Page 1
Math 141, Summer 2014-copyright Joe Kahlig
1. (a) vertex: (50, 70)
9. (a) not possible
(b) 70
(b) not possible
(c) none
(c) not possible
3x2
2. (a) R(x) = x ∗ p = 150x −
−150
vertex at x =
= 25
2 ∗ −3
plug into the price-demand function to get
the price.
Answer: $75
(b) break even when R(x) = C(x) or P (x) = 0.
−3x2 + 150x = 42x + 780
−3x2 + 150x − 780 = 0
either factor or use quadratic formula
Answer: 10 pairs and 26 pairs
3. (a) points (7, 1100) and (11, 900)
Answer: y − 1100 = −50(x − 7) or
y = −50x + 1450
(b) find y when x=0.
Answer: 1450
4. (a) Profit = Rev - Cost
P = A∗x−(5x+280) where A is the selling
price of the sandwich.
320 = A ∗ 150 − (5 ∗ 150 + 280)
A=9
Answer: $9
(b) solve 9x = 5x + 280
Answer: 70 sandwiches
5. use rref.
Answer: x=2, y=5, and z=0
6. use rref.
DVD Players: 40
"
b
1
"
16
0
(d)
(e)
2
0
5
11
#
10
3
#
10. Combine the matrices on the left side and you
get this:
"
2x-28
21-4w
2y-4z
z-8
#
"
=
5
4
7
6
#
since the matrices are equal, the corresponding
enteries are equal. i.e.
2x − 28 = 5
2y − 4z = 7
21 − 4w = 4
z−8=8
now solve for the variables.
Answer: x = 16.5, y = 31.5, z = 14, and w =
4.25
11. x = the number to type I cakes made.
y = the number to type II cakes made.
z = the number to type III cakes made.
Objective function:
P = 2x + 4y + 7z
Constraints:
2x + 4y + 2z ≤ 280
2x + y + 3z ≤ 230
x ≥ 3(y + z)
x, y, z ≥ 0
12. y + 700 = x + 500
z + 500 = 400 + y
300 + x = z + 600
price: 130
7. (a) x = 11, y = 44, and z = 7
13. use rref to get this matrix.
(b) no solution
"
(c) x = 2 − 5y + w
z = 3 − 7w
y, w =any number
8. points are in the form (x, p)
1
0
0
1
−1
2
−50
200
#
From this we know the parametric solution is
x = z − 50 y = 200 − 2z z =any number.
restrictions on Z: Z = 50, 51, 52, ......, 100
(200, 40) and (275, 60)
Check the back of the page for more problems.