Math Work Problems - Problem Set 1

Math 441 Problem Set 1 Summer 2011

Name:

1. A bus traveled from one city to another at an average rate of 50 mph. On the return trip, theaverage rate was 45 mph, and the elapsed time was 15 minutes longer. What is the total distancetraveled?Solution:

We have

D = 50t and D = 45

(t+

1

4

)where t is time in hours and D is the distance traveled in miles.

50t = 45

(t+

1

4

)50t = 45t+

45

4

5t =45

4t = 2.25hours

Substituting this value in for one of our equations for D,

D = 50 (t)

D = 50 (2.25)

D = 112.5

So, the total distance traveled is 112.5 miles.

2. It takes a boy 90 minutes to mow his father’s yard, but his sister can do it in 60 minutes. Howlong would it take them to do it if they worked together using two lawn mowers?Solution:

x

90+

x

60= 1 (yard)

We want both fractions to have the same denominator, so

2x

180+

3x

180= 1

2x+ 3x

180= 1

180

(5x

180

)= 180 (1) MPE

5x = 180

x = 36 minutes

So it will take them 36 minutes to mow their father’s yard if they mowed together.

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3. A rectangular plot of ground having dimensions 120 ft by 160 ft is surrounded by a walk ofuniform width. If the area of the walk is 1425 ft2, what is its width?Solution:

x 160 ft xx x

120 120

x xx 160 ft x

We are given that the area of the walk is 1425 ft2. The area of the walk describes

(Area of walk and plot)− (Area of the plot) = 1425 ft2

Collectively, the walk and plot has a length of x + 160 + x feet and a width of x + 120 + xfeet. So,

(2x+ 160) (2x+ 120)− (120 · 160) = 1425(4x2 + 240x+ 320x+ 19200

)− (19200) = 1425

4x2 + 560x = 1425

4x2 + 560x− 1425 = 0

From here, we can use the Quadratic Formula to solve for x.

−560±√

(560)2 − (4) (4) (−1425)

2 (4)=−560±

√336400

8

=−560± 580

8

x = 2.5 or x = −2.5

We want x > 0, so the width of the walk is = 2.5 feet.

4. A man puts a fence around a rectangular field and then subdivides it into three smaller rectanglesby placing two fences parallel to one of the sides. If the area of the field is 31, 250 yd2, and 1, 000 ydof fencing was used, find the dimensions of the field.Solution:

y

x

x

x

x

We are given that the area is 31, 250 yd2. We know

xy = A

=⇒ xy = 31250

2


We are also given that 1000 yards of fencing was used. So,

4x+ 2y = 1000

2y = 1000− 4x

y = 500− 2x

Substituting this value for y into our equation for area,

xy = 31250

x (500− 2x) = 31250

500x− 2x2 = 31250

−2x2 + 500x− 31250 = 0

−2(x2 − 250x+ 15625

)= 0

⇒ x2 − 250x+ 15625 = 0

We can use the Quadratic Formula to solve for x.

250±√

(250)2 − (4) (1) (15625)

2 (1)=

250± 0

2

x =250

2= 125 yd

We can now find the value for y using substitution.

y = 500− 2x

= 500− 2 (125)

= 250 yd

So the dimensions of the field are 125 yd× 250 yd

5. A man owns land that is bordered by a river. He wants to construct a rectangular pen that usesthe river as one of its sides to keep his animals from roaming around. If he requires that there bea divider (running perpendicular to the river) through the middle of the pen (to separate someof the animals) and he has 300 ft of fencing, what are the dimensions of the pen that enclose themaximum area?Solution:

y

x

x

x

3


The perimeter of the pen is

P = x+ x+ x+ y

= 3x+ y

We are given that the man has 300 ft of fencing, so

3x+ y = 300 ft

⇒ y = 300− 3x

The area of the pen is A = x · y, and substituting the value above into the equation for thearea gives us the following

A = x · y⇒ A = x (300− 3x)

⇒ A = 300x− 3x2

⇒ A = 3(100x− x2

)We will use the Second Derivative Test to find the dimensions that enclose the maximumarea.

