**Example 1:** A rectangular box with a square base and no top is to have a volume of 108 cubic inches. Find the dimensions for the box that require the least amount of material.

The function that is to be minimized is the surface area ( *S*) while the volume ( *V*) remains fixed at 108 cubic inches (Figure 1) .

**Figure 1 **The open‐topped box for Example 1.

Letting *x* = length of the square base and *h* = height of the box, you find that

with the domain of *f(x)* = (0,+∞) because *x* represents a length.

hence, a critical point occurs when *x* = 6. Using the Second Derivative Test:

and *f* has a local minimum at *x* = 6; hence, the dimensions of the box that require the least amount of material are a length and width of 6 inches and a height of 3 inches.

**Example 2:** A right circular cylinder is inscribed in a right circular cone so that the center lines of the cylinder and the cone coincide. The cone has 8 cm and radius 6 cm. Find the maximum volume possible for the inscribed cylinder.

The function that is to be maximized is the volume ( *V*) of a cylinder inscribed in a cone with height 8 cm and radius 6 cm (Figure ) .

**Figure 2 **A cross section of the cone and cylinder for Example 2.

Letting *r* = radius of the cylinder and *h* = height of the cylinder and applying similar triangles, you find that

Because *V* = π *r* ^{2} *h* and *h* = 8 −(4/3) *r*, you find that

with the domain of *f(r)* = [0,6] because *r* represents the radius of the cylinder, which cannot be greater that the radius of the cone.

Because *f(r*) is continuous on [0,6], use the Extreme Value Theorem and evaluate the function at its critical points and its endpoints; hence,

hence, the maximum volume is 128π/3 cm ^{3}, which will occur when the radius of the cylinder is 4 cm and its height is 8/3 cm.