Question

I need math hep plz :)

Answer 1

Management of a company manufacturing small refrigerators determines that the daily total cost of producing *x* units is given by the function \[C(x) = 0.0002x^3 -0.06x^2 + 120x+3000\] (dollars), where \[0 \le x \le 300\] a) Find the marginal Cost function C’(x)
b) Find the intervals on which the function C’(x) is increasing and those on which it is decreasing. Interpret your results.

Answer 2

Part *B* is the main issue here... check my solution plz

Answer 3

@Nnesha

Answer 4

@Hero

Answer 5

Did you ever try graphing the original function?

Answer 6

How do I figure out the intervals on which the C’(x) is increasing and those on whic it is decreasing.

Answer 7

It would have given you an idea of what is going on with the graph before you started doing all that work.

Answer 8

But i’m Not suppose to graph the equation tho

Answer 9

It’s waaayyyy up there....i think the graph doesn’t touch the x axis...

Answer 10

Graphing it won't stop you from doing the work. I meant to graph C(x) not C'(x)

Answer 11

Ok one second plz :)

Answer 12

Correction: Graphing it won't stop you from having to do the work, but it will change the way you might approach it and help you solve the problem more quickly.

Answer 13

Whoever said don't graph it wants you to spend countless hours wasting time.

Answer 14

Btw i did try to graph the original function as well as its derivative before... the following is the graph of the original

Answer 15

Notice the scale... i did zoom out as far as i could

Answer 16

The next step is to find the appropriate window for the graph

Answer 17

Which is?

Answer 18

Let me show you what it should look like when you choose the appropriate window. One second.

Answer 19

I recommend using Geogebra for all your graphing

Answer 20

Okie dokie

Answer 21

https://www.geogebra.org/classic/vcecfapt

Answer 22

How come my graph looks different than yours tho ?

Answer 23

\(\textbf{Appropriate Window}\)

Answer 24

It matters

Answer 25

K so How do I figure out the intervals on which the C’(x) is increasing and those on whic it is decreasing.

Answer 26

You mean when C(x) is increasing and decreasing.

Answer 27

Let me type out the question again..

Answer 28

Find the intervals on which the function C’(x) is increasing and those on which it is decreasing. Interpret your results

Answer 29

I see. To do that, you'll have to find the derivative of C'(x)

Answer 30

So you mean... C’’(x)??

Answer 31

Correct

Answer 32

Let’s do it...

Answer 33

And then set C''(x) = 0 then solve for x

Answer 34

You love dealing with decimals. I love avoiding them

Answer 35

Opposite attracts u know :)

Answer 36

I would rather have 12x = 1200 and then x = 1200/12 = 100

Answer 37

K so I i need a bit more dealing about what we just did... how do i write this interval and is it increasing or decreasing??

Answer 38

Well, you have the x value where C'(x) = 0

Answer 39

And you know the graph of C'(x) is concave up since \(a\) > 0

Answer 40

Yes

Answer 41

Which means find y when x = 100

Answer 42

How do we find y ?

Answer 43

And then you can describe the graph appropriately. The interval to the left of \((100,C'(100))\) will be decreasing. The interval to the right of \((100,C'(100))\) will be increasing.

Answer 44

Plug the x back into C'(x)

Answer 45

I see.. let’s do that.. :)

Answer 46

A smart thing to do is afterwards graph everything: C(x), C'(x) and C''(x) to make sure what you did algebraically is matching up geometrically.

Answer 47

It IS Okay to graph after you've done all the work.

Answer 48

It's for VERIFICATION

Answer 49

That way, if something isn't matching you can go back and fix any mistakes.

Answer 50

Is this correct?

Answer 51

For the decreasing interval it’s (0,100) and for the increasing interval, it’s (100, 300)

Answer 52

That’s the answer according to my textbook

Answer 53

You shouldn't have checked it until you were done

Answer 54

They gave the intervals to the left and right of the point

Answer 55

That's what you should have done

Answer 56

I thought i was suppose to get 0 when i plugged in 100 back into the first prime..but instead we got 114..so i panicked and wrote out the answer

Answer 57

The SLOPE of C"(x) is zero at the point (100,114). The coordinates of the point will not be zero.

Answer 58

We set C''(x) equal to zero because we wanted to find the x value for when the slope of C''(x) is zero.

