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Chemist116 March 2nd, 2019 01:05 PM

How to find the length of an iron bar when it is given as a function of its size but
 
I'm not very sure how to solve this problem, but it looks to be related with division of polynomials. Can somebody give me some help with this?.

The problem is as follows:
In an auto factory in Hsinchu, a technician is in charge of a robot which its function is to cut steel bars for a coupé's transmission. The size of the steel bar requested for safety purposes after testing is given by the function $B(x)=5x^2+mx+n$ inches. The robot cuts the bar in pieces measuring $(x+1)$ inches in length but quality control check reports that there is an excess of $10$ inches in the bar. The robot is reprogrammed and a new cut of the bar produces a piece of $x$ inches in length, but this time there is an excess of $20$ inches. If the initial length of the steel bar is $560$ inches. How many pieces of $(x+2)$ inches can be obtained as maximum?
The given alternatives in my book are as follows:

$\begin{array}{ll}
1.&\textrm{61 inches}\\
2.&\textrm{49 inches}\\
3.&\textrm{50 inches}\\
4.&\textrm{70 inches}\\
5.&\textrm{39 inches}\\
\end{array}$

As mentioned. I'm lost with this one. I believe it is related with the division of polynomials but I don't know which route should I go with.

If I try to use the fact that the length is given I could try to find the constants. But I'm left with two unknowns:

Let's say:

$B(x)=5x^2+mx+n=560$

$B(x)=5x^2+mx+n-560=0$

$x=\frac{-m\pm \sqrt{5^2-4\left(5\times \left(n-560\right)\right)}}{10}$

$x=\frac{-m\pm \sqrt{-20n+11225}}{10}$

In the passage there's mentioned about the remainder's. Hence I suspect that this is related with that.

The second part seems simple:

It says that each time that the robot divides the bar in cuts of $x$ then the remainder is $20$ inches, hence this would be like dividing the polynomial by $\left(x-0\right)$ hence by using the remainder theorem:

$B(x)=5x^2+mx+n$

$B(0)=n=20$

So, $n=20$

This leaves us the above equation with:

$x=\frac{-m\pm \sqrt{-20\times 20+11225}}{10}= \frac{-m\pm \sqrt{-40+11225}}{10}$

$x=\frac{-m\pm \sqrt{11185}}{10}$

But here's where I'm stuck at, since I don't really know If the short information I know is sufficient to solve this problem or If I'm using the information correctly moreover what can I do to find the maximum pieces that can be obtained with $\left(x+2\right)$. Can somebody help me? :help:

Chemist116 March 2nd, 2019 04:05 PM

I cannot make an edit of the previous question so I want to left as a comment that in the alternatives they should be listed as pieces not lengths. So instead of inches they should have said pieces. Hope this can make it more clear to anyone whom can help me with this one.

Denis March 2nd, 2019 05:37 PM

Quote:

Originally Posted by Chemist116 (Post 606405)

$B(x)=5x^2+mx+n-560=0$

$x=\frac{-m\pm \sqrt{5^2-4\left(5\times \left(n-560\right)\right)}}{10}$

$x=\frac{-m\pm \sqrt{-20n+11225}}{10}$

Your solution is incorrect; should be:

$x=\frac{-m\pm \sqrt{m^2-20n+11200}}{10}$

Chemist116 March 2nd, 2019 07:20 PM

Quote:

Originally Posted by Denis (Post 606416)
Your solution is incorrect; should be:

$x=\frac{-m\pm \sqrt{m^2-20n+11200}}{10}$

Yes I figured that out. However where to do from there?. It doesn't look that having that unknown can solve the question or could it be that there is a missing piece of information that I'm overlooking?. Needless to say that should I used that polynomial or just $B(x)=5x^{2}+mx+n$?. I'm still stuck or not sure about that part.

Denis March 3rd, 2019 04:19 AM

Quote:

Originally Posted by Chemist116 (Post 606418)
Yes I figured that out. However where to do from there?

Seems to me that since you're not able to solve a quadratic,
then "where to go from there" should be back to the basics.
You don't appear to be ready for this problem. Good luck.

skipjack March 3rd, 2019 05:56 AM

The correction of the wording of the problem doesn't go anything like far enough.

The wording should support the scenario described below.

(1). In the formula for B(x), x is the number of pieces to be cut.

(2). It is intended to turn out that m = -95 and n = 1010. Thus B(10) = B(9) = 560. Given the value of B(x) for two values of x, you don't need to solve a quadratic to determine the values of m and n.

(3). When 10 pieces are cut, they are each of length 55 inches, with 10 inches remaining.

(4). When 9 pieces are cut, they are each of length 60 inches, with 20 inches remaining.

(5). The wording should somehow mean that one has to determine the maximum number of pieces of length 11 inches that could be cut from a 560 inch bar (where the B(x) formula need not apply). Hence the answer is 50 (the third choice).

There are other possibilities. For example, if B(x) = 5x² - 45x + 660, B(5) = B(4) = 560 (corresponding to 5 pieces of length 110 inches each with 10 inch excess or 4 pieces of length 135 inches with 20 inch excess) and the maximum number of pieces of length 6 inches would be 93, which isn't amongst the choices given.

Hence the problem needs some additional wording that reduces the number of possibilities.

Chemist116 March 3rd, 2019 08:21 AM

Quote:

Originally Posted by Denis (Post 606443)
Seems to me that since you're not able to solve a quadratic,
then "where to go from there" should be back to the basics.
You don't appear to be ready for this problem. Good luck.

I do know how to solve a quadratic but as mentioned the mistake on my side was due I performed the computation in a rush. Given the unknowns I am not sure if what I did attempted was the way to go and solve this problem. But since that there are some clues which appears not very clear. I'll be reposting an improved version of this question.

