# Need Math Homework Help... Again

well a few weeks ago I asked for help with my maths homework, too me they have stepped it up a notch - i have been attending lectures and doing the readings but still kinda struggling as he hasnt given us examples too see the workings(the easiest way I learn new things). The homework is due tommorow and was hoping someone could help again, I’d be incredibly grateful

The cost function of a firm is given by C(Y) = (2/3)Y3 - 8Y2 + 48Y + 20, and the revenue function is given by R(Y) = 40Y - 3Y2.

(a) What is the profit function?
(b) Since marginal revenue is given by MR(Y) = â??R/â??Y, determine an expression for marginal revenue. Draw a graph of MR(Y).
(c) Since marginal cost is given by MC(Y) = â??C/â??Y, determine an expression for marginal cost. Draw a graph of MC(Y). At what value of Y does MC achieve a minimum?

ok for a) I got profit is equal too = -2Y^2+10Y-8

I’m unsure if I’ve taken it several steps too far as profit is equal too revenue minus cost…

assuming a is right I’m still stuck on b, havent even bothered with c

Ok, taking a break from my own work:

A) Yes, if profit is (revenue - cost), you must simply plug in these equations and find the resultant. I got:

P(Y) = -(2/3) Y^3 + 5Y^2-8Y-20

B) I take it marginal revenue is the partial derivative of R(Y)/dy, correct? If so:

dR/dy = 40-6Y

C) For marginal cost: dC/dY = 2Y^2 - 16Y + 48

I would think the 0 of the graph is at \$48, which means the minimum cost is \$48 (or whatever monetary unit)

Again, I could be wrong, but I’m pretty sure that is right.

Your marginal cost is equal to the derivative of the cost function. Your marginal revenue is equal to the derivative of the revenue function.

MC = 2Y^2 - 16Y + 48
MR = 40 - 6Y

Both graphs will be plotted similarily. Use Excel or Open Office Calc.

Good Luck.

so my profit function was right?

Assuming no fancy tricks are required and profit = revenue - costs you should simply be able to subtract the functions and obtain your profit function. When I do the subtraction using the equations given I come up with
P(y) = -(2/3)y^3 + 5y^2 -8y + 20

[quote]JLu wrote:
Assuming no fancy tricks are required and profit = revenue - costs you should simply be able to subtract the functions and obtain your profit function. When I do the subtraction using the equations given I come up with
P(y) = -(2/3)y^3 + 5y^2 -8y + 20[/quote]

Everyone’s favorite Spanish conqueror is correct and you are wrong, check the math again.

[quote]jtg987 wrote:
so my profit function was right?[/quote]

No, it was wrong… didn’t you read what PonceDeleon said? If P=R-C, his P is correct and yours was wrong.

[quote]PonceDeLeon wrote:
Ok, taking a break from my own work:

A) Yes, if profit is (revenue - cost), you must simply plug in these equations and find the resultant. I got:

P(Y) = -(2/3) Y^3 + 5Y^2-8Y-20

B) I take it marginal revenue is the partial derivative of R(Y)/dy, correct? If so:

dR/dy = 40-6Y

C) For marginal cost: dC/dY = 2Y^2 - 16Y + 48

I would think the 0 of the graph is at \$48, which means the minimum cost is \$48 (or whatever monetary unit)

Again, I could be wrong, but I’m pretty sure that is right.

[/quote]

everything looks right except the minimum for dC/dy. To find the minimum for dC/dy you need the second derivative of C, which is d^2C/dy^2=4y-16, which equals zero at y=4. I know nothing about economics, but if marginal cost is the first derivative of cost, then marginal cost is minimized when the second derivative of cost is zero, ie, when y=4.

As for graphing everything, if you don’t know how to graph the functions then just stick them into some sort of program that will graph and copy the graph (since I assume this is economics class I assume that’s all they want you to do anything).

On another note if you can’t subtract two simple polynomials (what you have to do to find the “profit” function in part A), you’re going to be in shit loads of trouble trying to compute derivatives (what you had to do in parts B and C) and in graphing functions… you are correct in saying that “they have stepped it up a notch”.

I suggest you either go to your prof during her office hours and ask her for help understanding whatever basics about derivatives you need for the course, or look for some type of tutor. I also suggest you don’t wait until the night before your homework is due to do it. Habits like that are probably why you’ve fallen behind in the first place. The sort of superficial understanding of derivatives that you need in order to complete an economics course shouldn’t be beyond anyone with a solid grasp of high school algebra.

It’s literally just learning some simple rules and applying them. For example the first derivative of y=ax^n where a and n are constants is y’=anx^(n-1). There are various rules for derivatives like this and one simply learns them and applies them to functions. A grasp of what it all means is useful, but that is simple anyway. In essence you are finding the “slope” of a function.

[quote]stokedporcupine8 wrote:
JLu wrote:
Assuming no fancy tricks are required and profit = revenue - costs you should simply be able to subtract the functions and obtain your profit function. When I do the subtraction using the equations given I come up with
P(y) = -(2/3)y^3 + 5y^2 -8y + 20

Everyone’s favorite Spanish conqueror is correct and you are wrong, check the math again. [/quote]

Yeah it’s -20 not +20 my bad.

[quote]PonceDeLeon wrote:
[/quote]

he didn’t give us the formula for marginal revenue/marginal cost nor explain how to do it. Another problem I have is I always think everything is a lot harder or I have to do more with it than I really need too. Also I haven’t looked at any maths in over 4 years so it’s a bit of a shock to the system still. Only reason I chose this subject is cause I thought it was more too do with how too manage a group of people sort of like the foundations of management subject I did about 3 semesters ago, this is called managerial economics.

