Consider a cone inscribed in a cube with side length a. The cone's height h and base radius r will be related to the cube's side length
step 2
The volume V of a cone is given by the formula V=31πr2h
step 3
To maximize the volume, we need to express V in terms of a single variable. Since the cone is inscribed in the cube, the height h plus twice the radius 2r equals the side length a of the cube, so h=a−2r
step 4
Substitute h into the volume formula to get V=31πr2(a−2r)
step 5
To find the maximum volume, take the derivative of V with respect to r and set it equal to zero: drdV=31π(2ra−6r2)
step 6
Solve drdV=0 for r to find the critical points
step 7
The critical point that maximizes the volume occurs when r=3a
step 8
Substitute r=3a back into the expression for h to find h=a−2(3a)=3a
step 9
Now substitute r=3a and h=3a into the volume formula to find the maximum volume: Vmax=31π(3a)2(3a)
step 10
Simplify the expression for Vmax to get Vmax=31π27a3=81πa3
Answer
Vmax=81πa3
Key Concept
Maximizing the volume of a cone inscribed in a cube
Explanation
The maximum volume of a cone inscribed in a cube is found by expressing the volume in terms of one variable, taking the derivative, finding the critical points, and then substituting back to find the maximum volume.