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#1 Re: Help Me ! » Minimum surface area » 2016-12-04 18:44:12

Ah, i now remember the author mentioning that about those sort of problems now. Then i will not dwindle on this and continue on with the book.

Thank you bobbym.

#2 Re: Help Me ! » Minimum surface area » 2016-12-04 11:08:33

bobbym wrote:

What did you get for an answer? Did you get 164.91 for the minimum surface area?

Sorry for my language barrier, i used the word "solved" to mean that i got to (6561w^3 + 917000)/2025w. I did not actually get an answer since i don't understand how to proceed further. At first i though that i would just get rid of the denominator and solve for w, but you mentioned something about derivatives, which i know nothing about. I tried to factor the numerator, but 6561 is not an integer multiple of 917000, so trying out thousands of compatible fractions would be a bit of a difficult task.

#3 Re: Help Me ! » Minimum surface area » 2016-12-04 07:31:17

bobbym wrote:


A) S = 2*h*w + 2 l*h + 2 w*l

Using the fact that h = (81 / 50) w we substitute and get

B) S = (131 l w)/25 + (81 w^2)/25

Solving for l in l*w*h = 140 we get

l = 7000/(81 w^2)

Substituting that into B) we get:

S = 36680/(81 w) + (81 w^2)/25

Now all you have to do is set the derivative of that equal to 0 and solve for w. What do you get now? The problem with this approach is that you will have to solve a cubic. Easy for a computer, hard for a human. And, 4.12 is correct for w rounded to 2 decimal places.

1) But i solved it the same way, i just did not use fractions (i might have made one error with the number in front of w^3)? The part i don't understand is (6561w^3 + 917000)/2025w. Am i supposed to factor the numerator into a quadratic, because that's the only thing i sort of know how to do with cubics? You mentioned derivatives, but the book does not even have a word like that in it. Is this something i am supposed to know by this point?

By the way, i am using this book (i'm currently at page 354)

2) Also, how can i improve the way i present my solutions (or rather attempts at them), so that helpful people like you could easier understand what the hell i was trying to do. I always try to show what i tried to do, but it seems to me like i'm doing such a big mess that nobody can understand what i wrote down. The obvious answer for this topic is that i could have used fractions, but other than that, where i am confusing?

#4 Re: Help Me ! » Minimum surface area » 2016-12-04 05:11:20

bobbym wrote:


You need to minimize the surface area 2(hw)+2(lh)+2(wl) subject to the constraints that lwh=140 and h=1.62w. This leads to a definite answer.

Oh and by the way, people see the legendary Sasquatch here often. It goes by the name Skunk Ape. No one really knows what the creature eats. Speculation says, it will grab a gator and snap it into two pieces as effortlessly as we could a twig. Others believe it kills bears for its grub. The darker side people think he will eat a human when it can. Nobody, but nobody, thinks it eats cereal.

Thanks. I tried it likes this now:

lwh = 140
h = 1.62w
1.62lw^2 = 140
l = 140/1.62w^2


So now i'm getting w = -6.54~. If i would take it as a positive i'm still no where near the 4.12~ listed in the book answers. Is my method still false, or have i made some arithmetic mistake?

#5 Help Me ! » Minimum surface area » 2016-12-04 03:39:06

Replies: 11

Hi. My snail speed math progression has carried me to a bunch of word exercises in the precalculus book that i'm trying to go through for most of this year, and i am pitifully terrible at word problems:

The box for the new Sasquatch-themed cereal, ‘Crypt-Os’, is to have a volume of 140 cubic inches. For aesthetic reasons, the height of the box needs to be 1.62 times the width of the base of the box. Find the dimensions of the box which will minimize the surface area of the box. What is the minimum surface area? Round your answers to two decimal places.

So my attempt at solving this is
1) find height function (i don't know another way of finding this without making width and length equal):
        h = 1.62x
        x^2 * 1.62x = 140
        h = 140/1.62x^2
2) find surface area function and plug in the height function:
        2x^2 + 4xh
        2x^2 + 4x(140/1.62x^2)
        2x^2 + 560x/1.62x^2
3) I feel like i should find the slant of this function, then find the vertex of this slant, which will give me x, then i should plug that number in surface area function, and get the minimal surface area. Of course none of this is working out for me, so could anyone lend me a helping hand here?

#6 Re: Help Me ! » Orders of Growth » 2016-10-13 07:01:30

Ah, thanks for the link. I've tried bigger numbers for the function in my example and it seems to be slowly approaching m^5, although of course i could only try out numbers up to 700~, after that my cpu is done for. A supercomputer would be handy right now smile.