NEEEEEEEEEERRRRRRRDS!!!

Sorry couldn’t resist. Carry on.

[quote]jtg987 wrote:
PonceDeLeon wrote:

he didn’t give us the formula for marginal revenue/marginal cost nor explain how to do it. Another problem I have is I always think everything is a lot harder or I have to do more with it than I really need too. Also I haven’t looked at any maths in over 4 years so it’s a bit of a shock to the system still. Only reason I chose this subject is cause I thought it was more too do with how too manage a group of people sort of like the foundations of management subject I did about 3 semesters ago, this is called managerial economics.[/quote]

Since marginal revenue and cost is the derivative of revenue and cost, there isn’t just some formula for it. It’s not like say profit, which is revenue minus cost (profit = revenue - cost). It’s a concept you need to understand and there’s a mathematical procedure for computing the quantity.

As I said, go talk to the prof or get a tutor.

[quote]stokedporcupine8 wrote:

everything looks right except the minimum for dC/dy. To find the minimum for dC/dy you need the second derivative of C, which is d^2C/dy^2=4y-16, which equals zero at y=4. I know nothing about economics, but if marginal cost is the first derivative of cost, then marginal cost is minimized when the second derivative of cost is zero, ie, when y=4. [/quote]

Fuck, I knew something was missing. I forgot about max/min calculations.

[quote]jtg987 wrote:
PonceDeLeon wrote:

he didn’t give us the formula for marginal revenue/marginal cost nor explain how to do it. Another problem I have is I always think everything is a lot harder or I have to do more with it than I really need too. Also I haven’t looked at any maths in over 4 years so it’s a bit of a shock to the system still. Only reason I chose this subject is cause I thought it was more too do with how too manage a group of people sort of like the foundations of management subject I did about 3 semesters ago, this is called managerial economics.[/quote]

You are one of the typical students who attempts the “brute force” method of solving things. Many times you will find that simply looking at a problem in a different context (or just using different definitions to define parts of the problem) will yield you a more elegant solution.

“Find the marginal cost of !#@\$^)(\$”

and thinking…

“Fuck! What was that formula again?!@”

“Ok, what does that formula actually mean? What is it used to determine?”

When you start understanding what a mathematical operation is supposed to do, you will learn the “language” such that you can express something in terms that are being requested.

For instance, if you know what a derivative is (not just HOW to derive, but what it is supposed to mean), it will drastically improve your understanding of mathematics.

Oh, and I strongly believe that you learn something more thoroughly after you’ve seen it from two ends:

For instance, understand how to graph a function, but also try to take a graphing program and increment a value within the function being graphed, so you can understand VISUALLY how the function is changing.

[quote]stokedporcupine8 wrote:

everything looks right except the minimum for dC/dy. To find the minimum for dC/dy you need the second derivative of C, which is d^2C/dy^2=4y-16, which equals zero at y=4. I know nothing about economics, but if marginal cost is the first derivative of cost, then marginal cost is minimized when the second derivative of cost is zero, ie, when y=4. [/quote]

I would take y=4 to mean the salary of the sweatshop worker is \$4 / hr and that is the cheapest labor the company can find

[quote]PonceDeLeon wrote:
stokedporcupine8 wrote:

everything looks right except the minimum for dC/dy. To find the minimum for dC/dy you need the second derivative of C, which is d^2C/dy^2=4y-16, which equals zero at y=4. I know nothing about economics, but if marginal cost is the first derivative of cost, then marginal cost is minimized when the second derivative of cost is zero, ie, when y=4.

I would take y=4 to mean the salary of the sweatshop worker is \$4 / hr and that is the cheapest labor the company can find :)[/quote]

That’s pretty good for a sweatshop worker. Probably better then what I’m pulling in right now too, lol.

In all seriousness though, I’m not sure I understand. For example, the “profit” function doesn’t make any sense as a representation of profit, since the function is decreasing almost everywhere for positive Y. So if the variable Y on the profit function P(Y) ranges over units sold or something like that, total profit goes down as each unit is sold? I’m guessing these examples are just completely fictitious, why would any company produce a product whose cost of production increased exponentially but whose revenue only linearly increased?

Perhaps I’m just misinterpreting the functions.

[quote]stokedporcupine8 wrote:
PonceDeLeon wrote:
stokedporcupine8 wrote:

everything looks right except the minimum for dC/dy. To find the minimum for dC/dy you need the second derivative of C, which is d^2C/dy^2=4y-16, which equals zero at y=4. I know nothing about economics, but if marginal cost is the first derivative of cost, then marginal cost is minimized when the second derivative of cost is zero, ie, when y=4.

I would take y=4 to mean the salary of the sweatshop worker is \$4 / hr and that is the cheapest labor the company can find

That’s pretty good for a sweatshop worker. Probably better then what I’m pulling in right now too, lol.

In all seriousness though, I’m not sure I understand. For example, the “profit” function doesn’t make any sense as a representation of profit, since the function is decreasing almost everywhere for positive Y. So if the variable Y on the profit function P(Y) ranges over units sold or something like that, total profit goes down as each unit is sold? I’m guessing these examples are just completely fictitious, why would any company produce a product whose cost of production increased exponentially but whose revenue only linearly increased?

Perhaps I’m just misinterpreting the functions. [/quote]

Because itâ??s a government run entity, silly.