Calc_Wordle.JPG
aaaaaa.jpg







Formal Definition

Optimization, or applied maximum and minimum, has no definite formal definition. It is solving for the absolute maximum or minimum of a function used to find real life applications. For example, a packaging company may want to make a box with X amount of volume, but with the minimum amount of packaging to minimize cost. Finding the dimensions of the box would be the proccess of optimization.

Step by step


Optimization almost always deals with word problems.
When solving word problems, it is important to:
a. read the problem thoroughly several times- sometimes you won't understand what the problem is after reading it once
b. draw a sketch of what the problem is asking for (if possible)
c. define your variables
d. identify useful equations and relations
e. make sure that you have an equation that you can differentiate within the Rules of Differentiation
f. use the intervals to confirm that the values are maximums or minimums


Source


Things to Remember


Make sure that you make your primary equation has only one independent variable.


Example 1: Fencing


You need to enclose a field with a fence. You have 500 feet of fencing material and a building is on one side of the field so you won’t need any fencing there. Determine the dimensions of the field that will enclose the largest area.

a: You need to maximize the area of the field, but you can't use more than 500 feet of fencing. Simple enough, isn't it?
b. Draw a sketch:
building.JPG
c: In this problem, x represents the length of the long end of the pen, while y represents the length of the short ends. Remember--- you have two y's, but only one x.
d: You need two equations for this problem: an optimization and a constraint.
You know you need to maximize the area, so you will use the area equation: A=xy
Your constraint is the amount of fencing that you can use, so we know that two y's and one x is 500 feet: x+2y=500.

e: You need to make your optimization equation a single variable, so you will solve the constraint equation for y=(500-x)/2
Then, you can substitute that y into the area equation- A=x(500-x)/2. Now you can differentiate
f: Differentiate like this:
A=x(500-x)/2
A=(500x-x^2)/2
A=250x-(1/2)x^2

A'=250-x
0=250-x
x=250



Example 2: Print

example taken from textbook

A rectangular page is to cointain 24 square inches of print. The margins at the top and bottom of the page are to be 1.5 inches, and th emargins on the left and right are to be 1 inch. What should the dimensions of the page be so that the least amount of paper is used?

a: You need to minimize the amount of paper being used, but you must remain within the margins given (1 and 1.5) while maintaining the 24 square inches of print in the center.
b. Draw a sketch:
pageprint.JPG
c. Area of the page: A=(x+3)(y+2)
The printed area: xy=24

d: Since you need to work with a single variable, solve for y for the printed area and substitute into the area of the page.
y=24/x
A=(x+3)(24/x+2)=30+2x+72/x

e: Since you already have your single variable equation, A=30+2x+72/x, you can now differentiate.
f: Differentiate like this:
A=30+2x+72/x
dA/dt=2-(72/x^2)

0=2-(72/x^2)
x=6 inches


Example 3: Wires

example taken from textbook

Two posts, one 12 feet high and the other 28 feet high, stand 30 feet apart. They are to be stayed by two wires, attached to a single stake, running from ground level to the top of each post. Where should the stake be placed to use the least amount of wire?

a. You need to find a way to minimize the amount of wire between the two poles.
b. Draw a sketch:

wire.JPG
c. Your y variable is the section of wire corresponding to the 12ft pole and the z variable is the wire corresponding to the 28ft pole. W is the amount of wire of y and z combined. You want to find the minimum amount for W. The distance between the two poles is 30ft, so the one section between the 12ft pole and stake is variable x and the section between the stake and the 28ft pole is 30-x.
d. W=y+z
y^2=12^2+x^2
so
y=sqrt(x^2+144) and
z=sqrt(x^2-60x+1684)

With these variables, you can substitute in to solve for W
Thus, W=sqrt(x^2+144) + sqrt(x^2-60x+1684)

e. W is differentiable, so,
dW/dx=(x/sqrt(x^2+144))+((x-30)/sqrt(x^2-60x-1684)
Set the equation equal to zero to find your intervals.

f. x=9 and x=-22.5 Since -22.5 is an impossible number (for this example), your intervals are from (0, 9)(9,30)
W(0)=53 W(9)=50 and W(30)=60
So, W(9) is the minimum value, optimizing the use of the wire.


Video




Practice Sites

Calculus homework: Optimization (.pdf)
Qrhetoric Calculus
Demos for Optimization
Optimization Problems by Ken Norris (.pdf)




This is the end of our wiki page.