The Feasible Region
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A linear equation
the graph of cos(x)=1 when x=0the graph of sin(x)=0 when x=0.But that only tells part of the story. The two graphs are out of sync by pi/2 radians (or 90°; also referred to as 1/4 wavelength or 1/4 cycle). One cycle is 2*pi radians (the distance for the graph to get back where it started and repeat itself.The cosine graph is 'ahead' (leads) of the sine graph by 1/4 cycle. Or you can say that the sine graph lags the cosine graph by 1/4 cycle.
plug the x coordinate in the x part of the equation and plug the y coordinate in the y's part of the equation and solve
(1) First draw the line y = -x + 5.To do that, find two points that lie on the line. Well, when x = 0, y = 5, so plot (0,5) on the plane. When x = 1, y = 4, so plot (1,4). Now draw the only straight line that goes through both of those points. Because the inequality allows for points to lie on the line itself (that's the "or equals to" part), you can make the line solid. If it were just "greater than" (and not equals to) you would draw a dotted line.(2) Shade the correct side of the line.This line divides the plane in two. One side is all the points that satisfy the inequality; on the other side of the line none of the points satisfy the inequality. We will shade in the side that satisfies the inequality. To figure out which side it is, pick a point not on the line, like (0,0). Plug it into your inequality:y >= -x + 50 >= 0 + g0 >= 5This is not true, so shade the side of the plane that does not contain the origin.
Before you start with limits, you should know that they are quite similar to finding the instantaneous rate of change. The limit of any given point (a) on the graph of a function would be the value the graph converges to at that point. The limit, in other words, is the slope of the tangent at a certain point on the graph. For example, take the graph of y = x [Which is the same as f(x) = x] Now, when you graph that function you get a perfectly diagonal line. You can just start at the point (0,0) on the graph and then for each point, go up 1, right 1. Do the same for the left part of the graph, going down 1 and left 1. Now that you got the graph, take ANY value of x. Say you take 5. Now what point is your FUNCTION approaching from EACH side. So its clear that your function is approaching a value of 5 on the y-axis when x=5, from each side i.e. the graph approaches 5 on the y-axis from the left and the right when x =5. Remember that for a limit to exist, the graph should always approach a certain point from BOTH directions, left and right. Consider the graph of y=x2. At x =5, y = 25. Now since the graph approaches the point 25, when x = 5 from both left and right sides, the limit as the graph approaches x=5 is 25!! Remember that it does NOT matter if the graph is defined at the point at which you are finding if the limit exists, what only matters is if the graph is approaching the point from both sides. So to say, you can have a hole at (5,25) and still have the limit as 25. Now there's a specific way of writing limits. Have a look at this image: http://upload.wikimedia.org/math/e/8/7/e879d1b2b7a9e19d16438c24fb8a7990.png Okay, I'll describe what the image states. All its saying is that as x approaches point 'p' on the function f(x), the limit is L. So, to say for the example I just did above, you have have '5' instead of 'p', and 'L' would be replaced by '25'. Now, say the limit at x=2, for the function f(x) is 10, but you actually have a hole at the point (2,10). And you have a DEFINED point at (2,12). IF your graph is still approaching the hole at (2,10) from both sides, then your limit will still exist. Moving on, suppose a point is x = 3 on a certain graph. So, in 'calculus terms', when the graph is approaching 3 from the left side it would be written like 3- while approaching from the right would be 3+.