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### slide73-A3

Course: MATH 131, Fall 2008
School: Ohio State
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7.2 Lecture Chapter Quiz QUIZ: Minimize Z=y-x, subject to the following constraints: x3 x + 3y 6 x - 3 y -6 x0 y0 Step one: construct feasible region for each inequality draw the corresponding line and decide on the side needed first inequality x 3 draw the line (vertical line); decide which side by using the origin: 0 3 false! The other side then x + 3y 6 Second inequality: corresponding line:...

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7.2 Lecture Chapter Quiz QUIZ: Minimize Z=y-x, subject to the following constraints: x3 x + 3y 6 x - 3 y -6 x0 y0 Step one: construct feasible region for each inequality draw the corresponding line and decide on the side needed first inequality x 3 draw the line (vertical line); decide which side by using the origin: 0 3 false! The other side then x + 3y 6 Second inequality: corresponding line: x+3y=6 or y=-x/3+2; draw the line (it passes through (0,2) and (6,0); decide on the side, by using the origin: 0 + 3 0 6 wrong, so it's the other side So far, our region looks like this: Third inequality: x - 3 y -6 corresponding line: x-3y=-6 or y=x/3+2; draw the line (passes through (0,2) and (-6,0); decide on the side - origin again: 0 - 3 0 -6 true! So it's the side WITH the origin in it 1 So far our region looks like this: For the last two inequalities (the ... nonnegativity conditions!) you can shortcut it (it's a standard technique; you won't lose any points in exams by using it!) by simply stating that we're interested in what happens only in the first quadrant; our region isn't in there, but we just crop it to the first quadrant component of it Feasible region: Step two: figure out whether the region is bounded, empty, unbounded ours is unbounded Step three: for unbounded regions we need to see which way the function Z gets minimized draw y-x=1 (same as y=x+1) draw y-x=0 (same as y=x) Graph: Find the minimum, if it exists it's not entirely clear whether the lines go towards the bounded part, or the unbounded part the corner (6,0) sure looks like a minimum, but is it really? Draw a line parallel to the first two, passing through it 2 Minimum or not minimum? As you can see, the boundary has a smaller slope (it's 0!), and so lines parallel to the first two can be drawn even further down, and will still intersect the feasible region ANSWER: no minimum is possible "Extra Credit": maximize Z for the same constraints since Z decreases going down, it increases going up, and so it goes towards the bounded part of our region; check the corners: easiest corner (6,0); Z=0-6=-6 next corner, intersection of x=3 and y=-x/3+2; since x=3, y=-3/3+2=-1+2=1; hence (3,1); Z=1-3=-2 last corner, intersection of x=3 and y=x/3+2; again, x=3, y=3/3+2=1+2=3; hence (3,3); Z=3-3=0 Largest value:0 in (3,3); that is where we obtain the maximum for Z (given the constraints), namely Z=0 Chapter 7.3 Multiple Optimum Solutions There is one situation which occurs for either bounded feasible regions OR for unbounded, as long as the profit/cost function varies accordingly towards the bounded part of it 3 It has to do with the case when, after analyzing all the corners, the maximum (minimum) happens at more than one corner (usually, though, no more than two) it's the case when the profit (cost) function, when graphed while passing through either one of the corners, COINCIDES with the boundary In this case we just say we have infinitely many solutions, all sitting on the line segment between the two corners observation: you can prepare yourself for such a case by noting the slopes for all the lines involved in constructing the feasible region, and comparing them with the slope of profit/cost lines; when they're equal, you'll probably get the above situation With this case, we now have analyzed all possibilities: one solution (a corner), no solution (empty feasible region, unbounded region AND profit/cost going towards unbounded part), many solutions (two corners) Chapter A.3 Average Rate Of Change The issue: a certain function f(x) varies as x, its variable, changes its value; our goal is to quantify the change Example: f(x)=3x+5; how much does f change as x goes from 1 to 10? f(1)=8, f(10)=35, so the change is 35-8=27 The problem here, though, is that this is not enough information about what's going on: if I tell you that the change was 27, but without pointing out from where to where, the information is pretty much useless so, let's try to put some more information in there 4 We could, in fact, append to the information about the change ALSO information about from where to where, but that would be too much hence we go for something in-between: compute relative the change of f with respect to the x, dividing the change in f by the change in x The keyword here is rate: how much does f(x) change when x changes - or rather, how much change we get per unit of x Definition: Let ( x1 , f ( x1 )) and ( x2 , f ( x2 )) be distinct pairs (that is, distinct x). The average rate of change of f from x1 to x2 is f ( x2 ) - f ( x1 ) x2 - x1 Examples: linear functions: f(x)=3x+6, from 0 to 3 f (3) - f (0) 15 - 6 = =3 3-0 3 same function, from 5 to 10 f (10) - f (5) 36 - 21 = =3 10 - 5 5 Coincidence? Same function, from x1 to x2 f ( x2 ) - f ( x1 ) (3x2 + 6) - (3x1 + 6) = = x2 - x1 x2 - x1 = 3 x2 - 3x1 + 6 - 6 3( x2 - x1 ) = =3 x2 - x1 x2 - x1 Observation: for linear functions, the average rate of change is same, regardless of the two pairs chosen. Note: the reverse is also true; in fact, the rate of change coincides with the slope of the linear function 2 Quadratic functions: f ( x) = x + 1 from 0 to 2 f (2) - f (0) 5 - 1 = =2 2-0 2 same function, from 5 to 10 f (10) - f (5) 101 - 26 = = 15 10 - 5 5 different rate! Since f is not a linear function 5 No need for an actual function - we can work with tables of data: stock portfolio Date (day in July) Value (in \$) 1 14200 3 14600 7 15200 10 14700 14 13900 17 14100 21 13700 24 14300 28 14700 31 14500 Same table, but from 10 to 28 value(28) - value(10) \$14700 - \$14700 \$0 = = = \$0 / day 28 - 10 18 days 18 (all dates assumed to be in July). Find average rate of change from July 3 to July 7 value(7) - value(3) \$15200 - \$14600 \$600 = = = \$150 / day 7-3 4 days 4 small note here: the rate of change can be zero sometimes, even though there was some change at some point; it's just that for certain two pairs, that particular rate was indeed 0 (think of a constant function!) Average driving speed: speed is (distance traveled)/(time elapsed) - as such, it's an average rate of change, if we view the distance as a function of time; assume you are 7 miles from home after driving 15 minutes, and 20 miles after driving 20 minutes; what is your average speed? That is, your average rate of change of distance with respect to time? Compute: dist ( 20) - dist (15) 11 - 7 miles 6 miles = = = 0.8 miles / min 20 - 15 5 min 5 min issue: for word problems sometimes we are required to use certain units - as above, we are asked to find the average speed in mph; in these instances, you can transform the values either before or after computation (before: 15mins=0.25h, 20mins=0.33h, etc; after, use 1min=1/60 h, etc) a more theoretical q...

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