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Math 414 Hw 3 Name ________________________________Score____/ 28
Complete both sides.
Hw 91: 2, 4, 6, 10, 12.
The problems below are called linear programming
problems. The inequalities are the constraints. The set S
of points which satisfy the constraints is the set of
feasible solutions. The function to be maximized or
minimized is the objective function.
Page 91.
v Sketch the set of feasible solutions and, on the graph,
give the coordinates of all extreme points and label the
lines with their constraint numbers.
v Draw the gradient vector for the objective function. If
the function is ax+by, the gradient vector is [a, b]. It
points in the direction of maximum increase.
v Make a table listing the value of the objective function
on each extreme point.
v Circle the maximum or minimum value rows. On the
graph circle the extreme point or set of points which
have this maximum or minimum. Label all boundary
lines.
The first three problems have been done for you.
Recommended 91: 1, 3, 5, 7.
2’(6). Minimize z = 5x-3y subject to the constraints
a: x+2y > 4,
x > 0,
b: x +3y < 6,
y > 0.
x
y
z
4(6). Maximize z = 2x+3y subject to the constraints
a: 3x+y < 6,
x > 0,
b: x + y < 4,
y > 0,
c: x +2y < 6.
`Minimize z = x + y subject to the constraints:
a: x +2y > 8, x > 0,
b: 2x + y > 8, y > 0,
c: x + y > 5.
Solution.
x
y
z
x
y
z
2.66 2.66 5.33
0
8
8
8
0
8
(0,8)
(b)
(2.66,2.66)
(a)
(c)
(8,0)
(y)
`Maximize z = x + y subject to the constraints:
a: x +2y > 8, x > 0,
b: 2x + y > 8, y > 0,
c: x + y > 5.
Solution. the picture and table are the same as above
but the region is unbounded in the direction of the
gradient and the answer is “no max”.
`Minimize z = x+y subject to the constraints:
a: x +2y > 8, x>0,
b: 2x + y > 8, y>0,
c: x + y < 5.
Solution. there are no points which satisfy all three
constraints. Thus the answer is “no feasible solutions”.
Added problem(4). Suppose f (x) is a linear function on a
convex set S whose boundary is the quadrilateral with
vertices c, b, q, d. Suppose f (a) = 1 is a local maximum;
¯¯, ca
¯ have the same
f (b) = -1 is a local minimum. ad
length; cp
¯ , pa
¯¯, pb
¯¯, have length 1. a, b, d, q are the
vertices of a parallelogram. Study Lecture 4's two theorems;
this is hard.
b
q
p
a
c
Then
f (c) =
f (d) =
d
f (p) =
f (q) =
`Sketch the set of feasible solutions.
Write “empty region” if there are no feasible solutions.
`On the graph, give the coordinates of all extreme
points. For 6, draw in the gradient vector and label each
constraint boundary line with the constraint label.
12’(6). Minimize w = 2x +z subject to:
a: x +y +z = 1
(The ‘=’ is not a typo.)
b: 2x +y +2z < 3
x>0, y>0, z>0
List all extremes and circle the minimum.
Label each constraint boundary plane with the constraint label.
`For 10, 12, make a table listing the values of the
objective function on the extreme points. In the table,
circle or shade the rows with the maximum or
minimum value or values. Label each boundary plane
with the constraint label.
`On the graph circle the extreme point (set of points)
which gives this maximum or minimum value.
x
y
z
w
There should be 3 edges.
Page 91
6(6). Maximize z = x +3y subject to:
a: x + 3y < 6
b: x +y > 4
x>0, y>0
z
y
4
3
2
y
1
x
1
1
2
3
4
5
6
x