#7 Help Me ! » Orders of Growth » 2016-10-13 05:45:39

Replies: 3

I'm reading the "structure and interpretation of computer programs" book (that i probably should not touch) and i'm trying to understand an explanation for an exercise.

Provided we have n types of coins 1, 5,10 ,25 ,50 (so if n = 3, then the type-of-coin(n) = 10), how many ways can we make change of m dollars (in cents)?

for f(m, n):
if m = 0 -> 1
if m < 0 or n = 0 -> 0
else -> f( m, (n - 1) ) + f( (m - type-of-coin(n)), n)
So the problem is "What are the orders of growth of the space and number of steps used by f(m, n) as m increases? ". The answer in says that "It's then straightforward to show that the number of steps required to evaluate f(m 5) is θ(m^5). "

I don't understand what θ(m^5) means here, because if i take 11 cents and put it into the function above, i get f(11, 5) = 4 in 55 steps. Now if i take 12 cents, i get that f(12, 5) = 4 in 63 steps. So how should i interpret θ(m^5) here, 11^5 is 161051, not 55?

#8 Re: Help Me ! » Factoring help » 2016-08-31 18:52:25

I have just figured it out, sorry for the useless post sad. I thought for some reason that you can only divide by the form (x-c) and not by the whole quadratic.

The answer usually dawns on me (if i am able to figure it out) when i make a thread trying to explain what i don't understand.

#9 Help Me ! » Factoring help » 2016-08-31 18:24:26

Replies: 3

The exercise:

Use polynomial long division to perform the indicated division. Write the polynomial in the form p(x) = d(x)q(x) + r(x).

(2x^3-x+1) / (x^2+x+1)

The answer given by the book is:

(2x^3-x+1) = (x^2+x+1)(2x-2)+(-x+3)

What i don't understand is how am i supposed to find q(x) and r(x) when (x^2+x+1) cannot be factored into smaller factors (which i could use for the long division, to find the quotient and the remainder).

#10 Re: Dark Discussions at Cafe Infinity » How to love the M? » 2016-07-04 22:15:12

bobbym wrote:

Hmmm, you know that rule. It is a reliable indicator but it is two faced. They say in pool you must hit a million balls, this only measures your determination to succeed. The wise men of the past understood that a human that truly puts all of himself into some discipline will definitely master it. This has only a few exceptions.

You only need a much smaller amount of knowledge than desire to use it. Perhaps you have not found the part of math that brings you joy. Once you do, you will make fast progress. Now mind you, we can not all be Isaac Newton, but we all can do math.

I definitely have moments where i really enjoy solving math problems and i can sometimes get a glimpse of how beautiful it can be, but most of the time i'm struggling with being frustrated and end up in throwing a tantrum. When i encounter a new subject it just seems like a steep wall that extends towards further than the eye can see. I would just love to get a quick peek at someones mind who really enjoys it and how they proceed in situation where they encounter something very challenging in math. And even if the going is good and i understand the exercises pretty well, i still get fatigued much faster than i would like to. It would be really interesting to experience a zen like state where one could push aside all other matters of the mind and just delve straight into pure mathematics for a day or two.

bobbym wrote:

The exceptions:

1) Bumpkins can not do math.

2) People who suffer massive head trauma like taking a direct hit in the head from a mortar shell will find math difficult.

I'm not a native speaker, what do you mean by "bumpkin"? Is this a metaphor for something else?

#11 Dark Discussions at Cafe Infinity » How to love the M? » 2016-07-01 05:11:20

Replies: 4

Did you always love math and if not, how did you come to love it?

I have a new found interest in math (used to hate it), but i feel like i still have to drag myself to meet my self set minimal quota of an hour per day (though i often get frustrated with some exercise and then spend the next 5 hours trying to understand it). I would love nothing more than to spend whole days sitting there, doing exercises, being fascinated with new concepts and ideas, but i feel like i get fatigued pretty fast. Looking at the 10000 hour rule, and how it's author came to that number, it seems like those who are the masters of their craft like nothing more than to constantly do that one thing, bordering on levels of autism. So that leaves me contemplating, if one does not reach anything near that level of enthusiasm, is it just an exercise in futility? Is there some mythical math cupid that could shoot an arrow up my math ars*?

#12 Re: Help Me ! » Inequality with two absolute values » 2016-07-01 04:27:25

Thanks, i guess i will refer to your example in this thread when i'm stuck with a similar problem.

#13 Re: Help Me ! » Inequality with two absolute values » 2016-07-01 02:54:36

I think i could do a few of these after looking at your answer for an hour, but i still don't understand what is it that i'm searching here.


