CHAPTER 5 section 1

Excel Companion Chapter 5

Chapter 5 Index

EC_5		NON-LINEAR OPTIMIZATION
	*MaxVolum.xls*	Classical Volume Maximization
	*ClsscMin.xls*	Three Classical Minimization Problems Minimum surface area of open and closed boxes Minimum cost of building open and closed boxes Minimum surface area of cylinder (tin can)
	*PointMP.xls*	Point Marginal Profit: Derivatives
	*EOQ.xls*	Economic Order Quantity
			Ordering and Holding Cost, Total Cost with Discount
	*Response.xls*	New Product Promotion
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Certain classical non-linear optimization problems are related to common two and three dimensional geometric shapes, such as rectangles, boxes, cylinders, or cones. Your objective will be to maximize or minimize some function which may represent an area, volume, or cost of building some two or three dimensional object. Frequently there is some condition that you must satisfy, such as the three dimensional shape must have a fixed volume or the amount of money to enclose the two dimensional shape is fixed. Usually the hardest part is to create the function to be optimized in terms of a single independent variable. For these classical geometric problems, an important first step is to draw a picture and label it with as few variables as possible.

In the first part of this chapter you will investigate and solve one maximization problem in Maxvolum.xls and three classic minimization problems in ClsscMin.xls. Each of the minimization models are three dimensional and include a condition ( constraint) that requires a box or cylinder to contain a fixed volume measured in cubic units.

Maximum Volume of an Open Box

Maxvolum.xls

Open the file Maxvolum.xls.

SITUATION: A flat sheet of cardboard is to be shaped into an open box by cutting identical squares from the four corners and then folding up the sides. The objective is to make the box which has the greatest volume. Suppose we start with a flat sheet which has dimensions 8.5 by 11 inches. If we remove squares with 2 inch sides from each corner, then the box will have volume

V(2) = (8.5 - 4)*(11-4)*2 = 63 cubic inches.

Will this box have the greatest possible volume?

QUESTIONS:
Complete the table and then answer # 1 - 4

x	lengthwidthheight	x	lengthwidthheight
0.50	37.500	2.50
0.75	49.875	2.75
1.00	58.500	3.00	37.500
1.25		3.25	29.250
1.50		3.50	21.000
1.75		3.75
2.00	63.000	4.00
2.25		4.25

1.	Explain why the table stops at x = 4.25.
2.	Sketch the graph of the volume function..
3.	Estimate the value of x that corresponds to the maximum volume.
	a) to the nearest tenth.
	b) to the nearest hundredth.
	c) to the nearest thousandth
4.	Use derivatives and algebra to find the x coordinate of the optimal point to six decimal places.
5.	The first derivative of the volume function has a second root between x = 5 and x = 6. What does this tells us about the graph of the volume function to the right of x =4.25?
6.	Modify the spreadsheet, or create your own spreadsheet from scratch, so that the user can enter the length and width of the flat sheet and obtain a useful table and graph. Enter at least three different pairs of values for length and width and find the value of x (to the nearest tenth) that maximizes the volume function in the appropriate interval
	a) L= W =	x=
	b) L= W =	x=
	c) L= W =	x=
*7.	Use differential calculus to investigate the relationship between the optimal value of x and the values of L and W in the function V(x) = (L-2x)(W-2x)x for x between 0 and W/2.

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