dA

dx= 3 (100− 2x)

= 300− 6x

Letting dAdx

= 0, we have

0 = 300− 6x

⇒ 6x = 300

⇒ x = 50

Now, we will find the second derivative to verify that our area is maximized.

dA

dx= 300− 6x

⇒ d2A

dx2= −6

So, we see that A′ = 0 when x = 50. We also see that the second derivative is the horizontalline given by the function f (x) = −6 which does contain the point x = 50. Further,

f (50) = −6 < 0

This means, according to the Second Derivative Test, that the area has a maximum atx = 50. Using substitution, we can now find the value for y.

y = 300− 3x

= 300− 3 (50)

= 150

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Thus, area will be maximum when x = 50 ft and y = 150 ft.

6. Two boys who are 224 meters apart start walking towards each other at the same time. If theywalk at a rate of 1.5 and 2 meters per second, respectively, when will they meet? How far willthey have walked?Solution:

We haveb1 = 1.5 m/sec and b2 = 2 m/sec

We can write their rates in terms of time t in seconds to find out when they will meet if theystart out 224 meters apart.

1.5t+ 2t = 224

3.5t = 224

t = 64 seconds

So, the two boys will meet in 64 seconds. After 64 seconds,

b1 = 1.5t

= 1.5 (64)

= 96 meters

b2 = 2t

= 2 (64)

= 128 meters

Boy 1 will have walked 96 meters, and Boy 2 will have walked 128 meters.

7. After playing 100 games, a major league baseball team has a record of 0.650. If it wins only 50%of its games for the remainder of the season, when will its record be 0.600?Solution:

Note that if the team wins only 50% of its games for the remainder of the season, the team’srecord will change for every 2 games played. The number of games needed to win for therecord to be 0.600 is given by the equation below.

0.600 (100 + 2x) = 65 + x

60 + 1.2x = 65 + x

0.2x = 5

x = 25

Thus the teams record will be 2 (25) = 50 games.

8. Twenty liters of a solution contains 20% of a certain chemical. How much water should be addedso that the resulting solution contains only 15% of the chemical?Solution:

5


The amount of chemical in the initial solution is

(0.20) (20L) = 4L of chemical

To find the amount of water needed to make 4L of the chemical compose only 15% of thesolution,

(0.15) (20 + x) = 4

3 + 0.15x = 4

0.15x = 1

x = 6.67L

So, we need to add 6.67 liters of water to the solution so that the solution contains 15% ofthe chemical.

9. A projectile is fied straight upward with an initial speed of 800 ft/sec. The number of feet sabove the ground is s = −16t2 + 800t.

(a.) When will the projectile be 3200ft above the ground?Solution:

Since s is the number of feet above the ground, we can substitute 3200 ft for s andsolve for t

3200 = −16t2 + 800t

16t2 − 800t+ 3200 = 0

16(t2 − 50t+ 200

)= 0

⇒ t2 − 50t+ 200 = 0

We can use the Quadratic Formula to solve for t.

50±√

(−50)2 − (4) (1) (200)

2 (1)=

50±√

1700

2

x = 45.61553 or x = 4.38447

Both values represent times where the projectile is 3200 ft above the ground, but weare looking for the time the projectile first reaches 3200 ft (when it’s on its way up).This time is t = 4.38 seconds.

(b.) When will it hit the ground?Solution:

Similarly to part (a.), we can substitute 0 ft for s and solve for t since s is the numberof feet above the ground.

−16t2 + 800t = 0

−16t (t− 50) = 0

⇒ −16t = 0 or t− 50 = 0

t = 0 or t = 50

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t = 0 represents the initial position of the projectile before it is shot upwards. Thus,the projectile will hit the ground at t = 50 seconds after it is shot upwards.