Never mind the help part, can anyone point me to an easy to understand, but in depth source that explains absolute values, absolute value equalities,  absolute value inequalities, absolute value functions to an ape like creature like me. I'm already stuck at this subject for a few weeks, even though i can solve most of the problems, i'm just going through the paces without really understanding what is happening.

#14 Re: Coder's Corner » Linux question » 2016-07-01 01:50:13

Initially i got tired of all the proprietary software that i was using in Windows. It was often neither clear, neither safe, neither free, so i kind of stumbled on a few open source programs that were just as good if not better than the proprietary ones. So after some investigating i found out that Linux in general is like that and so i tried it. Since then i have abandoned windows completely, because the whole linux ecosystem is completely usable if not better than windows (unless you work with adobe, cad programs etc) and it's free, open source and safe and you can do whatever you want with it, there are no restrictions. One can build it completely from the ground up, if he so wishes!

#15 Help Me ! » Inequality with two absolute values » 2016-07-01 01:16:51

Replies: 5

Hello, i just can't get out of the absolute value blackhole lol. The book i'm reading gave me this exercise, but no previous examples how to solve an inequality with two absolute values, so this is what i tried:

|x -3| - |2x +1| < 0

1) if x > 3

      (x-3) - (2x +1) < 0
      x > -4

2) if x < 3 and x >= 0

    -(x-3) - (2x + 1) < 0
     x > 2/3

3) if < 0

   -(x-3) + (2x +1) < 0
    x < -4

The answer given in the book is (-∞, -4) U (2/3, ∞), so all the "if" clauses are false (or at least the first one is). Why? And i'm kind of ashamed to admit, but i'm still not sure of what is it that i'm searching with these absolute value inequalities.

#16 Re: Help Me ! » Making a profit function » 2016-06-08 20:58:42

thickhead wrote:

Please note that x is the number of shirts either produced or sold in the corresponding equations.

This is the part that i got wrong, i thought x was different things in those equations (dollars in one, items in another). It seems i am not very good at reading and understanding word exercises. Either way, thank you for your help yet again!

#17 Re: Help Me ! » Making a profit function » 2016-06-08 20:26:32

Thanks for the pointer, even though i don't understand it fully, i now know where i made the mistake. When i use 2x + 26 (instead of my 2[30-2x]+26) in the profit function, it all works out great, but i don't understand why do we use 2x + 26? i thought that the input to this function is the amount of items and the output is cost in dollars. Is that not the case? Or is it the other way around in that the input to the 30 - 2x function is the amount of items and the output selling price? I am not that bright, i know.

#18 Help Me ! » Making a profit function » 2016-06-08 19:14:27

Replies: 6

Hello good people.

Today i was doing a word exercise and i was so happy because i thought that i finally managed to crack one of these on my own, but since that is impossible, when i checked the answers in the book, my answers were false. Would love to get some pointers on this exercise:

The cost, in dollars, to produce x “I’d rather be a Sasquatch” T-Shirts is C(x) = 2x + 26, x ≥ 0
and the price-demand function, in dollars per shirt, is p(x) = 30 − 2x, 0 ≤ x ≤ 15.

* Find the profit function P (x).
* Find the number of items which need to be sold in
* Find the maximum profit.
* Find the price to charge per item in order to maxi
* Find and interpret break-even points.

So to find the profit function, i thought that:

Profit(x) = Revenue(x) - Cost(x)

I thought i would set x to be the price, so that i could make the profit function out of the two functions provided in the exercise:

C(x) = 2x + 26, x here is the amount of items (am i interpreting this right?)
p(x) = 30 − 2x, x here is the price to find the amount of items that would sell (am i interpreting this right?)

I use the above to get:

R(x) = x(30-2x), revenue function where x is the price
C(x) = 2(30-2x)+36, cost function where x is the price

And then my profit function:

Profit(x) = (x(30-2x)) - (2(30-2x)+26),  where x is the price
Profit(x) = -2x^2 + 34x -96

Now with that profit function, to find the maximum profit at some price, i solve to get the vertex of the function:

Vertex formula = (-b/2a, f(-b/2a))
(17/2, Profit(17/2)) = (17/2, 48.5)

So i get that at a price of 8.5$ i maximize my profit at 48.5$. But of course that is wrong because the books says it should be (16$, 72$). Am i completely off the course here?

#19 Re: Help Me ! » Question about function transformations » 2016-05-23 05:44:14

bob bundy wrote:

Already looked it up in google, but thanks!

