Question

Marginal Revenue Functions Question...

Answer 1

Management of a company manufacturing small refrigerators determines that the daily total cost of producing *x* units is given by \[C(x) = 0.0002x^3 - 0.06x^2 + 120x +3000\] dollars, where \[0\le x \le 300\]

A) Find the marginal cost of function \[C’(x)\]. Ive done this part already
B) Find the intervals on which the functions c’x) is increasing and those on which it is decreasing. Interpret your results.

Answer 2

Wait. I think i took the derivative wrong.... brb

Answer 3

@Narad

Answer 4

Those zeros in the original equation seem suspicious....

Answer 5

How do i do the second part of the problem again ?

Answer 6

How do i figure out the intervals where c’(x) is increasing and decreasing???

Answer 7

Check the first term of the derivative of
\[C(x)\] and make a variation table of x, C'(x) and C(x)

Answer 8

What values do i plugin for the x in the table tho?

Answer 9

I don't see anything wrong with it

Answer 10

Like that?

Answer 11

Ok sorry. Let’s solve for the second part of the question

Answer 12

It's helpful to know how to convert in case you do need it though.

Answer 13

But basically all you need to do is set \(C'(x)\) = 0 and solve for \(x\)

Answer 14

What?? That;s it?

Answer 15

That will give you the points at which the slope of \(C(x)\) are zero.

Answer 16

Wait don't do that

Answer 17

Oh ok :/

Answer 18

Let me try another way

Answer 19

Instead multiply both sides by 10000

Answer 20

I don’t wanna use that method, it confuses me :/

Answer 21

U wanna multiply both sides because of that 0.0006???

Answer 22

If you multiply both sides by 10,000, you get:
\(6x^2 + 120000x + 1200000 = 0\)

Answer 23

Then you can divide both sides by \(6\)

Answer 24

Oops middle term is neg

Answer 25

No no no no no i don’t wanna do it like that... it confuses me.... let’s do it the conventional way plz

Answer 26

\(6x^2 + 120000x + 1200000 = 0\)

\(x^2 + 20000x + 200000 = 0\)

Answer 27

It's easier to apply the quadratic formula if you have it in this form

Answer 28

\(x = \dfrac{-20000 \pm \sqrt{20000^2 - 4(200000)}}{2}\)

Answer 29

I get it but that’s why i have a calculator for... plz don’t.....*~* plz let’s use the simpler method :/

Answer 30

Your hurting my math hero

Answer 31

This to me IS the simpler method

Answer 32

But even better to just complete the square

Answer 33

You do it however you want. Long as it is right. But for me, I like to avoid decimals as much as possible.

Answer 34

I’dn’t know what number to multiply the decimals in the original equation and that’ll throw me off the time time i try to apply your method

Answer 35

Actually I keep forgetting that darn middle term is supposed to be negative

Answer 36

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Hero
\(x = \dfrac{20000 \pm \sqrt{20000^2 - 4(200000)}}{2}\)
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 37

You don't have to do it the way that I say it. There's another way to do it

Answer 38

Move the decimal point four places to the right for each term

Answer 39

You can't do that until after you set \(C'(x) = 0\)

Answer 40

All you want is for the \(a\) value in the leading term to be an integer

Answer 41

K let’s try your way...wait let me draw real quick :)

Answer 42

If you prefer to work with decimals go for it. I avoid them at all costs unless absolutely necessary.

Answer 43

The middle term is 120000x

Answer 44

-120000x

Answer 45

Moving the decimal point four places to the right is what I actually do in reality. I only said multiply both sides by 10,000 because I'm helping you to understand the correct mathematical equivalent of shifting the decimal places.

Answer 46

You still forgot the x just like I forget negatives sometimes.

Answer 47

By the way you're supposed to set those expressions equal to \(x\)

Answer 48

My calculator can’t take in that many zeros... and i can’t solve it by calculator anymore..

Answer 49

If i were you, I would have divided both sides by six before applying the quadratic formula

Answer 50

You should have gotten \(x^2 - 20000x + 200000 = 0\)

Answer 51

And then apply the quadratic formula?

Answer 52

Correct.

Answer 53

My calculator still wouldn’t have been able to compute so many zeros...

Answer 54

You can also use exponents

Answer 55

Had we done it the conventional way...i’d Have finish my homework by now lol

Answer 56

\(x^2 + 2(10^4)x + 2(10^5)= 0\)

Answer 57

You're learning new methods so it will take slightly longer.

Answer 58

Once you're used to it, then you do it quicker

Answer 59

Try the exponents on your calculator and see what happens

Answer 60

You may want to adjust your calculator to more fixed decimals. Enable it to round up to six decimal places

Answer 61

Yeah that's wrong

Answer 62

That 4 X 2?

Answer 63

I don't know how you went from \(2(10^4)\) to just \(2^4\)

Answer 64

It’s 3 am here lol

Answer 65

I’m doing the question old fashion way for now...

Answer 66

I just wanted to see if using exponents would break your calculator

Answer 67

I'd be surprised if it did

Answer 68

Ok so.. once i have the x values...that would be my maximum right??? And what about where the function is decreasing?

Answer 69

You get plus AND minus values

Answer 70

So the question is *find the intervals where C’(x) is increasing and decreasing....interpret your results

Answer 71

Yes, let me explain

Answer 72

You have two values \(x_1\) and \(x_2\)

Answer 73

One of the values will be less than the other

Answer 74

Mhm

Answer 75

But both values represent where the slope of \(C'(x)\) is zero

Answer 76

Let's say \((x_1, f(x_1)\) is the lower point and \((x_2, f(x_2)\) is the higher point

Answer 77

Mhm

Answer 78

Then the function is increasing on the interval \((x_1, x_2)\)

Answer 79

And decreasing on \((-\infty, x_1)\) and \((x_2, \infty)\)

Answer 80

I'm just speaking hypothetically. I haven't investigated what is actually happening with the function.

Answer 81

Okie

Answer 82

But what I'm trying to show you is all you have to do is figure out which point is lower and which point is higher

Answer 83

And then you can figure things out from there

Answer 84

I see

Answer 85

Actually, I realize the function is quadratic

Answer 86

So there may only be one point where the function is lower or higher

Answer 87

Yes

Answer 88

Since \(a\) is positive, the function will be concave up meaning to the left of the point the function will be decreasing. To the right of the point, the function will be increasing

Answer 89

Yes

Answer 90

The only thing left for you to do now is figure out the intervals where the function is increasing and decreasing

Answer 91

Wait, what does *A* resemble in your last statement ?

Answer 92

\(ax^2 + bx + c\)

Answer 93

Oh icy now

Answer 94

Thanks a lot hero. I really appreciated your help bro. Thanks :)

Answer 95

You're welcome