ECON 1003 Review Paper

ECON 1003 Review Paper Dec 2014
1.
a. For what values of t is t 2 < 8t + 48 t 2 − 8t − 48 < 0
Consider
t 2 − 8t − 48 = 0
(t − 12)(t + 4) = 0 t = −4 or t = 12
When t = −5, (t − 12)(t + 4) = 17 which is greater than zero t < −4 is not is the solution set
When t = 13, (t − 12)(t + 4) = 17 which is greater than zero t < 12 is not is the solution set
The boundart point t = −4 and t = 12 are not set
Hence the solution is − 4 < t < 12

b. If the sum of n consecutive terms in an AP is 56, when the first term is − 160 and the n common difference is 12. Using the formula Sn = [2a + (n − 1)d], find the value of n?
2
n
56 = [−320 + (n − 1)12]
2
112 = −320n + 12n2 − 12n
12n2 − 332n − 112 = 0
3n2 − 83n − 28 = 0
3n2 − 84n + n − 28 = 0
3n(n − 28) + 1(n − 28) = 0 …show more content…
A certain diet consists of dishes A, B, and C. Each serving of A has 1 gram of fat, 2 grams of carbohydrate, and 4 grams of protein. Each serving of B has 2 grams of fat, 1 gram of carbohydrate, and 3 grams of protein.
Each serving of C has 2 grams of fat, 4 grams of carbohydrate, and 3 grams of protein. The diet allows
15 grams of fat, 24 grams of carbohydrate, and 30 grams of protein. Using either the ?????? ′ ? ???? or the ?????-?????? ???????????, find how many servings of each dish can be eaten?

1 …show more content…
By G-J Elimination
In matrix form Ax = b
1 2 2 a
15
(2 1 4) (b) = (24)
4 3 3 c
30
1 2 2 15
(2 1 4|24) r2 − 2r1 → R 2 and r3 − 4r1 → R 3
4 3 3 30
1 2
2 15
1
(0 −3 0 | −6 ) − r2 → R 2
3
0 −5 −5 −30
1 2
2 15
(0 1
0 | 2 ) r3 + 5r2 → R 3 and r1 − 2r2 → R1
0 −5 −5 −30
1 0 2 11
1
(0 1 0 | 2 ) − r3 → R 3
5
0 0 −5 −20
1 0 2 11
(0 1 0| 2 ) r1 − 2r3 → R1
0 0 1 4
1 0 03
(0 1 0|2)
0 0 14 a = 3, b = 2 and c = 4
By Cramer ′ s Rule
In matrix form Ax = b
1 2 2 a
15
(2 1 4) (b) = (24)
4 3 3 c
30
1 2 2
A = (2 1 4)
4 3 3
|A| = 1(3 − 12) − 2(6 − 16) + 2(6 − 4)
= 1(−9) − 2(−10) + 2(2)
= −9 + 20 + 4
= 15
2

15
A1 = (24
30

2 2
1