Hello Danneedshelp(a) You're right in saying that the shading for line (2) goes over all the shading for line (1), leaving the feasible region just the same as that obtained by line (1) together with .

This region extends infinitely far to the right and upwards, and so is unbounded. In other words, you can find values of and as large as you like and both inequalities will be satisfied. For example, if you want to make , just take , and both inequalities are satisfied.

It's very easy to decide which side of the line to shade. Choose a point not on the line (if the line doesn't pass through the origin, then choose the origin), and check to see whether the condition is satisfied at this point. If it is, shade that side; if not, shade the other side.

So for both (1) and (2), the values satisfy both inequalities. Therefore the origin is on the side to be shaded in each case.

(b) To maximise , draw the line (in other words, the line ) and then try to find the line parallel to that whose intercept with the axis is as large as possible, subject to at least one point on the line lying within the shaded region. This intercept is at giving a maximum value .

(c) We draw the same line again ( ), but this time, since , we need to make the intercept as far down as possible, to make a maximum. Since the region is unbounded to the right, there is no limit to the lowest intercept possible. Hence can be made as large as you like. (In the example that I gave above, , for instance.)

Look at the diagram I've attached.Q2: Minimize

subject to:

(1)

(2)

(3)

and .

For this question, I am also not sure where my feasible region is (or if one even exists). I have graphed all the lines and drawn arrows in the direction of the inequality signs, but I am not sure where to shade. I am assuming this means there is no feasible region.

Should I be checking to makie sure the inequalities hold or do I just assume they do and and shade above or below the lines accordingly?

Thanks in advanced for your time

disclaimer: My entire second week of class has been cancled due to snow days, so I only have one example of how to solve an LP in my notes and have yet to recieve my text book. Thats my excuse for these elementary questions....

Using the technique I outlined above, choosing the origin as my test point, I found the origin is on the 'wrong' side of line (1), but it's on the 'right' side of lines (2) and (3). Hence the only feasible area is the triangle I've shaded.

I've drawn in the optimum ' line' to make a minimum. This is the line that passes through , giving the minimum value of .

Grandad