What is the Linear Equation for Dollar Value of a Product Over Time?

[nycmathguy]

Homework Statement

Write a linear equation that gives the dollar value V of the product in terms of the year t.

Relevant Equations

y = mx + b

You are given the dollar value of a product in 2016 and the rate at which the value of the product is expected to change during
the next 5 years. Use this information to write a linear equation that gives the dollar value V of the product in terms of the year t. (Let t = 16 represent 2016.)1. 2016 Value = $3000
Rate = $150 decrease per year

Solution:

V(t) = dollar value V of the product in terms of the year t.

$150 decrease per year = -150t

So far, I have V(t) = -150t.

Stuck here...

2. 2016 Value = $200
Rate = $6.50 increase per year

V(t) = dollar value V of the product in terms of the year t.

$6.50 increase per year means 6.50t

So far, I have V(t) = 6.50t.

Stuck here...

 

[FactChecker]

Your idea for the increase and decrease is good and will also work if you use -150(y-2016) and 6.50(y-2016).
Now, what constants should you add to those to get the correct values when y=2016?

 

[symbolipoint]

The author of the book wrote this exercise in a very inefficient way.
Basically, you are given a point (variable t is 16 when V is 3000, or point (t,V) given as (16, 3000)), and the slope, 150 dollars increase in 1 year.

I would suggest directly going to the point-slope equation form, and then algebraically adjust V(t). That second (or first) step should look like this:

(V-3000)/(t-16)=150

The reminder is, you want to solve for V in terms of t.

 

[nycmathguy]

The author of the book wrote this exercise in a very inefficient way.
Basically, you are given a point (variable t is 16 when V is 3000, or point (t,V) given as (16, 3000)), and the slope, 150 dollars increase in 1 year.

I would suggest directly going to the point-slope equation form, and then algebraically adjust V(t). That second (or first) step should look like this:

(V-3000)/(t-16)=150

The reminder is, you want to solve for V in terms of t.

The book's answer for question 1 is

V(t) = −150t + 5400, 16 ≤ t ≤ 21

Where did 5400 come from?

What about the side condition?

No answers are given for even number questions. I have no idea what the equation and side condition is for question 2.

Help...

 

[symbolipoint]

The book's answer for question 1 is

V(t) = −150t + 5400, 16 ≤ t ≤ 21

Where did 5400 come from?

What about the side condition?

No answers are given for even number questions. I have no idea what the equation and side condition is for question 2.

Help...

The side condition I am unsure yet unless (until?) I work the the exercise more fully myself. But about the final equation,...,... !

 

[jbriggs444]

The book's answer for question 1 is

V(t) = −150t + 5400, 16 ≤ t ≤ 21

Where did 5400 come from?

The 5400 is the y intercept -- the value that would exist at t=0 (year 2000) if the valid range for t extended that far.

Regardless of what the 5400 represents, you can still plug in t=16 and see that 5400 - 150 * 16 = 3000. So if the -150 is right, the 5400 must also be right.

 

[nycmathguy]

The 5400 is the y intercept -- the value that would exist at t=0 (year 2000) if the valid range for t extended that far.

Regardless of what the 5400 represents, you can still plug in t=16 and see that 5400 - 150 * 16 = 3000. So if the -150 is right, the 5400 must also be right.

This is a key skill in mathematics. A skill I should focus on more than anything else.

 

[nycmathguy]

Your idea for the increase and decrease is good and will also work if you use -150(y-2016) and 6.50(y-2016).
Now, what constants should you add to those to get the correct values when y=2016?

Sorry but I don't get it. Can you set it up for me?

 

[nycmathguy]

The author of the book wrote this exercise in a very inefficient way.
Basically, you are given a point (variable t is 16 when V is 3000, or point (t,V) given as (16, 3000)), and the slope, 150 dollars increase in 1 year.

I would suggest directly going to the point-slope equation form, and then algebraically adjust V(t). That second (or first) step should look like this:

(V-3000)/(t-16)=150

The reminder is, you want to solve for V in terms of t.

The 5400 is the y intercept -- the value that would exist at t=0 (year 2000) if the valid range for t extended that far.

Regardless of what the 5400 represents, you can still plug in t=16 and see that 5400 - 150 * 16 = 3000. So if the -150 is right, the 5400 must also be right.

Question 1

3000 = -150(16) + C

3000 = -2400 + C

3000 + 2400 = C

5400 = C

V(t) = -150t + 5400

Question 2

200 = 6.50(16) + C

200 = 104 + C

200 - 104 = C

96 = C

V(t) = 6.50t + 96

Correct?

