 for a max/min of A
2r - 7/2 = 0
r = 7/4
so Ѳ = (7 - 2(7/4))/(7/4)
Ѳ = 2 (how nice)
so max area = (r^2)(Ѳ)/2
= (49/16)(2)/(7/4)
= 7/2