Multivariate Calculus - Bruce E. Shapiro

108 LECTURE 14. UNCONSTRAINED OPTIMIZATION
At (1,0), we have d = 4(1) cos(0) − 4 sin 2 (0) = 4 > 0. Since f xx = 2 > 0 this is
a local minimum.
At (0, −π/2) we have d = D(0, −π/2) = (4)(0) cos(−π/2) − 4 sin 2 (−π/2) =
−4(−1) 2 = −4 < 0 so this is a saddle point.
At (0, π/2) d = D(0, π/2) = (4)(0) cos(π/2) − 4 sin 2 (π/2) = −4(1) 2 = −4 < 0 so
this is also a saddle point.
Figure 14.2: Geometry for example 14.7.
Example 14.7 A rectangular metal tank with an open top is to hold 256 cubic feet
of liquid. What are the dimensions of the tank that requires the least material to
build?
Solution. Assume that the material required to build the tank is proportional to the
area of the cube. Let the box have dimensions x, y, and z as illustrated in figure
14.2.
The area of the bottom is xy; the area of each of the two small sides in the figure
is yz; and the area of each of the larger sides (in the figures) is xz. Therefore the
total area (since there is one bottom but two of each of the additional sides) is
A = xy + 2xz + 2yz
Furthermore, since the volume is 256, and since the volume of the box is xyz, we
have
256 = xyz
or
z = 256/xy
Combining this with our previous equation for the area,
A = xy + 2xz + 2yz
= xy + 2x(256/xy) + 2y(256/xy)
= xy + 512
y + 512
x
Revised December 6, 2006. Math 250, Fall 2006