Question

A cylinder is inscribed in a right circular cone of height 4.5 and radius (at the base) equal to 5.5 . What are the dimensions of such a cylinder which has maximum volume? Its radius is equation editorEquation Editor and its height is

Answer 1

Answer:

r = 3.667

h = 1.5

Step-by-step explanation:

Given:-

- The base radius of the right circular cone, R = 5.5

- The height of the right circular cone, H = 4.5

Solution:-

- We will first define two variables that identifies the volume of a cylinder as follows:

                                r: The radius of the cylinder

                                h: The height of cylinder

- Now we will write out the volume of the cylinder ( V ) as follows:

                                $V = \pi*r^2h$

- We see that the volume of the cylinder ( V ) is a function of two variables ( don't know yet ) - ( r,h ). This is called a multi-variable function. However, some multi-variable functions can be reduced to explicit function of single variable.

- To convert a multi-variable function into a single variable function we need a relationship between the two variables ( r and h ).

- Inscribing, a cylinder in the right circular cone. We will denote 5 points.

              Point A: The top vertex of the cone

              Point B: The right end of the circular base ( projected triangle )

              Point C: The center of both cylinder and base of cone.

              Point D: The top-right intersection point of cone and cylinder

              Point E: Denote the height of the cylinder on the axis of symmetry of both cylinder and cone.  

- Now, we will look at a large triangle ( ABC ) and smaller triangle ( ADE ). We see that these two triangles are "similar". Therefore, we can apply the properties of similar triangles as follows:

                              $\frac{AC}{AE} = \frac{BC}{DE} \\\\\frac{H}{H-h} = \frac{R}{r}$

- Now we can choose either variable variable to be expressed in terms of the other one. We will express the height of cylinder ( h ) in term of radius of cylinder ( r ) as follows:

                             $H- h = r\frac{H}{R} \\\\h = \frac{H}{R}*(R-r)$

- We will use the above derived relationship and substitute into the formula given above:

                            $V = \pi r^2 [ \frac{H}{R}*(R - r )]\\\\V = \frac{\pi H}{R}.r^2.(R-r)$

- Now our function of volume ( V ) is a single variable function. To maximize the volume of the cylinder we need to determine the critical points of the function as follows:

                            $\frac{dV}{dr} = \frac{\pi H}{R}*(2rR-2r^2 - r^2 )\\\\\frac{dV}{dr} = \frac{\pi H}{R}*(2rR-3r^2 ) = 0\\\\(2rR-3r^2 ) = 0\\\\2R -3r = 0\\\\r = \frac{2}{3}*R$

- We found the limiting value of the function. The cylinder volume maximizes when the radius ( r ) is two-thirds of the radius of the right circular cone.

- We can use the relationship between the ( r ) and ( h ) to determine the limiting value of height of cylinder as follows:

                          $h = \frac{H}{R} * ( R - \frac{2}{3}R)\\\\h = \frac{H}{3}$

- The dimension of the inscribed cylinder with maximum volume are as follows:

                         $r = \frac{2}{3}*5.5 = 3.667\\\\h = \frac{4.5}{3} = 1.5$

Note: When we solved for the critical value of radius ( r ). We actually had two values: r = 0 , r = 2R/3. Where, r = 0 minimizes the volume and r = 2R/3 maximizes. Since the function is straightforward, we will not test for the nature of critical point ( second derivative test ).