Page 1
Math 141, Summer 2014-copyright Joe Kahlig
1. (a) vertex: (26, −100)
"
9. (a)
(b) none
2x+4y
2+7y
#
(b) not possible
(c) −100

2. (a) R(x) = x ∗ p = 88x − 2x2
−88
vertex at x =
= 22
2 ∗ −2
plug into the price-demand function to get
the price.
Answer: $44
(b) break even when R(x) = C(x) or P (x) = 0.
−2x2 + 88x = 24x + 350
−2x2 + 64x − 350 = 0
either factor or use quadratic formula
Answer: 7 pairs and 25 pairs
3. (a) points (4, 2300) and (6, 2000)
Answer: y − 2300 = −150(x − 4) or
y = −150x + 2900
(b) find y when x=0.
Answer: 2900
4. (a) Profit = Rev - Cost
P = A∗x−(4x+520) where A is the selling
price of the sandwich.
680 = A ∗ 300 − (4 ∗ 300 + 520)
A=8
Answer: $7
(b) solve 8x = 4x + 520
Answer: 130 sandwiches
5. use rref.
Answer: x=0.5, y=0, and z = −2.5
6. use rref.
DVD Players: 60
price: 150
7. (a) x = 20, y = 22, and z = 10
m

(c)  k
3

1

4 
3
(d) not possible
"
(e)
4
6
12
1
#
10. Combine the matrices on the left side and you
get this:
"
2x - 28
21-4w
3y-4z
z-8
#
"
=
10
21
2
4
#
since the matrices are equal, the corresponding
enteries are equal. i.e.
2x − 28 = 10
3y − 4z = 2
21 − 4w = 21
z−8=4
now solve for the variables.
Answer: x = 19, y = 50/3, z = 12, and w = 0
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 = 5x + 3y + 2z
Constraints:
2x + 4y + 2z ≤ 120
2x + y + 3z ≤ 52
y ≥ 2(x + z)
x, y, z ≥ 0
12. w + 700 = v + 300
m + 800 = 880 + w
450 + v = m + 950
13. use rref to get this matrix.
(b) no solution
(c) x = 5 − 3y + 2w
z = 8 − 4w
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
(300, 25) and (350, 40)
Check the back of the page for more problems.