Find the dimensions of the largest right circular cylinder that can be inscribed in a sphere of radius 6 inches.
Draw a side view. If the cylinder has radius r, and height h, then
r^2+h^2 = 6^2
the volume of the cylinder is thus
v = πr^2h = πr^2√(36-r^2)
now just find dv/dr=0 and thus the maximum volume's radius.
To find the dimensions of the largest right circular cylinder that can be inscribed in a sphere, we need to consider the properties of the cylinder and the sphere.
1. Consider the cross-section of the inscribed cylinder: It will be a rectangle inscribed in a circle.
2. Since the cylinder is inscribed in a sphere, the diameter of the sphere is equal to the diagonal of the rectangle.
3. The diagonal of a rectangle can be found using the Pythagorean theorem, which states that the square of the diagonal is equal to the sum of the squares of the other two sides.
4. Let's assume the dimensions of the rectangle are length (L) and width (W). The diagonal (D) of the rectangle is equal to the diameter of the sphere, which is 2 times the radius (r) of the sphere.
5. Using the Pythagorean theorem, we have: (L^2 + W^2) = D^2 = (2r)^2
6. Substituting the value of the radius (r = 6 inches), we have: (L^2 + W^2) = (2*6)^2 = 36.
7. We want to find the largest rectangle possible within the sphere, which means we want to find the maximum value for LW while satisfying the equation (L^2 + W^2 = 36).
8. To find the maximum product of LW, we can use the method of constrained optimization.
9. By applying the AM-GM inequality, we know that (L^2 + W^2) divided by 2 is greater than or equal to the square root of L^2 * W^2.
10. Therefore, (L^2 + W^2) / 2 >= sqrt(L^2 * W^2).
11. Substituting (L^2 + W^2 = 36), we have: 36/2 >= sqrt(L^2 * W^2).
12. Simplifying, we get: 18 >= sqrt(L^2 * W^2).
13. Squaring both sides of the inequality, we obtain: 324 >= L^2 * W^2.
14. To maximize LW, we need to maximize L^2 * W^2. Therefore, we have L^2 * W^2 = 324.
15. The maximum value for L^2 * W^2 occurs when L^2 = W^2 = 162.
16. Taking the square root of both sides, we have L = W = √162 ≈ 12.7
Therefore, the dimensions of the largest right circular cylinder that can be inscribed in a sphere of radius 6 inches are approximately 12.7 inches for both the length and the width.
To find the dimensions of the largest right circular cylinder that can be inscribed in a sphere, we need to consider the relationship between the sphere and the cylinder.
Let's start by visualizing the problem. Imagine a sphere with a radius of 6 inches.
We know that the cylinder must fit inside the sphere. In other words, the diameter of the cylinder must be smaller than or equal to the diameter of the sphere.
The diameter of the sphere is twice its radius, so in this case, it would be 2 * 6 = 12 inches.
To inscribe a cylinder in the sphere, the cylinder's height must be equal to the diameter of the sphere. So, the height of the cylinder would also be 12 inches.
Therefore, the dimensions of the largest right circular cylinder that can be inscribed in a sphere with a radius of 6 inches are:
- Diameter: 12 inches
- Height: 12 inches