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Chapter 8: Problem 39

Given a right circular cone of base radius \(a\) and height \(h\), find the radius and the height of the right circular cylinder having the largest lateral surface area that can be inscribed in the cone.

Short Answer

Expert verified
The radius and the height of the largest inscribed cylinder within a right circular cone of base radius \(a\) and height \(h\) are \(\frac{a}{2}\) and \(\frac{h}{2}\), respectively.

Step by step solution

01

Problem visualization

Sketch a right circular cone and the inscribed cylinder. Label the radius and height of the cone as `a` and `h`, respectively, and the radius and height of the cylinder as `r` and `H`, respectively. Also label the distance from the vertex of the cone to the base of the cylinder as `x`.
02

Derive a relationship between r and x

Since the right circular cone and the inscribed cylinder share a common axis, we can use similar triangles to relate their dimensions. In the cone and the triangle formed by the vertex, the base center on the cone, and the point of the cone where the cylinder meets it, we can set up the following proportion: \[\frac{r}{x} = \frac{a}{h}\] We can solve for `r` in terms of `x`: \[r = \frac{a}{h}x\]
03

Express the height H in terms of x

Using the height H of the cylinder and the distance x from the vertex of the cone to the base of the cylinder, we can write down the following equation: \[H = h - x\]
04

Express the lateral surface area of the cylinder in terms of x

The lateral surface area A of a cylinder can be calculated as follows: \[A = 2\pi rH\] Now, substitute the expressions for `r` and `H` in terms of `x`: \[A = 2\pi \left(\frac{a}{h}x\right)(h - x)\] Simplify the equation: \[A = 2\pi ax - 2\pi \frac{a}{h}x^2\]
05

Find critical points of A

To find the critical points of A, we need to compute its first derivative with respect to x and set it equal to zero: \[\frac{dA}{dx} = 2\pi a - 4\pi \frac{a}{h}x\] Set the first derivative equal to zero and solve for x: \[2\pi a - 4\pi \frac{a}{h}x = 0\] \[x = \frac{h}{2}\]
06

Verify the maximum using the second derivative test

Compute the second derivative of A with respect to x: \[\frac{d^2A}{dx^2} = - 4\pi \frac{a}{h}\] Since \(\frac{a}{h} > 0\) and the second derivative is negative, we can conclude that the function A has a maximum at the critical point \(x = \frac{h}{2}\).
07

Compute r and H for the largest inscribed cylinder

Now, we can find the dimensions of the largest inscribed cylinder using the expressions for r and H in terms of x, and plugging in \(x = \frac{h}{2}\): \[r = \frac{a}{h}\left(\frac{h}{2}\right) = \frac{a}{2}\] \[H = h - \frac{h}{2} = \frac{h}{2}\] Thus, the radius and the height of the largest inscribed cylinder are \(\frac{a}{2}\) and \(\frac{h}{2}\), respectively.

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Most popular questions from this chapter

Let \(p, q \in \mathbb{R}\) satisfy \(0 \leq p

A twisted solid is generated as follows. A fixed line \(L\) in 3 -space and a square of side \(s\) in a plane perpendicular to \(L\) are given. One vertex of the square is on \(L\). As this vertex moves a distance \(h\) along \(L\), the square turns through a full revolution with \(L\) as the axis. Find the volume of the solid generated by this motion. What would the volume be if the square had turned through two full revolutions in moving the same distance along the line \(L ?\)

Let \(f:[a, b] \rightarrow \mathbb{R}\) be differentiable such that \(f^{\prime}\) is integrable on \([a, b] .\) Show that the average of \(f^{\prime}\) is equal to the average rate of change of \(f\) on \([a, b]\), namely \([f(b)-f(a)] /(b-a)\)

Let \(f:[a, b] \rightarrow \mathbb{R}\) be a function, \(n \in \mathbb{N}\), and \(P_{n}:=\left\\{x_{0}, x_{1}, \ldots, x_{n}\right\\}\) be any partition of \([a, b] .\) Define $$ \begin{gathered} R\left(P_{n}, f\right):=\sum_{i=1}^{n} f\left(x_{i-1}\right)\left(x_{i}-x_{i-1}\right) \\ M\left(P_{n}, f\right):=\sum_{i=1}^{n} f\left(\frac{x_{i-1}+x_{i}}{2}\right)\left(x_{i}-x_{i-1}\right) \\ T\left(P_{n}, f\right):=\frac{1}{2} \sum_{i=1}^{n}\left[f\left(x_{i-1}\right)+f\left(x_{i}\right)\right]\left(x_{i}-x_{i-1}\right), \end{gathered} $$ and $$ S\left(P_{n}, f\right):=\frac{1}{6} \sum_{i=1}^{n}\left[f\left(x_{i-1}\right)+4 f\left(\frac{x_{i-1}+x_{i}}{2}\right)+f\left(x_{i}\right)\right]\left(x_{i}-x_{i-1}\right) . $$ If \(f\) is a polynomial function of degree at most 1, then show that $$ R\left(P_{n}, f\right)=M\left(P_{n}, f\right)=T\left(P_{n}, f\right)=\int_{a}^{b} f(x) d x $$ and if \(f\) is a polynomial function of degree at most 2 , then show that $$ S\left(P_{n}, f\right)=\int_{a}^{b} f(x) d x $$

Use a result of Pappus to find (i) the volume of a cylinder with height \(h\) and radius \(a\) (ii) the volume of a cone with height \(h\) and base radius \(a .\)
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