Optimization Most often we will not be handed an objective function - - PDF document

Mt020.02 Slide 1 on 3/31/00

raj

Optimization Most often we will not be handed an objective function which should be optimized. In those cases we’ll have to rely on our wits to build

• ne.

Example: An open box is to be made from a sheet of tin eight inches square by cutting out identical squares from each corner and bending up the resulting flaps. Determine the dimensions of the largest box that can be made. Solution: <Draw a picture>, and label the aspects that will not vary with appropriate constants, and label those parts which can vary with variables. <Name the variables> and define them. Picture goes here:

Mt020.02 Slide 2 on 3/31/00

raj

Let x =the side length of the square which is removed from each corner. <Build the objective function> which describes the quantity to be optimized. Here we are to maximize the box’s volume. Notice that by changing x, we indirectly change the volume contained by the box. V, the volume, is a function of x. V(x) = (8 - 2x)2(x) = 64x - 32x2 + 4x3 <Limit the domain to conform to the model> Which values for x are we allowing? We must have 0 ≤ x ≤ 4, in order for our box model to make sense.

Mt020.02 Slide 3 on 3/31/00

raj

We are now at the point where we are used to being, namely our problem is Maximize V(x) = 64x - 32x2 + 4x3 over the interval [0,4]. Differentiate to get V’(x) = 64 - 64x + 12x2 which factors as V’(x) = 4(3x – 4)(x – 4). We see that the critical points of V lying in our special domain [0,4] are 4, and 4/3, so we add the endpoints into the mix and come up with

• ur candidate (x) basket {0 , 4/3 , 4} and when

we apply the original function (V) individually to items in this basket we get {0 , 37.926 , 0} showing that x = 4/3 is that x value yielding a maximal volume. So to build the biggest box, we should cut a square 1.33 inches on a side from each corner. Which values of x provide the smallest box?

Mt020.02 Slide 4 on 3/31/00

raj

Example: Imagine your job is to design a beer can which holds exactly 12 ounces of beer (this is just 336 cubic centimeters). Your beer can must be a cylinder, with a top and bottom, and you must use the minimum amount of aluminum in the manufacturing process. How should you design it? Solution: <Draw a picture>. Picture here: <Label it with constants and variables>. The can has two variables, r = the radius of the circular end disks, and h = the height of the cylindrical can itself. <Build the objective function> We are to minimize the total amount of aluminum used so

Mt020.02 Slide 5 on 3/31/00

raj

we begin with

• Alum. (A) used = (A for sides) + (A for ends)

= (2πr)h + 2(πr2) so we get A = 2πrh + 2πr2 = 2πr (h + r). Note that our objective function has two variables in it (r and h) and we only work with

• ne. We now need to work to express h in

terms of f, thereby eliminating h from our

• bjective function.

How can we link h and r in an equation to facilitate this goal? Well if h and r were truly independent, we could say increase them both

• simultaneously. What prevents us from doing

that? Our volume must remain at 336 cubic

• centimeters. Indeed the volume of our

proposed can is 336 = V = πr2h and this is the required link joining h and r. We solve it for h

Mt020.02 Slide 6 on 3/31/00

raj

and get h = 336 / (πr2), and now we can rewrite

• ur objective function

A = 2πr (h + r) = 2πr ((336 / (πr2)) + r). = 672 / r + 2 πr2 A(r) = 672 r-1 + 2 πr2 <objective fcn> <limits on domain?> We must restrict r to live in the interval (0 , ∞). Critical points of A occur where the derivative A’ = 0 or else where A’ d.n.e. Note there are no weird (A’ dne) critical pts on our domain. A’ will = 0 when 0 = A’(r) = -672 r-2 + 4πr or in

• ther words when r3 = 672 / (4π). This reduces

to r = 3.767 cm and then h = 7.535 cm (=2r). Why aren’t beer cans designed in these proportions?