Math Is Fun Forum

nycguitarguy

Member

Registered: 2024-02-24

Posts: 542

Cost of Transporting Goods

A trucking company transports goods between Chicago and New York, a distance of 960 miles.The company’s policy is to charge,for each pound, $0.50 per mile for the first 100 miles, $0.40 per mile for the next 300 miles, $0.25 per mile for the next 400 miles, and no charge for the remaining 160 miles. (a) Graph the relationship between the cost of transportation in dollars and mileage over the entire 960-mile route. (b) Find the cost as a function of mileage for hauls between 100 and 400 miles from Chicago. (c) Find the cost as a function of mileage for hauls between 400 and 800 miles from Chicago.

I need help setting up the correct function. I will do the rest.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,196

Re: Cost of Transporting Goods

I like to try and summarise the written information using (in this case) algebra.

0 < x ≤ 100                      cost per mile 0.5
100 < x ≤ 300                   cpm 0.4
300 < x ≤ 400                    cpm 0.25
400 < ≤ 960                       cpm no charge

But, beware. This doesn't show fully what a charge will be, nor does it give you the points for a graph.  Foe example, if the distance for a package is, say, 150 miles then the charge would be 100 x 0.5 + 50 x 0.4

So, to get the function you need to include charges for a previous stage and account for how many more miles have been travelled. I'll show what I mean for distances in the 100 - 300 group.

100 < x ≤ 300                   total charge = 100 x 0.5 + (x-100) x 0.4

I'll leave you to deal with the other groups similarly. That should enable you to make the graph.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

nycguitarguy

Member

Registered: 2024-02-24

Posts: 542

Re: Cost of Transporting Goods

I like to try and summarise the written information using (in this case) algebra.

0 < x ≤ 100                      cost per mile 0.5
100 < x ≤ 300                   cpm 0.4
300 < x ≤ 400                    cpm 0.25
400 < ≤ 960                       cpm no charge

But, beware. This doesn't show fully what a charge will be, nor does it give you the points for a graph.  Foe example, if the distance for a package is, say, 150 miles then the charge would be 100 x 0.5 + 50 x 0.4

So, to get the function you need to include charges for a previous stage and account for how many more miles have been travelled. I'll show what I mean for distances in the 100 - 300 group.

100 < x ≤ 300                   total charge = 100 x 0.5 + (x-100) x 0.4

I'll leave you to deal with the other groups similarly. That should enable you to make the graph.

Bob

Wow! Very informative. I will see what can be done from this point on.

nycguitarguy

Member

Registered: 2024-02-24

Posts: 542

Re: Cost of Transporting Goods

I like to try and summarise the written information using (in this case) algebra.

0 < x ≤ 100                      cost per mile 0.5
100 < x ≤ 300                   cpm 0.4
300 < x ≤ 400                    cpm 0.25
400 < ≤ 960                       cpm no charge

But, beware. This doesn't show fully what a charge will be, nor does it give you the points for a graph.  Foe example, if the distance for a package is, say, 150 miles then the charge would be 100 x 0.5 + 50 x 0.4

So, to get the function you need to include charges for a previous stage and account for how many more miles have been travelled. I'll show what I mean for distances in the 100 - 300 group.

100 < x ≤ 300                   total charge = 100 x 0.5 + (x-100) x 0.4

I'll leave you to deal with the other groups similarly. That should enable you to make the graph.

Bob

Bob,

I am having big time trouble with part A. I like the way you show your graphs using different colors. Can you graph part A for me?

Here is my effort for the algebra part of the problem.

(a) I came up with a crazy piecewise function, which is probably wrong.

C(x) = {0.50x, if 0 ≤ x ≤ 100....Part 1

C(x) = {50 + 0.40(x - 100), if 100 < x ≤ 400...Part 2
 
C(x) = {170 + 0.25(x - 400), if 400 < x ≤ 800...Part 3
 
C(x) = {270, if 800 < x ≤ 960...Part 4

(b) For hauls between 100 and 400 miles from Chicago, the cost as a function of mileage is:

C(x) = 50 + 0.40(x - 100), where 100 < x ≤ 400

(c) For hauls between 400 and 800 miles from Chicago, the cost as a function of mileage is:

C(x) = 170 + 0.25(x - 400), where 400 < x ≤ 800

I don't feel confident about my work here. I guesses all my side conditions.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,196

Re: Cost of Transporting Goods

I've added an extra column to my table:

0 < x ≤ 100                      cost per mile 0.5  max cost = 100 x 0.5 = 50
100 < x ≤ 300                   cpm 0.4               max cost = 200 x 0.4 = 80
300 < x ≤ 400                    cpm 0.25              max cost = 100 x 0.25 = 25
400 < ≤ 960                       cpm no charge

C(x) = {0.50x, if 0 ≤ x ≤ 100....Part 1

This looks ok.

C(x) = {50 + 0.40(x - 100), if 100 < x ≤ 400...Part 2

Not  x ≤ 400. The upper limit for part 2 is 300.

C(x) = {170 + 0.25(x - 400), if 400 < x ≤ 800...Part 3

Where did 170 come from?

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

nycguitarguy

Member

Registered: 2024-02-24

Posts: 542

Re: Cost of Transporting Goods

I've added an extra column to my table:

0 < x ≤ 100                      cost per mile 0.5  max cost = 100 x 0.5 = 50
100 < x ≤ 300                   cpm 0.4               max cost = 200 x 0.4 = 80
300 < x ≤ 400                    cpm 0.25              max cost = 100 x 0.25 = 25
400 < ≤ 960                       cpm no charge

C(x) = {0.50x, if 0 ≤ x ≤ 100....Part 1

This looks ok.

