Relating calculus application to similar triangles.

Question

I am having trouble with how to prove and show part (a). All tips are welcome.

Thank You.

Answer 1

Hint:

The figure is a section of your cone and cilinder. We have: $$ AM=r-R \qquad MN=H \qquad BC=h \qquad AB=r $$

and the triangle $AMN$ and $ABC$ are similar. So....

Answer 2

Considering the big triangle and the small triangle, We can write:

$$\frac{r}{R}=\frac{h}{h-H}$$

Therefore, we get: $$r(h-H)=hR$$ $$rH=hr-hR \longrightarrow H=\frac{h(r-R)}{r}$$

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Where, the big triangle can be specified by $2r$ and $h$, the small triangle can be specified by $r$ and $h-H$.