Unformatted text preview: 12 EXERCISES 1.4 . We graph the solution set for each inequality; the common region is the feasible set. For the
inequality 23; + 33; 2 2, we ﬁrst draw the line 29: + 33; = 2. Its a: and yuinteroepts are 9: = 1
and y = 2/3. Since the test point (0,0) does not satisfy 2s + 33; 2 2, we shade out below
the line (left ﬁgure below). The 93 and yintercepts of the line 23: + y = 8 are a: = 4 and
y = 8. Since the test point (0,0) satisﬁes 2s: + y S 8, we shade out above this line (right
ﬁgure below). Feasible Sell ' x ('3) The inequalitiw a: Z 0 and y 2 0 require us to consider only points in the ﬁrst quadrant,
and on the positive 9: and yaxas. We therefore shade out what remains of the seeond and
fourth quadrants (left ﬁgure above). This shows the feasible _set. All of the vertices are
obvious; they are shown in the right ﬁgure above. . We graph the solution set for each inequality; the common region is the feasible set. Fbr
the inequality 29: + y 2 50, we ﬁrst draw the line 29: + y = 50. Its a— and y—intercepts are
a: = 25 and y = 50. Since the test point (0,0) does not satisfy 29: + y 2 50, We shade out
below the line (left ﬁgure below). The a— and y—intercepts of the line 3 + 23; = 40 are a = 40
and y = 20. Sines the test point (0,0) does not satisfy 3+ 2y 2 40, we shade out below this
line also (right ﬁgure below). . (441.0) x The inequalities a: 2 0 and y 2 0 require us to consider only points in the ﬁrst quadrant, and
on the positive 3 and y~axes. We therefore shade out what remains of the second and fourth
quadrants (left ﬁgure above). This shows the feasible set. Two of the vertices are obvious.
To ﬁnd the third vertex, We solve for the point of intersection of the lines 23: + y = 50 and
a + 2y : 40. If we substitute a: = 40 — 2y into 22: + y = 50, we obtain' This material has been reproduced in accordance whiz copyright law by [he lhiversity ofManitoba Bookstore. Further reproduction. in cart or in full. is 2111'qu mnhihilmi _.f'_"“. ...
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 Fall '11
 Dr.Cornell
 Math

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