 

[symbolipoint]

V(t) = −150t + 5400, 16 ≤ t ≤ 21

Did you understand that domain region for t yet?

Remember, for year 2016, the author wants you to use t=16.
The amount of time passage was given as 5 years, which will be for t=21. Which is year 2021.
Now you should be comfortable with the reported domain of t between 16 and 21, inclusive.

 

[nycmathguy]

Did you understand that domain region for t yet?

Remember, for year 2016, the author wants you to use t=16.
The amount of time passage was given as 5 years, which will be for t=21. Which is year 2021.
Now you should be comfortable with the reported domain of t between 16 and 21, inclusive.

1. I am still confused by the horrible wording of this application.

2. I was able to find the equations for both problems only because I found similar problems online. I simply used the worked out online problems as a guide to help me solve the two questions in this thread and not so much due to my understanding of the situation at hand.

3. Writing 2 above made me sad. Read it again. I used similar problems (with different numbers) as a guide to help me find the two needed equations. Very pathetically sad. Maybe algebra 1 is calling?

 

[symbolipoint]

1. I am still confused by the horrible wording of this application.

2. I was able to find the equations for both problems only because I found similar problems online. I simply used the worked out online problems as a guide to help me solve the two questions in this thread and not so much due to my understanding of the situation at hand.

3. Writing 2 above made me sad. Read it again. I used similar problems (with different numbers) as a guide to help me find the two needed equations. Very pathetically sad. Maybe algebra 1 is calling?

Point #3: Possibly it is calling.

Points #1 and #2: Most of the time, textbooks write such exercises better, more neatly or more economically and are easier to follow. Recognizing the generalizations for such type of problems become easier to find, then.

 

[nycmathguy]

Point #3: Possibly it is calling.

Points #1 and #2: Most of the time, textbooks write such exercises better, more neatly or more economically and are easier to follow. Recognizing the generalizations for such type of problems become easier to find, then.

I am proud of the fact that I was able to use the online worked out problems as a guide to help me find the needed linear equations but this alone is not sufficient evidence that I know what the situation at hand is all about. A problem like this one should not be over my head.

 

[symbolipoint]

I am proud of the fact that I was able to use the online worked out problems as a guide to help me find the needed linear equations but this alone is not sufficient evidence that I know what the situation at hand is all about. A problem like this one should not be over my head.

A good way to write the exercise could be like this:

Use the given information to write a linear equation that gives the value V of a product in terms of the year; let t=16 for the year 2016.

The value was $3000 in year 2016.
The rate of change was $150 decrease every year.

--------

And you already knew you should use slope-intersect form to finish your function V(t).
You need to now understand that one of the points must be (16, 3000).

Directly going to formula for slope of a line first,
(V-3000)/(t-16)=-150

You want to have V in terms of t, and in slope-intercept form.
Simple Algebra steps:
V-3000=-150(t-16)
V=-150(t-16)+3000
V=-150t+2400+3000
V=-150t+5400---------------------DONE, as far as the linear equation is concerned.

 

[nycmathguy]

A good way to write the exercise could be like this:

Use the given information to write a linear equation that gives the value V of a product in terms of the year; let t=16 for the year 2016.

The value was $3000 in year 2016.
The rate of change was $150 decrease every year.

--------

And you already knew you should use slope-intersect form to finish your function V(t).
You need to now understand that one of the points must be (16, 3000).

Directly going to formula for slope of a line first,
(V-3000)/(t-16)=-150

You want to have V in terms of t, and in slope-intercept form.
Simple Algebra steps:
V-3000=-150(t-16)
V=-150(t-16)+3000
V=-150t+2400+3000
V=-150t+5400---------------------DONE, as far as the linear equation is concerned.

I get it now. I used the problem below as a guide. See attachment.

 

[nycmathguy]

A good way to write the exercise could be like this:

Use the given information to write a linear equation that gives the value V of a product in terms of the year; let t=16 for the year 2016.

The value was $3000 in year 2016.
The rate of change was $150 decrease every year.

--------

And you already knew you should use slope-intersect form to finish your function V(t).
You need to now understand that one of the points must be (16, 3000).

Directly going to formula for slope of a line first,
(V-3000)/(t-16)=-150

You want to have V in terms of t, and in slope-intercept form.
Simple Algebra steps:
V-3000=-150(t-16)
V=-150(t-16)+3000
V=-150t+2400+3000
V=-150t+5400---------------------DONE, as far as the linear equation is concerned.