C(x) = {50 + 0.40(x - 100), if 100 < x ≤ 400...Part 2

Not  x ≤ 400. The upper limit for part 2 is 300.

C(x) = {170 + 0.25(x - 400), if 400 < x ≤ 800...Part 3

Where did 170 come from?

Bob

The number 170 is a typo. I got lost somewhere along the way. I now know that it takes lots of practice to create a piecewise function from given information in a word problem. I could try again later.

nycguitarguy

Member

Registered: 2024-02-24

Posts: 542

Re: Cost of Transporting Goods

I've added an extra column to my table:

0 < x ≤ 100                      cost per mile 0.5  max cost = 100 x 0.5 = 50
100 < x ≤ 300                   cpm 0.4               max cost = 200 x 0.4 = 80
300 < x ≤ 400                    cpm 0.25              max cost = 100 x 0.25 = 25
400 < ≤ 960                       cpm no charge

C(x) = {0.50x, if 0 ≤ x ≤ 100....Part 1

This looks ok.

C(x) = {50 + 0.40(x - 100), if 100 < x ≤ 400...Part 2

Not  x ≤ 400. The upper limit for part 2 is 300.

C(x) = {170 + 0.25(x - 400), if 400 < x ≤ 800...Part 3

Where did 170 come from?

Bob

I think I got it now. It took me 2 hours to work this out.

Part A

To graph the relationship between the cost of transportation (in dollars) and mileage  over the entire 960-mile trip, I need to calculate the total cost for each segment of the trip based on the given pricing policy.

My breakdown of the total distance of 960 miles placed in different segments.
Let me know if this correct.

First 100 miles:

$0.50 x 100 = $50

Next 300 miles:

$0.40 x 300 = $120

Next 400 miles:

$0.25 x 400 = $100

Remaining 160 miles = no charge.

Is this right?

The total cost C for the entire trip is:

C = $50 + $120 + $100 + $0 = $270

Bob, is the graph here a step function with the cost increasing at each segment change?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,196

Re: Cost of Transporting Goods

Is this right? No. You're charging too much.

For part 2, 100 miles has already been charged so 0.4 only applies for 200 miles (300 minus 100)

Same for the remaining parts. Subtract the miles that have already been charged to determine how many more miles to charge at the next rate.

A step function graph looks like a staircase, flat parts getting higher.

This graph has sloping lines with gradients 0.5, 0.4, 0.25, 0 so the sections go up in sloping lines except for the last which is flat.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

nycguitarguy

Member

Registered: 2024-02-24

Posts: 542

Re: Cost of Transporting Goods

Is this right? No. You're charging too much.

For part 2, 100 miles has already been charged so 0.4 only applies for 200 miles (300 minus 100)

Same for the remaining parts. Subtract the miles that have already been charged to determine how many more miles to charge at the next rate.

A step function graph looks like a staircase, flat parts getting higher.

This graph has sloping lines with gradients 0.5, 0.4, 0.25, 0 so the sections go up in sloping lines except for the last which is flat.

Bob

This problem is giving me too much trouble.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,196

Re: Cost of Transporting Goods

I'm like the Canadian Mounties; I never give up. I'll post more help if you want.  Just say.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

nycguitarguy

Member

Registered: 2024-02-24

Posts: 542

Re: Cost of Transporting Goods

I'm like the Canadian Mounties; I never give up. I'll post more help if you want.  Just say.

Bob

Ok. Lead the way. Can you at least provide the right function?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,196

Re: Cost of Transporting Goods

0 < x ≤ 100                      cost per mile 0.5  max cost = 100 x 0.5 = 50
100 < x ≤ 300                   cpm 0.4               max cost = 200 x 0.4 = 80
300 < x ≤ 400                    cpm 0.25              max cost = 100 x 0.25 = 25
400 < ≤ 960                       cpm no charge

So:

0<x≤100                         C = 0.5x
100<x≤300                      C = 50 + 0.4(x-100)
300<x≤400                      C = (50+80) + 0.25(x-300)
400x≤960                         C = (50 + 80 + 25)

I've put the 'already paid' amounts in brackets so you can see where the numbers comes from.  You can complete the additions in your answer.

Plot these points: (0,0) (100,50) (300, 130) (400, 155) (960, 155) and join them with straight lines.

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

nycguitarguy

Member

Registered: 2024-02-24

Posts: 542

Re: Cost of Transporting Goods

0 < x ≤ 100                      cost per mile 0.5  max cost = 100 x 0.5 = 50
100 < x ≤ 300                   cpm 0.4               max cost = 200 x 0.4 = 80
300 < x ≤ 400                    cpm 0.25              max cost = 100 x 0.25 = 25
400 < ≤ 960                       cpm no charge

So:

0<x≤100                         C = 0.5x
100<x≤300                      C = 50 + 0.4(x-100)
300<x≤400                      C = (50+80) + 0.25(x-300)
400x≤960                         C = (50 + 80 + 25)

I've put the 'already paid' amounts in brackets so you can see where the numbers comes from.  You can complete the additions in your answer.

Plot these points: (0,0) (100,50) (300, 130) (400, 155) (960, 155) and join them with straight lines.

I will keep working on this one and return here if need be to continue our discussion.