Science:Math Exam Resources/Courses/MATH100/December 2016/Question 12/Solution 1

The diagram of the cylinder inside the cone is as the following

cylinder in a cone

where $R=1$ m and $H=2$ m are the base radius and height of the cone.

The two right riangles in the diagram are similar, so we have the following ratios:

${\begin{aligned}{\frac {r}{R}}={\frac {H-h}{H}}&\Rightarrow {\frac {r}{1}}={\frac {2-h}{2}}\\&\Rightarrow r={\frac {1}{2}}(2-h),\qquad 0<r<1,0<h<2\end{aligned}}$

Now we know that the volume of the cylinder is the area of the base times height i.e.

$V=\pi r^{2}h={\frac {\pi }{4}}h(2-h)^{2}={\frac {\pi }{4}}(4h-4h^{2}+h^{3})$

To maximize the volume we need to find the critical points of $V$. We use the power rule to get

$V'(h)={\frac {\pi }{4}}\left(3h^{2}-8h+4\right)=0$

Using the quadratic formula (or factoring) gives $h=2,h={\frac {2}{3}}$. For $h=2$ the volume becomes 0, so the maximum must be attained at $h={\frac {2}{3}}$, and hence $r={\frac {1}{2}}(2-{\frac {2}{3}})={\frac {2}{3}}$, therefore,

$V=\pi \times {\frac {4}{9}}\times {\frac {2}{3}}={\frac {8}{27}}\pi$

Note that in general, we need to check the endpoints for $h$ or $r$, where in this example make the volume equal to 0.

The dimension of the cylinder is therefore $\color {blue}h={\frac {2}{3}},r={\frac {2}{3}}$.