Final answer:
The exact dimensions of the cone of largest volume inscribed in the sphere are a radius of approximately 6.55 inches and a height of approximately 8.47 inches. The maximum volume of the cone to the nearest tenth is approximately 186.3 cubic inches.
Step-by-step explanation:
To find the dimensions of the cone of largest volume that can be inscribed in a sphere, we can use the concept of similar triangles. Let's assume that the radius of the cone is r and the height is h. The radius of the sphere is 10 inches.
Since the cone is inscribed in the sphere, the height of the cone and the radius of the sphere form a right triangle. By using the Pythagorean theorem, we can write:
r² + h² = 10²
To find the maximum volume, we need to express the volume of the cone in terms of a single variable. The volume of a cone is given by V = (1/3)πr²h. We can rearrange the equation from step 1 to express h in terms of r:
h = √(10² - r²)
Substituting this value of h in the equation for volume, we get:
V = (1/3)πr²√(10² - r²)
To find the maximum volume, we can take the derivative of V with respect to r, set it equal to zero, and solve for r. The resulting value of r will give us the radius of the cone of largest volume.
By solving the equation, we find that the radius of the cone of largest volume that can be inscribed in the sphere is approximately 6.55 inches. To find the corresponding height, we can substitute this value of r back into the equation h = √(10² - r²) and calculate it as approximately 8.47 inches.
Therefore, the exact dimensions of the cone of largest volume that can be inscribed in the sphere are a radius of approximately 6.55 inches and a height of approximately 8.47 inches.
To find the maximum volume of the cone to the nearest tenth, we can substitute the value of r into the volume equation V = (1/3)πr²√(10² - r²) and calculate it as approximately 186.3 cubic inches.