a cone of height h and radius r is constructed from a flat, circular disk of radius 4 inches. by removing a sector AOC of arc length x inches. and then connecting the edges OA and OC. what are length x will produce the ocne of maximum volume, and what is that volume?
1. show that
r=(8pie-x)/2pie, h= radical 16-r^2,and
V(x)=pie/3((8pie -x)/(2pie))^2 *
radical (16-(8pie-x/2pie)^2)
2. shwo that the natural domain of V is 0<equal x <equal 16pie.
3. explain why the restriciton 0<equal x <equal 8pie makes sense in the problem situation.
4. use graphical methods to find where the cone has its maximum volume, and what that volume is.
5. confirm your findings in part 4 analytically. (Hint: use V(x)=(1/3)peir^2h, h^2+r^2=16, and the chain rule)
To solve this problem step-by-step:
1. To show the relationship between x and the cone's dimensions, we can start by noting that the radius of the cone is equal to the remaining radius of the disk after removing sector AOC. Since the whole disk has a radius of 4 inches, the remaining radius is given by:
r = 4 - (x/(2π))
Thus, we have r = (8π - x)/(2π).
Next, we calculate the height h of the cone. The length of OC serves as the slant height of the cone, and by Pythagoras' theorem, we have:
h^2 + r^2 = 4^2
Simplifying this equation, we get:
h^2 = 16 - r^2
Taking the square root of both sides gives:
h = √(16 - r^2)
2. Now, let's determine the natural domain of the volume function V(x). We know that x represents the arc length, which is a positive value. Additionally, the arc length cannot exceed the circumference of the original disk. The circumference is given by 2π(4) = 8π. Therefore, the domain of x is 0 ≤ x ≤ 8π.
3. The restriction 0 ≤ x ≤ 8π makes sense in the problem situation because x represents the arc length being removed. If x were greater than 8π, it would mean removing an arc longer than the circumference of the original disk, which is not possible.
4. To find where the cone has its maximum volume graphically, we can plot the volume function V(x) as a function of x within the domain 0 ≤ x ≤ 8π. This can be done using graphing software or a graphing calculator.
5. To confirm the findings from part 4 analytically, we can differentiate the volume function V(x) with respect to x and find where the derivative equals zero. This will give us critical points, and we can determine whether they correspond to maximum volume by checking the second derivative. Alternatively, we can rewrite the volume function in terms of r and h, substitute the relationship between r and x, and use the chain rule to find the derivative with respect to x. By setting this derivative equal to zero and solving for x, we can find the value of x that maximizes the volume.
To answer this question, we need to follow the steps provided. Let's go through each step one by one.
Step 1: Showing the Formulas
To find the formulas for the radius, height, and volume of the cone, we need to understand the construction process. From the given information, we have a circular disk with a radius of 4 inches. We remove a sector AOC of arc length x inches and then connect the edges OA and OC.
Let's begin by finding the radius (r). We need to find the remaining portion of the circumference after removing the sector AOC. The total circumference of the circular disk is 2πr, so the remaining circumference is (2πr - x). Since the remaining circumference will become the circumference of the base of the cone, we can equate it to 2πr.
Therefore, (2πr - x) = 2πr
By simplifying the equation, we get:
x = 2πr - 2πr = 0
This means that the value of x must be zero. Therefore, we can conclude that the cone with maximum volume is obtained when sector AOC is not removed. Hence, x should be 0.
Now, let's find the formulas for the radius (r), height (h), and volume (V) when x is 0.
r = (8π - x) / (2π) = (8π - 0) / (2π) = 4 inches
For the height (h), we know that the height of the cone is given by the Pythagorean theorem, as the cone is formed by connecting the edges OA and OC. Therefore, we have:
h = √(16 - r^2) = √(16 - 4^2) = √(16 - 16) = √0 = 0 inches
The volume (V) of a cone is given by the formula V = (1/3)πr^2h. Plugging in the values of r and h, we get:
V(0) = (1/3)π(4)^2(0) = 0
So, when x is 0, the length of the cone will be zero (no cone at all) and the volume will also be zero.
Step 2: Natural Domain of V
To determine the natural domain of V, we need to consider the possible values of x. The given information states that a sector AOC of arc length x inches is removed from the circular disk. Since the circumference of the original disk is 2πr, the maximum value of x can be at most the circumference of the disk.
Therefore, the natural domain of V is 0 ≤ x ≤ 2πr.
Now, let's calculate the maximum value for x:
x ≤ 2π(4)
x ≤ 8π
Hence, the natural domain of V is 0 ≤ x ≤ 8π.
Step 3: Restriction 0 ≤ x ≤ 8π
The restriction 0 ≤ x ≤ 8π makes sense in the problem situation because it represents the values of x that are physically possible. In this scenario, removing a sector AOC means cutting a portion of the circular disk. The maximum value for x would occur when the sector AOC covers the entire circumference of the base.
Since the maximum value for x is 8π (which is less than 2πr), the restriction 0 ≤ x ≤ 8π ensures that the arc length x remains within the bounds of the circular disk.
Step 4: Graphical Method to Find Maximum Volume
To find where the cone has its maximum volume graphically, we can plot the volume function V(x) and analyze its behavior.
Plot the function V(x) = (1/3)π((8π - x)/(2π))^2√(16 - ((8π - x)/(2π))^2)
By analyzing the graph, we can observe the value of x where the volume reaches its maximum point. This will give us the arc length x that produces the cone of maximum volume.
Step 5: Analytical Confirmation
To confirm our findings analytically, we can differentiate the volume function V(x) with respect to x and find the critical points.
Using the volume formula V(x) = (1/3)π((8π - x)/(2π))^2√(16 - ((8π - x)/(2π))^2), we can apply the chain rule and differentiate V(x) to find the critical points.
Once we have the critical points, we evaluate the volume function at those points and compare the values to determine which critical point corresponds to the maximum volume of the cone.
There is a much easier approach to this problem than the one outlined in your steps
Clearly the radius of 4 inches of the original circle to be cut becomes the slant height of the cone.
then r^2 + h^2 = 16 ---> r^2 = 16-j^2
Volume = (1/3)pi(r^2)h
= (1/3)pi(16-h^2)h
= (1/3)pi(h) - (1/3)pi(h^3)
volume' = (1/3)pi - pi(h^2) = 0 for a max volume
then h = 1/√3 inches
and r^2 = 16 - 1/3
= 47/3
r = √(47/3)
so max volume = (1/3)pi(47/3)(1/√3) = 47√3pi/27 or appr. 9.472 cubic inches
back to the circle
let the central angle of the cut-out piece be α radians
then x = 4α
Circumference = 8pi
so the circumference of the cone = 8pi - x
but the circumference of the cone is 2pi(r)
so 2pi(r) = 8pi - x
r = (8pi - x)/(2pi)