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Homework 1
1
Calculus
Homework 1
Due Date: September 26 (Wednesday)
1. A doughnut shop sells a dozen doughnuts for $4.50. Beyond the fixed cost of $220 per day,
it costs $2.75 for enough materials and labor to produce each dozen doughnuts. During
the month of January, the daily profit varies between $60 and $270. Between what levels
(in dozens) do the daily sales vary?
Solution:
Revenue: R = 4.5x.
Cost: C = 2.75x + 220.
Profit = P = R − C = 1.75x − 220. So
60 ≤ 1.75x − 220 ≤ 270
160 ≤ x ≤ 280
Therefore, the daily donut sales vary between 160 dozens donuts per day and 280 dozens
donuts per day.
2. The revenue for selling x units of a product is x
R = 115.95x
and the cost of producing x unit is
C = 95x + 750.
To obtain a profit, the revenue must be greater than the cost. For what values of x will
this product return a profit?
Solution:
To make a profit, we must have
R>C
115.95x > 95x + 750
750
≈ 35.7995 . . .
x>
20.95
x ≥ 36.
So this product will return a profit if x ≥ 36 units.
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Calculus H415611 Fall 2012
3. A utility company has a fleet of vans. The annual operating cost C (in dollars) of each
van is estimated to be
C = 0.35m + 2500
where m is the number of miles driven. What number of miles will yield an annual operating cost that is less than $13,000?
Solution:
C = 0.35m + 2500 < 13, 000
0.35m < 10, 500
m < 30, 000
So, the number of miles driven must be less than 30,000.
4. The acceptable weights for a 20-once cereal box are given by
|x − 20| < 0.75
where x is measured in ounces. Determine the high and low weights for the cereal box.
Solution:
|x − 20| < 0.75 ⇔ −0.75 < x − 20 < 0.75 ⇔ 19.25 < x < 20.75
The lowest and highest acceptable weights for a 20-ounce cereal box are 19.25 ounces and
20.75 ounces.
5. The American Kennel Club has developed guidelines for judging the features of various
breeds of dogs. To not receive a penalty, the guidelines specify that the weights for male
collies must satisfy the inequality
w − 67.5 7.5 ≤ 1
where w is the wights (in pounds). Determine the interval on the real number line in
which these wights lie?
Solution:
w − 67.5 7.5 ≤ 1 ⇔ |w − 67.5| ≤ 7.5 ⇔ 60 ≤ w ≤ 75.
the guidelines specify that the weights for male collies lies between 50 pounds and 65
pounds.
6. The period of pendulum is
r
L
32
where T is the period (in seconds) and L is the length (in feet) of the pendulum. Find
the period of a pendulum whose length is 4 feet.
Solution:
r
r
r
L
4
1
π
T = 2π
= 2π
= 2π
= √ ≈ 2.22 seconds.
32
32
8
2
T = 2π
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7. Find all real zeros of the polynomial.
(a) x2 − 5x + 6
(b) 3x2 + 5x + 2
(c) x3 − x2 − 4x + 4
(d) 2x3 + x2 + 6x + 3
Solution:
(a)
x2 − 5x + 6 = 0
(x − 2)(x − 3) = 0
x = 2, 3.
(b)
3x2 + 5x + 2 = 0
(3x + 2)(x + 1) = 0
2
x = − , −1.
3
(c)
x3 − x2 − 4x + 4 = 0
x2 (x − 1) − 4(x − 1) = 0
(x − 1)(x2 − 4) = 0
x = 1, ±2.
(d)
2x3 + x2 + 6x + 3 = 0
x2 (2x + 1) + 3(2x + 1) = 0
(2x + 1)(x2 + 3) = 0
1
x=− .
2
Note: (x2 + 3) has no real roots.
8. The profit P (in dollars) from sales is given by
P = −200x2 + 2000x − 3800
where x is the number of units sold per day (in hundreds). Determine the interval for x
such that the profit will be greater than $1000.
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Calculus H415611 Fall 2012
Solution:
−200x2 + 2000x − 3800 > 1000
0 > 200x2 − 2000x + 4800
0 > 200(x − 4)(x − 6).
The roots are 4 and 6. By testing the points inside and outside the interval [4, 6], we
found the expression is defined when 4 < x < 6.
9. Simplify the expression.
(a)
√
2−t
√
− 1+t
2 1+t
(b)
√
x+1
√
x
−
√
x
√
x+1
2(x + 1)
(c)
−x2
2x
+
3/2
(2x + 3)
(2x + 3)1/2
(d)
3
−x
+
2(3 + x2 )3/2 (3 + x2 )1/2
Solution:
(a)
√
2−t
√
− 1+t
2 1+t
√
2−t
= √
− 1+t·
2 1+t
(2 − t) − 2(1 + t)
√
=
2 1+t
−3t
= √
2 1+t
√
2 1+t
√
2 1+t
(b)
√
x+1
√
x
−
√
x
√
x+1
2(x + 1)
x+1
x
1
= √ √
−√ √
x x+1
x x + 1 2(x + 1)
1
= √
2 x(x + 1)3/2
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(c)
−x2
2x
+
3/2
(2x + 3)
(2x + 3)1/2
−x2
2x(2x + 3)
=
+
3/2
(2x + 3)
(2x + 3)3/2
3x(x + 2)
=
(2x + 3)3/2
(d)
−x
3
+
2
3/2
2(3 + x )
(3 + x2 )1/2
−x + 3 · 2 · (3 + x2 )
=
2(3 + x2 )3/2
6x2 − x + 18
=
2(3 + x2 )3/2
10. A retailer has determined that the cost C (in dollars) of ordering and storing x units of
a product is
900, 000
C = 6x +
x
(a) Write the expression for cost as single fraction.
