Presentation on theme: "3.7 Modeling and Optimization"— Presentation transcript:
1 3.7 Modeling and Optimization Buffalo Bill’s Ranch, North Platte, NebraskaGreg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, 1999
2 A Classic ProblemYou have 40 feet of fence to enclose a rectangular garden along the side of a barn. What is the maximum area that you can enclose?There must be a local maximum here, since the endpoints are minimums.
3 A Classic ProblemYou have 40 feet of fence to enclose a rectangular garden along the side of a barn. What is the maximum area that you can enclose?
4 Procedure for Solving Optimization Problems Assign symbols to all given quantities and quantitiesto be determined.Write a primary equation for the quantity to bemaximized or minimized.Reduce the primary equation to one having ONEindependent variable. This may involve the use ofsecondary equation.Determine the domain. Make sure it makes sense.Determine the max or min by differentiation. (Find the first derivative and set it equal to zero)If the maximum or minimum could be at the endpoints, you have to check them.
5 We can minimize the material by minimizing the area. Example:What dimensions for a one liter cylindrical can will use the least amount of material?MotorOilWe can minimize the material by minimizing the area.We need another equation that relates r and h:area ofendslateralarea
6 Example 5:What dimensions for a one liter cylindrical can will use the least amount of material?area ofendslateralarea
7 A manufacturer wants to design an open box having a square base and a surface area of 108 in2.What dimensions will produce a box with maximumvolume?Since the box has a squarebase, its volume isV = x2hhNote: We call this the primaryequation because it gives aformula for the quantity wewish to optimize.xxThe surface area = the area of the base + the area of the 4 sides.S.A. = x2 + 4xh = 108We want to maximize the volume,so express it as a function of justone variable. To do this, solvex2 + 4xh = 108 for h.
8 3x2 = 108 Substitute this into the Volume equation. To maximize V we find the derivative and it’s C.N.’s.3x2 = 108We can conclude that V is a maximum when x = 6 andthe dimensions of the box are 6 in. x 6 in. x 3 in.
9 Find the points on the graph of y = 4 – x2 that are closest to the point (0,2).What is the distance from thepoint (x,y) and (0,2)?4321(x,y)(x,y)We want this dist. to be minimized.
10 = 2x(2x2 – 3) = 0 C.N.’s imaginary dec. dec. inc. inc. 1st der. test dec.dec.inc.inc.1st der. testminimumminimumTwo closest points.