A conical drinking cup is made from a circular piece of paper of radius R inches by cutting out a sector and joining the edges CA and CB. Find the radius, height and volume of the cone of greatest volume that can be made this way.
I used to cut open a paper drinking cup in class when I was teaching this particular question.
It was easy to see that the resulting arc was equal to the circular base of the cup and the "slant" height of the cup was the radius of the partial circle.
Try this, and see if you can do the question.
To find the radius, height, and volume of the cone of greatest volume that can be made, we need to determine the dimensions that will maximize the volume of the cone.
Let's start by visualizing the problem. We have a circular piece of paper with radius R, and we need to cut out a sector from it to form a cone. To do this, we select two points, A and B, on the circumference of the circle, and join them with a straight line segment to form the lateral surface of the cone.
To maximize the volume of the cone, we need to find the dimensions that will maximize the height and radius of the cone.
Let's denote the radius of the cone as "r" and the height as "h".
From the problem statement, we can see that the slant height of the cone will be the radius of the circular piece, which is R.
Using Pythagoras' theorem, we can write:
h^2 + r^2 = R^2
Now, we need an equation for the volume of the cone. The volume of a cone can be calculated using the formula:
V = (1/3) * π * r^2 * h
To simplify the equation, we need to eliminate one of the variables. Let's solve the equation h^2 + r^2 = R^2 for h:
h^2 = R^2 - r^2
h = √(R^2 - r^2)
Substituting this value of h into the volume equation, we get:
V = (1/3) * π * r^2 * √(R^2 - r^2)
Now, using calculus, we can find the maximum volume by taking the derivative of V with respect to r and setting it equal to zero:
dV/dr = (1/3) * π * [(2r√(R^2 - r^2)) + (r^2 / √(R^2 - r^2))] = 0
Simplifying the equation, we get:
2r√(R^2 - r^2) + (r^2 / √(R^2 - r^2)) = 0
To solve this equation, we can multiply both sides by √(R^2 - r^2) to get rid of the denominator:
2r(R^2 - r^2) + r^2 = 0
Expanding and rearranging the equation, we get:
2rR^2 - 3r^3 = 0
Factoring out an r, we get:
r(2R^2 - 3r^2) = 0
This equation tells us that r = 0 or 2R^2 - 3r^2 = 0
Since r cannot be zero (as it represents the radius of the cone), we solve the equation 2R^2 - 3r^2 = 0 for r.
2R^2 - 3r^2 = 0
3r^2 = 2R^2
r^2 = (2/3)R^2
r = √((2/3)R^2)
r = R * √(2/3)
Now, to find the height, we can substitute this value of r into the equation h = √(R^2 - r^2):
h = √(R^2 - (R * √(2/3))^2)
h = √(R^2 - (2/3)R^2)
h = √((1 - 2/3)R^2)
h = √((1/3)R^2)
h = R * √(1/3)
h = R/√3
Finally, to determine the maximum volume, we substitute the values of r and h into the volume formula:
V = (1/3) * π * (R * √(2/3))^2 * (R/√3)
V = (1/3) * π * (2/3)R^3 * R/√3
V = (2/9) * π * R^4 / √3
Therefore, the radius of the cone of greatest volume is R * √(2/3), the height is R/√3, and the volume is (2/9) * π * R^4 / √3.