A right triangle of hypotenuse L is rotated about one of its legs to generate a right circular cone
find the largest volume that such a cone could occupy
Let cone height be h, which is one of the two legs. The base radius will be
R = sqrt(L^2 - h^2)
Volume = (1/3)pi*R^2*h
= pi*(L^2 - h^2)*h/3
Volume is a maximum when dV/dh = 0
dV/dh = pi*L^2 - pi*3*h^2 = 0
L^2 = 3 h^2
Max volume = (1/3) pi*(L^2-L^2/3)*sqrt(1/3)L
= 2*pi*L^3/(9*sqrt3)
To find the largest volume that a cone generated by rotating a right triangle of hypotenuse L can occupy, we need to maximize the volume of the cone.
Let's denote the length of one leg of the right triangle as x. Therefore, the length of the other leg will be L - x (since L is the hypotenuse).
The volume V of a cone is given by the formula:
V = (1/3) * π * r^2 * h,
where r is the radius of the base of the cone and h is the height of the cone.
In this case, the height of the cone h will be equal to the length of the leg x. And the radius of the base r can be found using the Pythagorean theorem. The radius r is the distance from the center of the base to the hypotenuse, which is half the length of the hypotenuse. Therefore, r = (L - x) / 2.
Now we can substitute these values into the formula for the volume:
V = (1/3) * π * [(L - x) / 2]^2 * x.
To find the maximum volume, we need to find the value of x that maximizes V. We can do that by finding where the derivative of V with respect to x is zero.
Taking the derivative of V with respect to x and setting it equal to zero:
dV/dx = (1/3) * π * [2 * (L - x) * x - x^2] = 0.
Simplifying the equation:
2 * (L - x) * x - x^2 = 0.
Expanding and rearranging:
2Lx - 2x^2 - x^2 = 0.
Combining like terms:
3x^2 - 2Lx = 0.
Factoring out an x:
x(3x - 2L) = 0.
Since the volume cannot be zero, we have:
3x - 2L = 0.
Solving for x:
3x = 2L.
x = 2L/3.
So the value of x that maximizes the volume is x = 2L/3.
Now we substitute this value of x back into the volume formula to find the maximum volume:
V = (1/3) * π * [(L - 2L/3) / 2]^2 * (2L/3).
Simplifying,
V = (1/3) * π * [L/6]^2 * (2L/3).
V = (1/3) * π * L^3 / 27.
Therefore, the largest volume that the cone could occupy is (1/3) * π * L^3 / 27.