Problem 41 - Calculus Early Transcendentals

Question

Slant height and cones Among all right circular cones with a slant height of \(3,\) what are the dimensions (radius and height) that maximize the volume of the cone? The slant height of a cone is the distance from the outer edge of the base to the vertex.

Step-by-Step Solution

Answer

Answer: The dimensions that maximize the volume of the right circular cone with a slant height of 3 units are radius r = 1 unit and height h = 2√2 units.

Step 1: Understanding the Problem and Variables

In this problem, we have a right circular cone with a fixed slant height of 3 units (slant height = \(3\)). The dimensions that we need to find are the radius (\(r\)) and the height (\(h\)) of the cone. The volume of a right circular cone with radius \(r\) and height \(h\) is given by the formula: \(V = \frac{1}{3}\pi r^2 h.\)

Step 2: Define the Constraint Equation

Since the slant height is given as a fixed value, we can use the Pythagorean theorem to establish a relationship between the variables \(r\), \(h\), and the slant height, \(3\). The Pythagorean theorem states that for a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. Therefore, in this case, we have: \( r^2 + h^2 = 3^2 \)

Step 3: Express the Height in Terms of the Radius

From the constraint equation, we can express the height \(h\) in terms of the radius \(r\): \( h^2 = 3^2 - r^2 \Rightarrow h = \sqrt{9 - r^2}\)

Step 4: Rewrite the Volume Formula in Terms of Radius

Replace the height \(h\) in the volume formula with the expression obtained in Step 3: \(V(r) = \frac{1}{3}\pi r^2 (\sqrt{9 - r^2})\)

Step 5: Find the Radius that Maximizes the Volume

To maximize the volume of the cone, we need to find the critical points (\(r\)) of the volume function \(V(r)\). To do this, find the derivative of \(V(r)\) with respect to \(r\) and set it equal to zero. Then, solve for \(r\). \(V'(r) = \pi r \left(\sqrt{9 - r^2}\right) + \frac{1}{3}\pi r^2 \left(-\frac{r}{\sqrt{9 - r^2}}\right) = 0\) After simplification, we have: \(3\sqrt{9 - r^2} - r^2 = 0\) Solving for \(r\), we get: \( r = 1 \)

Step 6: Find the Height that Maximizes the Volume

Now that we have the radius \(r\) that maximizes the volume, we can find the corresponding height \(h\) using the expression obtained in Step 3: \( h = \sqrt{9 - r^2} = \sqrt{9 - 1^2} = \sqrt{8} = 2\sqrt{2} \)

Step 7: State the Solution

The dimensions that maximize the volume of the right circular cone with a slant height of \(3\) are: - Radius: \(r = 1\) - Height: \(h = 2\sqrt{2}\)