A right triangle with hypotenuse 5 inches is revolved around one of its legs to generate a right circular cone. Find the radius and height of the greatest volume cone.
v = 1/3 pi r^2 h
r^2 + h^2 = 25
v = 1/3 pi (25-h^2)h = 1/3 pi (25h - h^3)
a max/min occurs when dv/dh = 0:
25 - 3h^2 = 0
h = 5/√3
r = 5√2/√3
It's a max because v'' < 0
To find the radius and height of the cone with the greatest volume, we need to maximize the volume of the cone. The volume of a right circular cone can be calculated using the formula: V = (1/3) * π * r^2 * h, where V is the volume, r is the radius, and h is the height.
Given that the hypotenuse of the right triangle is 5 inches, we can use the Pythagorean theorem to find the lengths of the legs. Let's say the legs of the triangle are a and b.
Using a^2 + b^2 = c^2, where c is the hypotenuse:
a^2 + b^2 = 5^2
a^2 + b^2 = 25 ...(1)
Since we are revolving the triangle around one of its legs, let's assume that leg a is the axis of revolution. This means the cone's height (h) will be equal to leg b.
Now, we need to express the radius (r) in terms of a and b. In a right circular cone, the radius is equal to the length of the hypotenuse of the cross-section triangle at any particular height.
In this case, when the cone is a maximum volume, the height (h) of the cone will be equal to leg b of the right triangle. Let's substitute leg b for h. From equation (1), we can solve for leg a in terms of leg b:
a^2 = 25 - b^2
a = √(25 - b^2)
Now, let's consider a cross-sectional triangle of the cone when it has a height of b. This cross-sectional triangle will have one leg a (which is the radius of the cone) and the height h (which is equal to b).
Using the Pythagorean theorem for this cross-section triangle:
r^2 + h^2 = a^2
r^2 + b^2 = (√(25 - b^2))^2
r^2 + b^2 = 25 - b^2
2b^2 = 25 - r^2
We can solve this equation for b^2:
2b^2 + b^2 = 25 - r^2 + b^2
3b^2 = 25 - r^2
b^2 = (25 - r^2)/3 ...(2)
Now, we can substitute equation (2) into the volume formula to express the volume V in terms of r:
V = (1/3) * π * r^2 * h
V = (1/3) * π * r^2 * b
Substituting b^2 from equation (2) into the volume formula:
V = (1/3) * π * r^2 * ((25-r^2)/3)
V = (π/9) * r^2 * (25-r^2)
To find the maximum volume V, we need to find the critical points by taking the derivative of V with respect to r and setting it equal to zero:
dV/dr = (π/9) * (2r * (25-r^2) - r^2 * 2r)
dV/dr = (π/9) * (50r - 3r^3 - 2r^2)
Setting dV/dr equal to zero:
(π/9) * (50r - 3r^3 - 2r^2) = 0
Simplifying the equation:
50r - 3r^3 - 2r^2 = 0
To solve this equation, we can apply the Rational Root Theorem to find potential rational solutions. The possible rational roots for our equation are factors of the constant term (0), divided by factors of the leading coefficient (3). The possible rational roots are ±1, ±2, ±5, and ±10.
By testing these possible rational solutions, we find that the only solution is r = 5.
Now that we have the value of r, we can substitute it back into equation (2) to find b^2:
b^2 = (25 - r^2)/3
b^2 = (25 - 5^2)/3
b^2 = (25 - 25)/3
b^2 = 0
So, b = 0.
Since b represents the height of the cone, we find that the height of the cone with the maximum volume is 0 inches.
Therefore, in this case, the radius of the greatest volume cone is 5 inches, and the height is 0 inches.