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# suppose that the cost of producing r radios is ...

## Tags: Mathematik

00:18 Uhr, 26.01.2023

Hi,dear friends and tutors . I need your help with this problem. I have attached an inmage of the problem.

I know that my independent variable is the number of radios $r\phantom{\rule{0.12em}{0ex}}$ and that I am looking for the rate of change of the cost $c\phantom{\rule{0.12em}{0ex}}$ (with respect to $r\phantom{\rule{0.12em}{0ex}}\right)$ so
I know the average rate of change formula.

$\frac{f\phantom{\rule{0.12em}{0ex}}\left(b\phantom{\rule{0.12em}{0ex}}\right)-f\phantom{\rule{0.12em}{0ex}}\left(a\phantom{\rule{0.12em}{0ex}}\right)}{b\phantom{\rule{0.12em}{0ex}}-a\phantom{\rule{0.12em}{0ex}}}$

From this I can derive this as I am looking for the rate of change of the cost $c\phantom{\rule{0.12em}{0ex}}$ (with respect to $r\phantom{\rule{0.12em}{0ex}}\right)$.
$\frac{c\phantom{\rule{0.12em}{0ex}}\left(r\phantom{\rule{0.12em}{0ex}}2\right)-c\phantom{\rule{0.12em}{0ex}}\left(r\phantom{\rule{0.12em}{0ex}}1\right)}{\left(r\phantom{\rule{0.12em}{0ex}}2-r\phantom{\rule{0.12em}{0ex}}1\right)}$

but I am stuck here. I don't know wht the values of $r\phantom{\rule{0.12em}{0ex}}2$ and $r\phantom{\rule{0.12em}{0ex}}1$ should be
I thought that as they give intervals in the multiple choice, eg, between $20$ and $25$ radios
I could assign values $r\phantom{\rule{0.12em}{0ex}}2=25r\phantom{\rule{0.12em}{0ex}}1=20$

$\frac{c\phantom{\rule{0.12em}{0ex}}\left(25\right)-c\phantom{\rule{0.12em}{0ex}}\left(20\right)}{25-20}$ radios)

but I $\text{'}m\phantom{\rule{0.12em}{0ex}}$ confused because when I simplify that it gives me 1
and if I do the same thing with the other intervals it also amounts to 1

Für alle, die mir helfen möchten (automatisch von OnlineMathe generiert):
"Ich möchte die Lösung in Zusammenarbeit mit anderen erstellen."

00:22 Uhr, 26.01.2023

the image did not attach. let me try again.

01:58 Uhr, 26.01.2023

Hello,

in general your approach is right. But somehow you misinterpret the formula. For the interval $\left[20,25\right]$ the corresponding term is

$\frac{300+15\cdot 25-0.3\cdot 2{5}^{2}-\left(300+15\cdot 20-0.3\cdot 2{0}^{2}\right)}{25-20}=1.5$

Hint: The average change in the interval $\left[60,65\right]$ is $-22.5$.

greetings
pivot

02:19 Uhr, 26.01.2023

thanks a lot, my good friend. Ich wünsche Dir das Beste!.

02:22 Uhr, 26.01.2023

You're welcome. Thank you. I wish the same for you.

02:30 Uhr, 26.01.2023

I did the same calculation that you did, the same way for interval $\left[60,65\right]$ and got $97.5$ as a result.

02:32 Uhr, 26.01.2023

$\frac{\left(300+15\left(65\right)-0.3{\left(65\right)}^{2}-300+15\left(60\right)-0.3{\left(20\right)}^{2}\right)}{\left(65-60\right)}$

02:32 Uhr, 26.01.2023

$\frac{\left(300+15\left(65\right)-0.3{\left(65\right)}^{2}-300+15\left(60\right)-0.3{\left(20\right)}^{2}\right)}{\left(65-60\right)}$

02:57 Uhr, 26.01.2023

Oh, sorry, I made a mistake and instead of $60$ wrote ${20}^{2}$ in the numerator. that is why I got that. Ignore it.

03:10 Uhr, 26.01.2023

I rectified the mistake and still don't get what you get. I get $-94.5$. can you pls tell me if you can see were I go wrong?.Thanks, pivot.

04:36 Uhr, 26.01.2023

Would you please put the picture into the right position and, if possible, enlarge it!
It takes big effort to read it.
It is very little apart from that, very unpleasant to help you if the data are that
Some don't react, because they are not willing to read exercises presented in such a way.

06:11 Uhr, 26.01.2023

@Rasta

You've forgotten the brackets around the whole expression you subtract. I use square brackets to highlight it.

$\frac{\Delta y\phantom{\rule{0.12em}{0ex}}}{\Delta x\phantom{\rule{0.12em}{0ex}}}=\frac{300+15\cdot 65-0.3\cdot 6{5}^{2}-\left[300+15\cdot 60-0.3\cdot 6{0}^{2}\right]}{65-60}$

The term on the LHS denotes the difference quotient.

Then remember: If you have a negative sign in front of a bracket and you then remove the brackets, the signs are exchanging. That means, that $-\left(a\phantom{\rule{0.12em}{0ex}}-b\phantom{\rule{0.12em}{0ex}}+c\phantom{\rule{0.12em}{0ex}}\right)=-a\phantom{\rule{0.12em}{0ex}}+b\phantom{\rule{0.12em}{0ex}}-c\phantom{\rule{0.12em}{0ex}}$, for instance.

Remark: If you use a calculator the brackets must not be removed.

17:51 Uhr, 26.01.2023

I rectified that part about the signs for interval $\left[60-65\right]$ Thanks.

$300+15\left(65\right)-0.3\cdot {65}^{2}-\left[300+15\cdot 60-0.3\cdot {60}^{2}\right]$

I distributed the negative sign like you advised.

so,
left hand side of the numerator produced $7.5$
right side produced $-120$
divided by $65-60$
$\frac{-112.5}{5}$

interval $\left[60,65\right]=-22.5$

for interval $\left[5,10\right]$ the average rate of change of cost was= $10.5$

and for interval $\left[30,35\right]$ the average rate of change of cost= $-22.5$

for interval $\left[20,25\right]$ the average rate of change of cost= $1.5$

so interval $\left[5,10\right]$ has the greatest average rate of change of cost to produce a radio.

17:54 Uhr, 26.01.2023

Yes, you're right,walbus. I will keep it in mind for next time. Sometimes it is a challenge to get the webpage to accept a certain dimension for an image. But next time it will be better. thanks a lot.

18:34 Uhr, 26.01.2023

The rates are right, except for the interval [30,35]

18:56 Uhr, 26.01.2023

thanks,
$\frac{\left(\left(300+15\cdot 35-0.3\cdot {35}^{2}-\left(300+15\cdot 30-0.3\cdot {30}^{2}\right)\right)\right)}{35-30}$

$-4.5$
Corrected.

19:07 Uhr, 26.01.2023

That's right.

Now we have to interpret "the greatest average rate of change".
In my option it is -22.5. Sure it is negative. But a negative change is a change as well, only in the opposite direction. The cost decrease on average by 22.5 per unit in the interval $\left[30,35\right]$

19:08 Uhr, 26.01.2023

DoppelpoSt.

19:23 Uhr, 26.01.2023

I was going to ask you that. Thanks. so it is a change anyways. The cost changed for the better. the cost of producing a radio went down so it is negative. thanks a lot

21:53 Uhr, 26.01.2023

>>I was going to ask you that.<<
Good idea.

You're welcome.
Finally set a check mark. Thanks.
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