Question

Prove that the radius of the right circular cylinder of maximum curved surface inscribed in a cone is half of the radius of the cone.

Hint:

In optimization problems involving inscribed shapes, the key is almost always to find a relationship between the dimensions of the inner and outer shapes using similar triangles (for cones/pyramids) or trigonometry (for spheres). This allows you to reduce the function you're optimizing to a single variable.

Answer 1

Step 1: Understanding the Concept:
This is an optimization problem that falls under the application of derivatives. We need to find the dimensions of an inscribed cylinder that maximize its curved surface area. The key is to express the quantity to be maximized (curved surface area) as a function of a single variable and then use calculus to find the maximum.
Step 2: Key Formula or Approach:
1. Define the dimensions of the cone (radius \(R\), height \(H\)) and the inscribed cylinder (radius \(r\), height \(h\)).
2. Use similar triangles from a cross-sectional view to establish a relationship between \(r\), \(h\), \(R\), and \(H\).
3. Write the formula for the curved surface area (CSA) of the cylinder, \(S = 2\pi rh\).
4. Use the relationship from step 2 to express \(S\) as a function of a single variable, say \(r\).
5. Find the first derivative of \(S\) with respect to \(r\) (\(\frac{dS}{dr}\)) and set it to zero to find critical points.
6. Use the second derivative test (\(\frac{d^2S}{dr^2}\)) to confirm that the critical point corresponds to a maximum.
Step 3: Detailed Explanation or Calculation:
Let the cone have a fixed radius \(R\) and height \(H\). Let the cylinder inscribed within it have a variable radius \(r\) and height \(h\).
Consider a vertical cross-section of the cone and cylinder. This forms two similar right-angled triangles. The larger triangle has height \(H\) and base \(R\). The smaller triangle, which is above the cylinder, has height \(H-h\) and base \(r\).
By the property of similar triangles: \[ \frac{H-h}{H} = \frac{r}{R} \] We can express \(h\) in terms of \(r\) (and the constants \(R, H\)): \[ 1 - \frac{h}{H} = \frac{r}{R} \implies \frac{h}{H} = 1 - \frac{r}{R} \implies h = H\left(1 - \frac{r}{R}\right) \] The curved surface area (CSA) of the cylinder is given by \(S = 2\pi rh\).
Substitute the expression for \(h\) into the formula for \(S\) to get \(S\) as a function of \(r\): \[ S(r) = 2\pi r \left[ H\left(1 - \frac{r}{R}\right) \right] \] \[ S(r) = 2\pi H \left(r - \frac{r^2}{R}\right) \] To find the maximum area, we differentiate \(S(r)\) with respect to \(r\) and set the derivative to zero: \[ \frac{dS}{dr} = \frac{d}{dr} \left[ 2\pi H \left(r - \frac{r^2}{R}\right) \right] \] \[ \frac{dS}{dr} = 2\pi H \left(1 - \frac{2r}{R}\right) \] Set \(\frac{dS}{dr} = 0\) for critical points: \[ 2\pi H \left(1 - \frac{2r}{R}\right) = 0 \] Since \(2\pi H \neq 0\), we must have: \[ 1 - \frac{2r}{R} = 0 \implies 1 = \frac{2r}{R} \implies 2r = R \implies r = \frac{R}{2} \] To confirm this is a maximum, we use the second derivative test: \[ \frac{d^2S}{dr^2} = \frac{d}{dr} \left[ 2\pi H \left(1 - \frac{2r}{R}\right) \right] \] \[ \frac{d^2S}{dr^2} = 2\pi H \left(0 - \frac{2}{R}\right) = -\frac{4\pi H}{R} \] Since \(R\) and \(H\) are positive dimensions, \(\frac{d^2S}{dr^2}\) is negative. A negative second derivative indicates that the function has a local maximum at the critical point.
Step 4: Final Answer:
The curved surface area of the inscribed cylinder is maximum when its radius \(r\) is equal to \(\frac{R}{2}\), which is half the radius of the cone. This completes the proof.