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

MATH100 December 2016

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Question 12
Consider a cone with height $2$ metres and whose circular base has radius $1$ metre. Find the dimensions of the circular cylinder of largest volume that can be contained in the cone. (The base of the cylinder lies at the base of the cone.)

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
First, draw a diagram of a cone with a cylinder inside such that the base of the cylinder lies on the base of the cone.

Hint 2
Use similar triangles to find a relation between the radius of the cylinder, $r$ and its height $h$ .

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $\color {blue}h={\frac {2}{3}},r={\frac {2}{3}}$.