Chemist116 March 3rd, 2019 09:06 AM

Quote:

Originally Posted by skipjack (Post 606450)
The correction of the wording of the problem doesn't go anything like far enough.

The wording should support the scenario described below.

(1). In the formula for B(x), x is the number of pieces to be cut.

(2). It is intended to turn out that m = -95 and n = 1010. Thus B(10) = B(9) = 560. Given the value of B(x) for two values of x, you don't need to solve a quadratic to determine the values of m and n.

(3). When 10 pieces are cut, they are each of length 55 inches, with 10 inches remaining.

(4). When 9 pieces are cut, they are each of length 60 inches, with 20 inches remaining.

(5). The wording should somehow mean that one has to determine the maximum number of pieces of length 11 inches that could be cut from a 560 inch bar (where the B(x) formula need not apply). Hence the answer is 50 (the third choice).

There are other possibilities. For example, if B(x) = 5x² - 45x + 660, B(5) = B(4) = 560 (corresponding to 5 pieces of length 110 inches each with 10 inch excess or 4 pieces of length 135 inches with 20 inch excess) and the maximum number of pieces of length 6 inches would be 93, which isn't amongst the choices given.

Hence the problem needs some additional wording that reduces the number of possibilities.

Because there are some ambiguities I am reposting a better translated version of the original question. I'm so sorry if this can caused some confusion, but I do hope that this time it can settle down any doubts. I'm wondering if the edit does it changes the result you found?.

Okay here's the question.
In an auto factory in Hsinchu, a technician is in charge of a robot which its function is to cut steel bars for a coupé's transmission. The size of the steel bar requested for safety purposes after testing is given by the function $B(x)=5x^2+mx+n$ inches. The robot cuts the bar in pieces measuring $(x+1)$ inches in length. The remainder of this division results in $10$ inches as reported by quality control inspection. Then, the robot is reprogrammed and a new cut of the bar produces a piece of $x$ inches in length, but this time the remainder of the division is of $20$ inches. If the initial length of the steel bar is $560$ inches. How many pieces of $(x+2)$ inches can be obtained as maximum?.
Needless to say that the alternatives remain the same.

From going by the remainder theorem. I suspect that for the first part it would end like this:

Division by $\left(x-\left(-1\right)\right)$

$B(-1)=5(-1)^2+m(-1)+n=10$

Therefore:

$5-m+n=10$

$-m+n=5$

Then for $\left(x-0\right)$:

$B(0)=5(0)^2+m(0)+n=20$

So: $n=20$ and $m=n-5=20-5=15$

Both of course measuring inches.

So the function is given as:

$B(x)=5x^2+15x+20$

The initial length of the bar is given at $560$, so I suppose that what I need to find is the "$x$".

$5x^2+15x+20=560$

$5x^2+15x-540=0$

Therefore:

$x=\frac{-15 \pm \sqrt{15^2+20\times 540}}{10}$

Solving this equation would yield:

$x=\frac{-15 \pm \sqrt {11025}}{10}=\frac{-15 \pm 105}{10}$

From this:

$x_{1}=\frac{-15+105}{10}=9$

$x_{2}=\frac{-15-105}{10}=-12$

Since the length of the bar cannot be negative then I choose $9$:

Therefore the cuts will be of $x+2=9+2=11$.

But this presents a problem since this $11$ inches is not divisible by $560$ so there will be remaining different than zero in this division, hence:

$\frac{560}{11}=50+\frac{10}{11}$

In other words a remainder of $10$ inches.

So the maximum pieces which can be obtained from that cut will be $50$ or the third alternative. Which corresponds to what you have obtained. I'm not sure if this method it is the right one, but checks with your answer so I think this was the intended choice. Can you take a look on what I did skipjack? . :)

Denis March 3rd, 2019 11:28 AM

Representative diagram (u,v,w = number of pieces):

---x---|---x---| ............................ |---x---|--20--| = 560 : ux + 20 = 560

---x+1---|---x+1---| .................. |---x+1---|-10-| = 560 : v(x+1) + 10 = 560

----x+2----|----x+2----| ............ |----x+2----|-y-| = 560 : w(x+2) + y = 560

Usually "a picture is worth 1000 words"!

Chemist116 March 3rd, 2019 12:05 PM

Quote:

Originally Posted by Denis (Post 606457)
Representative diagram (u,v,w = number of pieces):

---x---|---x---| ............................ |---x---|--20--| = 560 : ux + 20 = 560

---x+1---|---x+1---| .................. |---x+1---|-10-| = 560 : v(x+1) + 10 = 560

----x+2----|----x+2----| ............ |----x+2----|-y-| = 560 : w(x+2) + y = 560

Usually "a picture is worth 1000 words"!

Just if $u,v$ and $w$ were the same in value I could conclude that $y=0$, but this doesn't seem to be the case. Without having any other clue on that. I can't guess what would be the values for the variables you mention. Other than finding the factors for $560$ which are $7 \times 8 \times 10$. But to produce something with a remainder of $20$ the only choice would be $9\times 6 \times 10$, now from there which would I choose as $x$ and which as $u$ for that I would jump to the next equation and see what it checks, this would be $9$ for $x$ and $v=55$ and finally $w=50$ as well. I don't know if you intended to use these clues to get to the answer. By using this I got to $y=10$ inches the remainder. I'm still not sure if this is what you intended to be the method for obtaining the answer from your equations.

Now the tricky part was to get $9$. Honestly I didn't like the part of doing several trials to get that specific number. I was looking for a more methodically way to find it and hence this question was asked in a section of the book devoted to algebra, the intended methodology I believe it was to use the remainder theorem. Of course this doesn't mean that there are other ways to find a solution, but probably due my limitations maybe this method isn't the one for me.


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