Math 1090-001
Midterm 1
09/23/2014
Instructions: Follow the given instruction for every exercise. You have
60 minutes to complete the exam. Show proof of your work. Notice that
the space left for each question is sufficient, but possibly not necessary,
to answer the question. The number in the square next to the problem
deignates the number of points that problem is worth. The last question
is worth 5 extra credits.
First and Last Name:
UnID:
1. Consider the function
√
f (x) = (4 + x2 ) −2x + 6.
5
(a) Find the domain of f ;
5
(b) Evaluate f at x = 3, and simplify your answer;
5
(c) Evaluate f at x = − 32 , and simplify your answer.
Solution: The function f is defined only when the square root has non-negative
argument, i.e. when −2x + 6 ≥ 0. This happens when −2x ≥ −6, i.e. x ≤ 3
(remember the rules of switching signs..) Then the domain of f is the set (−∞, 3].
Plug 3 instead of x in the function: the square root is then zero, hence f (3) = 0
1
Plug − 32 instead of x:
3
3
f (− ) = (4 + (− )2 )
2
2
15
3
25
9 √
75
−2(− ) + 6 = (4 + ) 9 = 3 = .
2
4
4
4
r
2. Write the equation of the line passing through the point (2, 3) and perpendicular to
the line of equation 2y − x = 6.
Solution: The given line has equation y = x2 + 3, hence a perpendicular line
has slope equal to −2. The equation we are looking for will then have the form
y = −2x + b. Now, plug the coordinates of the point in this last equation:
3 = −2 · 2 + b
(0.1)
3 = −4 + b
(0.2)
b=7
(0.3)
Then the equation needed is y = −2x + 7.
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3. Solve the following linear inequality and draw its solution set. Solve the linear equation.
10
(a)
10
(b)
x
+ 3 < 12;
−4
5
(2x + 5) = 25.
3
Solution: For the first one, multiply everything by −4 and remember to switch
the sign: you get
x + 3 · (−4) > 12 · (−4)
(0.4)
x − 12 > −48x > −36.
(0.5)
For the second equation, multiply everything by
3
5
2x + 5 = 15
2x + 5 = 25 ·
3
5
and get
(0.6)
(0.7)
2x = 10
(0.8)
x = 5.
(0.9)
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20
4. Solve the following linear system of three equation in three variables:


z = 3y
x + 2y − z = 6


2x + 7y − 2z = 8
Solution: From the first equation, we
( get z = 3y. Substitute that in the other
x + 2y − 3y = 6
two equations, and get the system:
Now solve this system:
2x + 7y − 6y = 8
from the first equation you have x − y = 6, hence y = x − 6. Plug this in the
14
second equation: from 2x + y = 8 we get 3x − 6 = 8, hence x =
, substitute
3
−4
back and get y = x − 6 = 3 and then z = 3y = −4.
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15
5. Assume that the supply for a certain textbook is given by the equation 35p−20q = 350,
and the demand is determined by 3p + 6q = 300. Find the equilibrium point for the
market.
(
35p − 20q = 350
Solution: You need to solve the system:
which is equivalent to
3p + 6q = 300
solve a system obtained by multiplying both the members of either of the equations
by
( a constant number. So, divide the first line by 5 and the second by 3. You get
7p − 4q = 70
. Now, eliminate the p by multiplying the second line by −7 and
p + 2q = 100
summing up the lines: the sum is −18q = −630. This gives that q = 35. Then
substitute back the value of q in one equation to get p: say p + 2 · q = 100. You
get p = 100 − 2 · 35 = 30. The equilibrium point is then the point (35, 30).
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6. An industry can produce up to 1000 vehicles, which are either cars or trucks. According to some regulation, they have to produce 200 cars more than trucks. The profit
they make for a sold car is $1, 500, while selling a truck gives a profit of $2, 500. Let x
be the number of cars produced and sold, and y the number of trucks produced and
sold. Then:
4
(a) Write down the profit function P , and the constraints that the quantities must
satisfy. What kind of problem is this, and which is the objective function?
4
(b) use the constraints to get a linear system of inequalities, compute and draw the
feasible region on the plane in the figure;
4
(c) find and label the vertices of the feasible region;
3
(d) what is the proportion of cars and trucks that maximizes the profit?
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Solution: The profit P is given by P (x, y) = 1, 500x + 2, 500y. We have the
following contraints:
x≥0
(0.10)
y≥0
(0.11)
x + y ≤ 1000
(0.12)
x ≥ y + 200.
(0.13)
We are facing a problem of linear programming, and want to maximize the profit
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function.
The feasible region will be the intersection of 4 half-planes: x ≥ 0 and y ≥ 0 are
two. Then we have y ≤ −x + 1000 that tells us to consider the half-plane below
the line y = −x + 1000 (if you test the point (0, 0) this belong to the correct half
plane). Finally, the last constraint tells us that the half-plane to consider is the
one below the line y = x − 200 (also in this case, test (0, 0)). The region is a
triangle, with two vertices on the x-axis, and one above the x-axis.
The vertices on the x-axis are found plugging y = 0 in the equations of the lines
they lie on. We get (200, 0) from 0 = x − 200 and((1000, 0) from 0 = −x + 1000.
y = −x + 1000
The other point is found solving the linear system
. The solution
y = x − 200
of the system is (600, 400).
Now, to determine which vertex is the one that realizes the maximum of the
function, just plug the coordinates of the vertexes in the function P (x, y): after a
quick computation you realize the maximum is achieved in (600, 400). Hence, the
optimal proportion of production is to make 600 cars and 400 trucks.
7. Solve the following linear equation:
7
4
+6=
x−2
x−2
Solution: Multiply both members by x − 2, and remember that this implies that
x = 2 is not an acceptable solution for this, just in case. You get 4 + 6x − 12 = 7
5
which is 6x = 15. This gives 15
6 = 2 . This solution is acceptable.
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Question
Points
1
15
2
15
3
20
4
20
5
15
6
15
7
0
Total:
100
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Score