A sphere of radius \(r\) is inscribed in a right circular cone (Figure \(1a\)). Find the minimum volume of the cone.
Example 2
Find the cylinder with the smallest surface area (Figure \(2a\)).
Example 3
Given a cone with a slant height \(l\) (Figure \(3a\)). Find the largest possible volume of the cone.
Example 4
Determine the largest volume of a cylinder inscribed in a cone with height \(H\) and base radius \(R\) (Figure \(4a\)).
Example 5
A coffee filter has a conical shape (Figure \(5a\)). It is made from a circle sector with the central angle \(\alpha\) such that the volume of coffee held in the filter is maximized. Determine the central angle \(\alpha.\)
Example 6
Find a cone with the largest volume inscribed in a sphere of radius \(R\) (Figure \(6a\)).
Example 7
A piece of cardboard is a rectangle of sides \(a\) and \(b.\) An equal square is cut out from each corner and the sides are folded up to make an open-top rectangular box (Figure \(7a\)). How large the square should be to make the box with the largest possible volume?
Example 8
A Chinese tea bowl has the shape of a spherical cap with the radius \(r\) and the height \(h\) (Figure \(8a\)). Determine the ratio \(\frac{h}{r}\) that maximizes the volume of the bowl for a fixed surface area.
Example 1.
A sphere of radius \(r\) is inscribed in a right circular cone (Figure \(1a\)). Find the minimum volume of the cone.
Solution.
The volume of a cone is given by the formula
\[{V = \frac{1}{3}\pi {R^2}H},\]
where \(R\) is the radius of the base and \(H\) is the height.
Figure 1a.
The triangles \(\triangle CMO\) and \(\triangle CKB\) are similar. Then
Find the cylinder with the smallest surface area (Figure \(2a\)).
Solution.
The optimal shape of a cylinder at a fixed volume allows to reduce materials cost. Therefore, this problem is important, for example, in the construction of oil storage tanks (Figure \(2a\)).
Figure 2a.
Let \(H\) be the height of the cylinder and \(R\) be its base radius. The volume and total surface area of the cylinder are calculated by the formulas
This value of \(R\) corresponds to the minimum surface area \(S\left( R \right),\) as when passing through this point the derivative changes sign from minus to plus.
In other words, the height of the cylinder with the smallest surface area should be equal to its diameter, that is an axial section of this cylinder has the form of a square.
Example 3.
Given a cone with a slant height \(l\) (Figure \(3a\)). Find the largest possible volume of the cone.
The solution \({x_1} = 0\) corresponds to a zero-volume cylinder and has no physical meaning. When passing through the point \({x_2} = \frac{{2R}}{3}\) the derivative changes sign from positive to negative. Therefore, \(x = \frac{{2R}}{3}\) is the point of maximum of the function \(V\left( x \right).\) For this base, the height of the cylinder is given by
A coffee filter has a conical shape (Figure \(5a\)). It is made from a circle sector with the central angle \(\alpha\) such that the volume of coffee held in the filter is maximized. Determine the central angle \(\alpha.\)
Solution.
Figure 5a.Figure 5b.
The arc length of the sector is given by
\[P = l\alpha ,\]
where \(l\) is the radius of the sector, and the angle \(\alpha\) is measured in radians.
Comparing the two figures we can write the relationship
The first solution \(\alpha = 0\) is trivial. The second point \(\alpha = \sqrt {\frac{8}{3}} \pi \) corresponds to the maximum of the function \(V\left( \alpha \right)\) (by the First Derivative Test).
Find a cone with the largest volume inscribed in a sphere of radius \(R\) (Figure \(6a\)).
Solution.
Consider the axial cross-section of the cone inscribed in the sphere (Figure \(6a\)).
Figure 6a.
We introduce the following notations: \(H\) is the height of the cone, \(r\) is the base radius of the cone, \(\alpha\) is the angle between the radius of the sphere and the base of the cone. The base radius and height of the cone are connected with the radius of the sphere by the following relationships:
\[r = R\cos \alpha ,\;\;\;H = R\sin \alpha + R.\]
In such a case, the volume of the cone can be written as
As it can be seen, the solution is \(\sin \alpha = \frac{1}{3}.\) We can check that with increasing \(\alpha\) and passing through this point the derivative changes sign from plus to minus, i.e. here we reach the maximum volume of the cone.
which is \(\frac{{8}}{{27}}\) of the volume of the sphere.
Example 7.
A piece of cardboard is a rectangle of sides \(a\) and \(b.\) An equal square is cut out from each corner and the sides are folded up to make an open-top rectangular box (Figure \(7a\)). How large the square should be to make the box with the largest possible volume?
Solution.
Figure 7a.
Let \(x\) be the side of the square. The three sides of the rectangular box are equal to \(a - 2x,\) \(b - 2x\) and \(x\). Then the volume of the box is given by
\[{x_{1,2}} = \frac{{4\left( {a + b} \right) \pm \sqrt {16\left( {{a^2} + ab + {b^2}} \right)} }}{{24}} = \frac{{a + b \pm \sqrt {{a^2} + ab + {b^2}} }}{6}.\]
We should take the solution with the minus sign as \(x \lt a\) and \(x \lt b.\) Hence,
\[x = \frac{{a + b - \sqrt {{a^2} + ab + {b^2}} }}{6}.\]
Example 8.
A Chinese tea bowl has the shape of a spherical cap with the radius \(r\) and the height \(h\) (Figure \(8a\)). Determine the ratio \(\frac{h}{r}\) that maximizes the volume of the bowl for a fixed surface area.
Solution.
Figure 8a.Figure 8b.
The outer surface area of the bowl is expressed in the form