Answer 59

Nothing about that implies that the coordinates of the point on C'(x) will be zero.

Answer 60

Sooo what do we do now?

Answer 61

One min. Let me check something

Answer 62

Yeah, I don't know why they have those intervals. Mentally, I got a quadratic function for C'(x). And that the interval of decrease is \((-\infty, 100)\) and the interval of increase is \((100, \infty)\). I don't know why they're getting something other than that

Answer 63

I graphed to confirm this

Answer 64

Do you want me to post the answe maswer?

Answer 65

Ah, I see, the original function is restricted from \(0 \le x \le 300\) so that explains it

Answer 66

So the book is correct. We always have to pay attention to those restrictions

Answer 67

@Ballery1

Answer 68

Yes, I just said the book is correct and we have to pay attention to the restriction given for the original function \(0 \le x \le 300\). Do you understand that the original function is restricted to that interval in this case?

Answer 69

Going forward we have to consider the interval restriction of the original function. Let me graph it for you.

Answer 70

yes... the x values have to be within 0 and 300... i see now

Answer 71

so basically the company made units between 0 and 300?

Answer 72

Restricting the graph to that interval causes this when we graph: https://www.geogebra.org/classic/ktqxmkrh

Answer 73

So it looks something completely different

Answer 74

Yes correct

Answer 75

Then you can graph C'(x) to verify it as well https://www.geogebra.org/classic/x8pzyk2w

Answer 76

All of that you should do after you have done algebraic work.

Answer 77

And then after, then you can check the back of the book.

Answer 78

k so that curve and the point represents when the slope of C’’(x) is zero true?

Answer 79

I did see that. Something seems off about it.

Answer 80

C'(x) is supposed to be the slope of C(x)

Answer 81

did i do something wrong in my calculations?

Answer 82

I don't think it works the way I'm imagining. We did find the correct x value where the behavior of the function changes.

Answer 83

I don't think we did anything wrong besides forget to restrict the interval

Answer 84

ok but couldn’t we just figureout the point of vertex of the original graph by factoring the first derivative tho ??

Answer 85

does the first derivative factor tho ??

Answer 86

Vertex is usually used describing the lowest point on a parabola. I have not seen it used to describe a point on a cubic function.

Answer 87

ok

Answer 88

There is no point on the cubic function with a slope of zero

Answer 89

Which would explain why when you tried to find the points, you got negative values for the discriminant.

Answer 90

It is for these reasons, that when you come across a situation like this when you get something other than what you expect, the logical thing to do is to graph the function to see what is going on with it.

Answer 91

yes but why can’t we get the decreasing point (0,100) tho??

Answer 92

of the 1st derivative that is

Answer 93

That's not a point. That's the interval for which C'(x) is decreasing. If you look at its graph, it is decreasing on that interval.

Answer 94

Maybe I should post it again so it's clear. But do not confuse intervals with points. That would be bad

Answer 95

https://www.geogebra.org/classic/quwu8gkr

Answer 96

I graphed all three functions: C(x), C'(x), and C"(x). You'll have to scroll to adjust the window to see the others. They're not going to graph well in the same window.

Answer 97

did u draw the second prime tho?

Answer 98

I did.

Answer 99

i wanna see if the 1st prime is constant?

Answer 100

I did say these are not going to graph well. No the first prime is not constant. Why are you getting that idea. C'(x) is a parabola.

Answer 101

i thought that’s the range of the unit proucts between 0 and 300

Answer 102

The domain of the units of products is \(0 \le x \le 300\)

Answer 103

Its a domain not a range. I don't know why you're confusing yourself more here

Answer 104

And then on the interval is split to two intervals \(0 \le x \le 100\) is the interval for which C'(x) is decreasing and \(100 \le x \le 300\) is the interval where C'(x) is increasing.

Answer 105

k so the point 100,114 on the marginal curve is what?

Answer 106

It's just the point where the slope of C'(x) is zero

Answer 107

ok.. got it.

Answer 108

I hope so. You said a lot of confusing stuff

Answer 109

btw when i said *range* is between...0 and 300... it wasn’t in math terms...i was talking about the range as in the domain of x between 0 and 300

Answer 110

I recommend making a "Things You Learned" list of do's and don'ts when tackling a problem like this.

