Solution:
Using the diagram:
Step 1: Using similar triangles and simplifying:
[tex]\begin{gathered} \frac{H}{R}=\frac{H-h}{r} \\ Making\text{ }h\text{ }subject \\ h=\frac{H}{R}(R-r) \end{gathered}[/tex]Step 2: Use the Volume of the smaller cone
[tex]\begin{gathered} V(r)=\frac{1}{3}\pi r^2h \\ Substituting\text{ }h \\ V(r)=\frac{1}{3}\pi r^2[\frac{H}{R}(R-r)] \\ V(r)=\frac{\pi H}{3R}(Rr^2-r^3) \end{gathered}[/tex]Step 3: Differentiating to get maximum value
[tex]\begin{gathered} V^{\prime}(r)=\frac{\pi H}{3R}r(2R-3r) \\ At\text{ maximum V'\lparen r\rparen=0} \\ r=0\text{ }or\text{ 3r=2R} \\ r=\frac{2R}{3} \end{gathered}[/tex]Step 4: Equate radius, r in the height formula from step 1
[tex]\begin{gathered} h=\frac{H}{R}(R-r) \\ h=\frac{H}{R}(R-\frac{2R}{3}) \\ h=\frac{H}{R}\cdot\frac{R}{3} \\ h=\frac{H}{3} \end{gathered}[/tex]
Final answer:
the inner cone has a maximum volume when:
[tex]h=\frac{H}{3}[/tex]