Mathematics 1090
EXAM III-Solution
Spring 2005
1. Find the maximum of the following objective functions under constraints (draw the
feasible region):
a) f (x, y) = 2x + 3y, constraints x ≥ 0 , y ≥ 0 , x + 2y ≤ 4 , 2x + y ≤ 5.
The corners of the (closed and bounded) feasible region are (0, 0), (0, 2), (2.5, 0) and
the last one is found by solving:
(
(
(
x + 2y = 4
x = 4 − 2y
x=2
⇐⇒
⇐⇒
2x + y = 5
8 − 4y + y = 5
y=1
Then the last corner is (2, 1). We evaluate the function at all corners: f (0, 0) =
0, f (0, 2) = 6, f (2.5, 0) = 5, f (2, 1) = 7, then the maximum is 7 at (2, 1).
b) g(x, y) = 2x + 3y, constraints x ≥ 0 , y ≥ 0 , x + 4y ≤ 8 , 3x + 5y ≤ 15
The corners of the (closed and bounded) feasible region are (0, 0), (0, 2), (5, 0) and the
last one is found by solving:
(
(
(
x + 4y = 8
x = 8 − 4y
x = 20/7
⇐⇒
⇐⇒
3x + 5y = 15
24 − 12y + 5y = 15
y = 9/7
Then the last corner is (20/7, 9/7). We evaluate the function at all corners: f (0, 0) =
0, f (0, 2) = 6, f (5, 0) = 10, f (20/7, 9/7) = 67/7 ≈ 9.6, then the maximum is 10 at
(5, 0).
2. A small car factory produces two different models of cars, the sport model and the
family model. If x is the number of sport cars and y is the number of family cars, the
profit is given by
P (x, y) = 5x + 3y
To build a sport car, 30 hours of labour is needed, and 15 hour for the family model.
Knowing that the factory cannot build more than a total of 20 cars a day, and that
there are only 45 workers (each working 10 hours a day) to build all the cars, find the
right combination of sports and family model that the factory has to build and sell to
maximize its profit. Calculate also this maximum profit.
The constraints are: x ≥ 0, y ≥ 0, and the ones given by the limitations written in
the text:
(
Maximum of produced cars a day is 20: x + y ≤ 20
45 workers working 10 hours a day:
30x + 15y ≤ 45 × 10 = 450
the second inequality comes from the fact that you need 30 hours to build one sport
car, so you need 30x hours for x sport cars, and similarly you need 15y hours to build
y family cars.
Then the corners of the (closed and bounded) feasible region are: (0, 0), (0, 20), (15, 0)
and the last one is found by solving:
(
(
(
x = 10
x = 20 − y
x + y = 20
⇐⇒
⇐⇒
y = 10
40 − 2y + y = 30
30x + 15y = 450
Then the last corner is (10, 10). We evaluate the function at all corners: P (0, 0) =
0, P (0, 20) = 60, P (15, 0) = 75, P (10, 10) = 80, then the maximum profit is achieved
when producing 10 sport cars and 10 family cars a day, and this maximum profit is
80.
3. Solve for x in the following equations:
a) 5(1.3)2x + 2 = 4
5(1.3)2x + 2 = 4 ⇐⇒ 5(1.3)2x = 2 ⇐⇒ (1.3)2x = 2/5 ⇐⇒ 2x log 1.3 = log(2/5)
log(2/5)
⇐⇒ x =
2 log 1.3
b) 5 ln(2x − 5) − 3 = 10
5 ln(2x − 5) − 3 = 10 ⇐⇒ 5 ln(2x − 5) = 13 ⇐⇒ ln(2x − 5) = 13/5 ⇐⇒ e ln(2x−5) = e13/5
⇐⇒ 2x − 5 = e13/5
⇐⇒ x = (e13/5 + 5)/2
4. You want to invest $8000 on an account which has an interest rate of 6%, compounded
monthly. Your goal is to have $15000 on the account, how long must you wait.
The future value is given by the formula:
S = 8000(1.005)t
where t is the time in months.
a) After 4 years, that is for t = 4 × 12 = 48, there is
S = 8000(1.005)48 ≈ $10164
b) We know that S = 15000 so that gives us the following equation (plugging S =
15000 in the formula above):
15000 = 8000(1.005)t ⇐⇒ 15/8 = 1.005t ⇐⇒ log(15/8) = t log 1.005
log(15/8)
≈ 126
⇐⇒ t =
log 1.005
You must wait 10 years and a half.