Answer 111

That would be essential to do when preparing for a test. If you don't, you might make a critical mistake.

Answer 112

i will lol so what’s next?? we have the slope of zero of a marginal curve at point (100,114)...whats next?

Answer 113

I don't know what's next. What's left for you to do for the problem?

Answer 114

how do i algebraically figure out the invreasing internval of the first prime (0,100)??? we figured out the 100...soo how do i find the x??

Answer 115

increasing interval*

Answer 116

(0,100) is the interval of decrease. (100,300) is the interval of increase.

Answer 117

but i need to show algebraically tho ?

Answer 118

The function is given on the interval \(0 \le x \le 300\)

Answer 119

That's the thing to make the mental note of.

Answer 120

Somewhere along that interval the function is increasing and decreasing. To find the intervals, we had to find C"(x) which we did. Then we set C''(x) equal to zero and solved for x which we got x = 100.

Answer 121

That told us that at the point \((100, C'(100))\) the slope is zero. (the function is neither increasing or decreasing at that point.

Answer 122

i think i see it now... so that point (100,114) is the point where the slope of the derivative is 0. so by that logic... the invertal from (0 to 100) is the interval where the derivative was decreasing. and the interval where the derivative is increasing is poast the (100,114) so 100 to 300 on the RHS is the increasing interval....

Answer 123

The next thing is to find the intervals of increase and decrease. To figure that out algebraically, you have to understand two things for C'(x): That it is a quadratic function with an expression of the form \(ax^2 + bx + c\) and that \(a > 0\). Because \(a > 0\), we know that the graph of the function is concave up. Which means that on the interval \(0, 100), the function will be decreasing and on the interval (100, 300) it will be increasing.

Answer 124

am i right hero??

Answer 125

Another way to do it is find the point \((0, C'(0))\) and point \((300, C'(300))\)

Answer 126

When you find those points I will graph to show you what happens but you don't need to graph them to understand what happens.

Answer 127

Look at this: https://www.geogebra.org/classic/urux4qfg

Answer 128

Yes that's the graph but suppose you can't graph and you're taking a test. Then you do it the way I am explaining

Answer 129

yes master!

Answer 130

You're posting too much stuff

Answer 131

Check this out. We know the bounds of the interval of C'(x) is \(0 \le x \le 100\)
Which means we can find \(C'(0)\) and \(C'(300)\) and look at the y values of those.

Answer 132

do u want me to algebraically figure out the values of C’(0) and C’(300)? master

Answer 133

You will see that the point (0, C'(0) is higher than (100, C'(100)) and then
(300, C'(300)) is also higher than (100, C'(100))

Answer 134

Which means (100, C'(100)) is the lowest point on that interval. The only way that happens is if the function is decreasing from (0,100) and increasing from (100, 300).

Answer 135

If you wrote that on a test, the teacher has no choice but to give you the highest score.

Answer 136

All you do is show you know what you're talking about.

Answer 137

true true... k thanks a lot man. i freakin love you!! lol

Answer 138

That's the point of a test

Answer 139

wait, so in terms of units...what does it mean?

Answer 140

the more products the company makes, the cheaper it costs them?

Answer 141

If C'(x) is decreasing on (0,100) it means the marginal cost went down on that interval.

Answer 142

If C'(x) is increasing on the interval (100,300) it means the marginal cost increased for those number of units.

Answer 143

k so up to the point of (100,114) the marginal cost is 0

Answer 144

Negative

Answer 145

The point (100,114) is the point where the marginal cost stopped increasing or decreasing.

Answer 146

so from zero to 100 units ..the marginal cost is decreasing... ?? from that point onwards..it costs the company more to make products...aka it gets expensive and more expensive fbeyounf the (100,114) point mark

Answer 147

k i think i got it bro... thanks :) ily <3 :)

Answer 148

You should describe it the way I suggested. Which is more mathematically. So use words like "on the interval", "marginal cost" "increasing and decreasing" providing those intervals of increase and decrease using the function notation when describing points etc. And add to your list of do's and don'ts to remember to make sure you recognize if (a, b) is referring to a point or an interval.

Answer 149

Your do's and don'ts list should be pretty lengthy for this

Answer 150

for shore

Answer 151

most importantly i need to remember that not all functions are going to have roots and algebraic solutions..