(b) Which order size should the retailer place: 240 units, 387 units, or 480 units? Explaining your reasoning.
Solution:
(a)
C = 6x +
900, 000
6x2 + 900, 000
6(x2 + 150, 000)
=
=
.
x
x
x
(b) When x = 240 units.
6(2402 + 150, 000)
C=
≈ 5190.00.
240
When x = 387 units.
C=
6(3872 + 150, 000)
≈ 4647.58.
387
C=
6(4802 + 150, 000)
≈ 4755.00.
480
When x = 480 units.
The retailer should place an order of 387 units, because this cost is the lowest.
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Calculus H415611 Fall 2012
11. Find the value(s) of y such that the distance between the points is 8.
(a) (0, 0), (3, y)
(b) (5, 1), (5, y)
Solution:
(a)
d=
p
d=
p
(5 − 5)2 + (y − 5)2
p
(y − 1)2
(y − 1)2
y
(3 − 0)2 + (y − 0)2 = 8
p
9 + y2 = 8
y 2 = 55
√
y = ± 55.
(b)
=8
=8
= 64
= −7, 9.
12. The table shows the number of ear infections treated by doctors at HMO clinics of three
different sizes: small, medium, and large.
Number of doctors
Cases per small clinic
Cases per medium clinic
Cases per large clinic
0 1
0 20
0 30
0 35
2 3 4
28 35 40
42 53 60
49 62 70
(a) On the same coordinate plane, show the relationship between doctors and treated ear
infections using three line graphs, where the number of doctors is on the horizontal
axis and the number of ear infections treated is on the vertical axis.
(b) Compare the three relationships.
Solution:
(a) aa
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(b) The larger the clinic, the more patient a doctor can treat.
13. Sketch the graph of the equation.
(a) y = x2 − 3
(b) y =
√
x+1
(c) y =
1
x−3
(d) x = 4 − y 2
Solution:
(a) a
(b) a
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Calculus H415611 Fall 2012
(c) a
Homework 1
(d) a
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Calculus H415611 Fall 2012
14. Find the points of intersection (if any) of the graphs of the equations.
(a) y = −x + 7, y = 1.5x − 8
(b) y = x3 , y = 2x.
(c) y = x3 − 2x2 + x − 1, y = −x2 + 3x − 1
Solution:
(a) Set the two equation equal to each other.
−x + 7 =1.5x − 8
−5x = − 30
x =6
Substitute x = 6 into one of the equations.
y = −6 + 7 = 1.
The point of intersection is (6, 1).
(b) Set the two equation equal to each other.
x3 =2x
x(x2 − 2) =0
√
x =0, ± 2.
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√
√
The corresponding y values are 0, −2 2, and 2 2, so the points of intersection are
√
√
√ √
(0, 0), (− 2, −2 2), ( 2, 2 2).
(c) Set the two equation equal to each other.
x3 − 2x2 + x − 1 = −x2 + 3x − 1
x(x + 1)(x − 2) =0
x =0, −1, 2.
The corresponding y values are -1, -5, and 1, so the points of intersection are
(0, −1), (−1, −5), (2, 1).
15. A 2010 Honda Accord costs $28,695 with a gasoline engine. A 2010 Toyota Camry costs
$29,720 with a hybrid engine. The Accord gets 21 miles per gallon of gasoline and the
Camry gets 34 miles per gallon of gasoline. Assume that the price of gasoline is $2.719.
(a) Show that the cost Cg (in dollors) of driving the Honda Accord x miles is
Cg = 28, 695 +
2.719x
21
and the cost Ch (in dollars) of driving the Toyota Camry x miles is
Ch = 29, 720 +
2.719x
34
(b) Find the break-even point. That is, find the mileage at which the hybrid-powered
Toyota Camry becomes more economical than the gasoline-powered Honda Accord.
Solution:
(a) The cost Cg to drive x miles is the cost of the car itself plus the cost of gasoline per
mile. The cost of gasoline per gallon divided by the number of gallons per mile.
Cg = 28, 695 +
2.719x
.
21
Similarly, the cost Ch to drive x miles is the cost of the car itself plus the cost of
gasoline per mile.
2.719x
Ch = 29, 720 +
34
(b) To find the break-even point, set the cost equations to each other.
2.719x
2.719x
= 29, 720 +
21
34
35.347x = 731, 850
x ≈ 20, 705 miles.
28, 695 +
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Calculus H415611 Fall 2012
16. The demand and supply equation for an MP3 player are given by
p = 190 − 15x
p = 75 + 8x
Demand Equation
Supply Equation
where p is the price (in dollars) and x represents the number of units (in hundreds of
thousands). Find the equilibrium point for this market.
Solution:
190 − 15x = 75 + 8x
x = 5.
Equilibrium point is (5, 115).
17. The graph shows the cost and revenue equations for a product.
(a) For what numbers of units sold is there a loss for the company?
(b) For what number of units sold does the company break even?
(c) For what numbers of units sold is there a profit for the company?
Solution:
(a) If less than 100,000 units are sold, the company losses money.
(b) If 100,000 units are sold, the company breaks even.
(c) If more than 100,000 units are sold, the company make a profit.