I also used the following example as a guide.
See attachments.

 

[Mark44]

You are given the dollar value of a product in 2016 and the rate at which the value of the product is expected to change during
the next 5 years. Use this information to write a linear equation that gives the dollar value V of the product in terms of the year t. (Let t = 16 represent 2016.)1. 2016 Value = $3000
Rate = $150 decrease per year

Using the suggested meaning of t, and the fact that the slope is -150 (the value decreases $150 per year), and a point on the line is (16, 3000), the point-slope form of the equation of the line is
$$V - 3000 = -150(t - 16), 16 \le t \le 21$$
Expanding the above yields
$$V = -150t + 2400 + 3000$$
or, writing the equation as a function of t and including the domain: $$V(t) = -150t + 5400, 16 \le t \le 21$$

If you don't remember this form of the equation of a line, it can be used when you know a point on the line $(x_0, y_0)$ and the slope $m$ of the line.
$y - y_0 = m(x - x_0)$

1. I am still confused by the horrible wording of this application.

?
The wording seems clear-cut and straightforward to me. What is it about this problem that seems to confusing to you, and why do you feel that the wording is so horrible?

2. I was able to find the equations for both problems only because I found similar problems online. I simply used the worked out online problems as a guide to help me solve the two questions in this thread and not so much due to my understanding of the situation at hand.

3. Writing 2 above made me sad. Read it again. I used similar problems (with different numbers) as a guide to help me find the two needed equations. Very pathetically sad. Maybe algebra 1 is calling?

I don't know what to think about item 2. Granted it's been 28 years since you took the same class (and received an A- for your efforts), but it seems to me that you should have retained at least some of what you learned back then. The learning curve ought to be a lot less steep the second time around.

Part of the problem might be an inability to see past the details of the various problems to the similarities, as well as not understanding the importance of knowing which tools (i.e., equations/formulas) are applicable.

For any problem about lines, there are a couple of formulas that are very useful:
Point-slope -- $y - y_0 = m(x - x_0)$
Slope-intercept -- $y = mx + b$
Which one to use depends on what information is given in the problem. Note that these equations are equivalent: each one can be derived from the other.

Also, knowing which one to use ties into why we are so insistent on filling in the Relevant Equations section for homework posts.

And you already knew you should use slope-intersect form to finish your function V(t).

Technically, that's the slope-intercept form.

 

[symbolipoint]

it seems to me that you should have retained at least some of what you learned back then. The learning curve ought to be a lot less steep the second time around.

Why do so few of us understand that? That is a common part of study, restudy, review.

 

[nycmathguy]

Why do so few of us understand that? That is a common part of study, restudy, review.

I should retain material that I learned back in 1993, the same year that Boy Meets World made its debut. Really? Noticed that I responded to you and not the other guy.

I am really getting tired of all the belittling nonsense here. I will be searching for another math site or FB math group where self-study students are not criticized for making an effort to review material learned long ago. Unfortunately, this site has wasted my time.

 

[berkeman]

I should retain material that I learned back in 1993, the same year that Boy Meets World made its debut. Really? Noticed that I responded to you and not the other guy.

I am really getting tired of all the belittling nonsense here. I will be searching for another math site or FB math group where self-study students are not criticized for making an effort to review material learned long ago. Unfortunately, this site has wasted my time.

That's a pretty offensive comment, given all of the patient help that you've received here. Yes, some of us have gotten frustrated by some of your posts and habits, but I think the bulk of the helpful posts in your threads have been overly patient and helpful.

Best wishes at the next discussion forum where you post.

 

[nycmathguy]

That's a pretty offensive comment, given all of the patient help that you've received here. Yes, some of us have gotten frustrated by some of your posts and habits, but I think the bulk of the helpful posts in your threads have been overly patient and helpful.

Best wishes at the next discussion forum where you post.

Is that right? I'm the problem, huh? Ok..

 

[MidgetDwarf]

Is that right? I'm the problem, huh? Ok..

You are very argumentative when people take the time out of their day to answer your questions. Notice that this is a volunteer site. Ie., the members here are not being payed to help you. However, they do so due to a deep interest of math/physics and other science branches. They have been more than cordial with you, and not to mention some of your snarky responses. Moreover, you are given advice numerous times, and flat out ignore it. So yes, you are the problem.

 

[berkeman]

(Not any more...)