#20 Re: Help Me ! » Question about function transformations » 2016-05-23 05:24:44

For some reason i was thinking about this whole problem algorithmically. This is how my thinking was going - i have a number as an input to the function, i change that number with +2 (i make an equation to find x), now i plug in the changed input into the function, which gives me a different output which i also change with -3. In the midst of it, it never occurred to me that i had to change the placement of the parabola in the cartesian plane by not changing it's dimensions. Looking at the graph of the side by side comparison helped me to at least start to understand what was it that i was trying to do here, i also had to look up what "translation vector" meant, so i probably skipped a topic somewhere, as this is the first time i heard about it. 

Thank you for the help guys, i'm still fuzzy about this whole function transformation thing, but the rays of light are starting to get through (i hope).

#21 Help Me ! » Question about function transformations » 2016-05-23 02:44:19

Replies: 5

Good day good people.

I have a function: f(x) = x^2
Its point:              (-2, 4)
Second function: g(x) = (x+2)^2 - 3
Its point:              (-4, 1)

So the second function is a transformation of the first function:

g(x) = (x+2)^2 - 3  =   f(x+2) - 3

I understand why -2 (x in point one) becomes -4 (x in point two):

x + 2 = -2
x = -4

Now i don't understand (conceptually) why 4 (y in point one) becomes 1 (y in point two). Since i have changed the input in the transformation function from x to (x+2), why doesn't the output reflect this change and i'm still left with 4, from which i then subtract 3? Shouldn't 4 become 16 (because i have -4 in my input now) from which i then subtract 3? I know the right sequence of actions here, i just don't understand why. I feel like i'm thinking about this in a completely wrong way.

#22 Re: Help Me ! » Absolute number equation help » 2016-05-16 07:14:32

thickhead wrote:

It is not the matter of if clause. you have to have a clear start especially for terms involving modulus or absolute as we call it. |x| to be taken as -x only if x is negative. That must be followed till solution is obtained. Here you started with |x|=-x and got 2 values 4 and -3 .You have to choose only the consistent one. mathematics is very easy if you follow simple logic and do not mix up things.

I understand that, when i am learning a new subject things sometimes get a bit overwhelming, so in this case writing a clear if clause helped me to get my bearings and what it was that i was trying to do. Thank you for the answers!

#23 Re: Introductions » Want to get better at math » 2016-05-16 07:04:41

bob bundy wrote:

Throughout my life I learnt no music nor how to play an instrument.  When I retired I decided to build my own electric guitar.  Still struggling to play but I did discover some interesting maths in those frets.  If you like, I'll explain in easy to follow steps.

That's pretty cool, i went the easy way and bought a midi keyboard initially, then a full piano after i got better at playing. I think you can get pretty proficient in 2-3 years. I began from knowing absolutely nothing about music 4 years ago, to being able to play the instrument and improvise quite comfortably (i didn't have a teacher), although i still have ways to go. Would love to share some ideas on the subject with you!

bobbym wrote:

I did, so can you. There happens to be an easy way to do just about everything in the world, unfortunately the guys who know it do not want to tell anyone.

Thanks, i have thought much about this (learning and how to think about it) and have come to at least one reliable method - every time i quit, i always come back and try again (even if it takes a while). It works somewhat decently as a strategy so far.

bobbym wrote:

A snail? Remember Achilles and the tortoise or the tortoise and the hare? I would suggest you trade that snail in for a tortoise. This is not a race, go at your own pace. I used to be fast and made a lot of mistakes, now I am slow and make a lot of mistakes. There is a moral in there somewhere.

Hm... I understand the importance of really understanding the basics, but if i move too slow i may never reach my math dreams (obtaining a grasp on calculus and going beyond that!). I'm trying to figure out a good balance between going my own pace, some compromises and pushing myself to do more.

bobbym wrote:

Post those problems you are having.

Already did one and got a fast answer!

#24 Re: Help Me ! » Absolute number equation help » 2016-05-15 22:18:03

Oh, thank you, i forgot all about the "if" clauses.

#25 Help Me ! » Absolute number equation help » 2016-05-15 19:52:08

Replies: 4


I'm doing some exercises and i'm stuck at one:

|x| = 12 - x^2

I need to solve this equation. Well, when i try to solve this, i get two quadratic equations:

-1(x+4)(x-3) = 0


(x-4)(x+3) = 0

So the answers i get are:  x = -4, x = 4, x = 3, x = -3, however the book says only x = 3 and x = -3 are viable answers. I understand why this is by plugging in 3 and -3 in the first equations, however i don't understand how i should come to this conclusion computationally?

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