The ebook Elementary Calculus is based on material originally written by H.J. Keisler. For more information please read the copyright pages.

Home Partial Differentiation Maxima and Minima Examples Example 2: Maximum Volume
 
del.icio.us Reddit Facebook StumbleUpon MySpace Twitter Search the VIAS Library  |  Index

Example 2: Maximum Volume

For a package to be mailed in the United States by parcel post, its length plus its girth (perimeter of cross section) must be at most 84 inches. Find the dimensions of the rectangular box of maximum volume which can be mailed by parcel post.

Step 1

Let x, y, and z be the dimensions of the box, with z the length. We wish to find the maximum of the volume

V = xyz given the side condition length + girth = z + 2x + 2y = 84. We eliminate z using the side condition and express Fas a function of x and y.

z = 84 - 2x - 2y, V = xy(84 -2x- 2y).

Since x, y, and z cannot be negative the domain is the closed triangle 0 ≤ x, 0 ≤ y, 0 ≤ 84 - 2x - 2y. This is the same as the closed region 0 ≤ x ≤ 42, 0 ≤ y ≤ 42 - x. The region is sketched in Figure 11.7.6.

11_partial_differentiation-461.gif

Figure 11.7.6

Step 2

11_partial_differentiation-462.gif

Step 3

84y - 4xy - 2y2 = 0, 84x - 2x2 - 4xy = 0.

Since x > 0 and y > 0 at all interior points, we have

84 - 4x - 2y = 0, 84 - 2x - 4y = 0.

There is one critical point

x = 14, y = 14, V = (84 - 28 - 28) · 14 · 14 = 2(14)3.

Step 4

On all three of the boundary lines

x = 0, y = 0, 84 - 2x - 2y = 0

we have

V = (84 - 2x - 2y)xy = 0.

Therefore the maximum value of V the boundary of D is 0.

CONCLUSION

The maximum of V is at x = 14, y = 14, where V = 2(14)3 (Figure 11.7.7). The box has dimensions

x = 14, y =14, z = 28.

11_partial_differentiation-463.gif 11_partial_differentiation-464.gif
Figure 11.7.7 (a) (b)

 
Home Partial Differentiation Maxima and Minima Examples Example 2: Maximum Volume

Last Update: 2006-11-15