This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
Step 1: Define the volumes and variables. Let R be the radius of the sphere. The volume of the sphere is V_S = (4)/(3) R^3. Let the inscribed cone have radius r and height h. The volume of the cone is V_C = (1)/(3) r^2 h. Step 2: Relate the cone's dimensions to the sphere's radius. Consider a cross-section of the sphere and the inscribed cone. Let the center of the sphere be at the origin (0,0). Let the height of the cone be h. The vertex of the cone is at (0, R). Let the center of the base of the cone be at (0, R-h). The radius of the cone's base, r, forms a right-angled triangle with the sphere's radius R and the distance from the sphere's center to the cone's base center. Let k be the distance from the center of the sphere to the center of the cone's base. Then h = R + k. From the Pythagorean theorem, r^2 + k^2 = R^2, so r^2 = R^2 - k^2. Since the cone's vertex is at the top of the sphere, its height h is R+k. The variable k represents the distance from the center of the sphere to the base of the cone. k can range from 0 (base passes through the center) to R (base is a point at the bottom of the sphere). Step 3: Express the cone's volume in terms of a single variable. Substitute r^2 = R^2 - k^2 and h = R+k into the volume formula for the cone: V_C = (1)/(3) (R^2 - k^2)(R+k) Factor R^2 - k^2 as (R-k)(R+k): V_C = (1)/(3) (R-k)(R+k)(R+k) V_C = (1)/(3) (R-k)(R+k)^2 We need to maximize V_C with respect to k. The domain for k is 0 k R. Step 4: Find the critical points by differentiation. To find the maximum volume, we differentiate V_C with respect to k and set the derivative to zero. Let f(k) = (R-k)(R+k)^2. Using the product rule, (d)/(dk)(uv) = u'v + uv': (dV_C)/(dk) = (1)/(3) [ (-1)(R+k)^2 + (R-k) · 2(R+k) · 1 ] (dV_C)/(dk) = (1)/(3) (R+k) [ -(R+k) + 2(R-k) ] (dV_C)/(dk) = (1)/(3) (R+k) [ -R-k + 2R-2k ] (dV_C)/(dk) = (1)/(3) (R+k) (R-3k) Set (dV_C)/(dk) = 0: (1)/(3) (R+k) (R-3k) = 0 Since R > 0 and k 0, R+k cannot be zero. Therefore, R-3k = 0, which implies k = (R)/(3). This value of k is within the valid range 0 k R. Step 5: Verify that this critical point corresponds to a maximum. We can check the sign of the derivative around k = (R)/(3): If k < (R)/(3), then R-3k > 0, so (dV_C)/(dk) > 0 (volume is increasing). If k > (R)/(3), then R-3k < 0, so (dV_C)/(dk) < 0 (volume is decreasing). This confirms that k = (R)/(3) corresponds to a local maximum. Step 6: Calculate the maximum volume of the cone and compare it to the sphere's volume. Substitute k = (R)/(3) back into the expressions for h and r^2: h = R+k = R + (R)/(3) = (4R)/(3) r^2 = R^2 - k^2 = R^2 - ((R)/(3))^2 = R^2 - (R^2)/(9) = (8R^2)/(9) Now, calculate the maximum volume of the cone: V_C = (1)/(3) r^2 h = (1)/(3) ((8R^2)/(9)) ((4R)/(3)) V_C = (1)/(3) (32R^3)/(27) = (32)/(81) R^3 Now, compare this to the volume of the sphere V_S = (4)/(3) R^3: V_C = (32)/(81) R^3 = (8 × 4)/(27 × 3) R^3 = (8)/(27) ((4)/(3) R^3) V_C = (8)/(27) V_S Thus, the volume of the largest cone that can be inscribed in a sphere of radius R is (8)/(27) of the volume of the sphere. That's 2 down. 3 left today — send the next one.