(c.) What is its maximum height?Solution:

We can find the maximum height of the projectile by applying the first derivative test.Note that s is continuous on R.We first need to find s′ (t) to determine the critical values (if any) for s.The derivative of s

s′ (t) = −32t+ 800

s′ (t) = 0 gives us

−32t+ 800 = 0

−32t = −800

t = 25 seconds

is 0 when x = 25 seconds.

s′ (10) = −32 (10) + 800

= 480

s′ (30) = −32 (30) + 800

= −160

Since s is continuous on R, differentiable on Z, and s′ (t) changes from positive tonegative at t = 25, s has a relative maximum at t = 25 by the First Derivative Test.

s (25) = −16 (25)2 + 800 (25) = 10000 ft

Therefore, the maximum height is 10, 000 feet.

10. Find the one’s digit of the number 47327.Solution:

Since we only care about the one’s digit, we can use modular arithmetic with 10 being ourmodulus (because we only care about numbers 0 to 9).

47 ≡ 7 (mod 10)

So we now want to find the remainder of 7327 with a modulus of 10.

7327 ≡ x (mod 10)

Euler’s Generatlization tells us that using the Euler-phi function, a number raised to thepower of φ (modulus m) where the Euler-phi function is the number of reduced residues modm.

7φ(10) ≡ 1 (mod 10) φ (10) = 4 (1, 3, 7, 9 rel. prime to 10)

74 ≡ 1 (mod 10)

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So, we will take 327 and divide it by 4

7327 ≡ 74·81+3

≡ 74·81 · 73 exponential rule

≡(74)81 · 73

≡ (1)81 · 73 shown above

≡ 1 · 73

≡ 73 = 343 ≡ 3 (mod 10)

≡ 3

Thus, the one’s digit of the number 47327 will be 3.

11. The cost, in dollars, of producing x yards of a certain fabric is C (x) = 1200+12x−0.1x2+0.0005x3

and the company finds that if it sells x yards, it can charge p (x) = 19−0.00021x dollars per yardfor the fabric. Determine the number of yards that should be sold to maximize profit. What isthe price per yard of fabric that maximizes profit?Solution:

Profit = Revenue− Cost

P (x) = R (x)− C (x)

Revenue = (price) · (x yards sold)

R (x) = p (x) · x= (19− 0.00021x)x Given p (x)

= 19x− 0.00021x2

Then,

P (x) = R (x)− C (x)

=(19x− 0.00021x2

)−(1200 + 12x− 0.1x2 + 0.0005x3

)= −0.0005x3 + 0.09979x2 + 7x− 1200

P ′ (x) to find the critical points

P ′ (x) = −0.0015x2 + 0.19958x+ 7

P ′ (x) = 0 gives us−0.0015x2 + 0.19958x+ 7 = 0

We can use the Quadratic Formula to solve for x.

−0.19958±√

(0.19958)2 − (4) (−0.0015) (7)

2 (−0.0015)=−0.19958±

√0.08183

−0.003

x = −28.8277 or x = 161.88

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We obviously want to sell a positive amount of yards i.e. x > 0, so our critical value isx = 162.Now, we’ll find P ′′ (x) to check if x = 162 results in a maximum profit.

P ′′ (x) = −0.003x+ 0.19958

P ′′ (162) = −0.003 (162) + 0.19958

= −0.28642 < 0

So, since P ′ (162) = 0, P ′′ (x) is linear and contains x = 162, and P ′′ (162) < 0, profit ismaximized at x = 162 yards of fabric according to the Second Derivative Test.

Further, the price per yard of fabric that maximizes profit

p (162) = 10− 0.00021 (162)

= 18.96598

= $18.97 per yard of fabric

12. Find the inverse of the matrix (3 716 13

)Solution:[

3 7... 1 0

16 13... 0 1

]13R1→R1

=⇒

[1 7

3

... 13

0

16 13... 0 1

]−16R1+R2→R2

=⇒

[1 7

3

... 13

0

0 -733

... -163

1

]

- 373R2→R2

=⇒

[1 7

3

... 13

0

0 1... 16

73- 373

]- 73R2+R1→R1

=⇒

[1 0

... -1373

773

0 1... 16

73- 373

]So, [

3 716 13

]-1=

[-1373

773

1673

- 373

]

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Math Work Problems - Problem Set 1

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