interval notation word problem

[uniquaa]

Please help me with the following problem:

Young’s rule for determining the amount of a medicine dosage for a child is given by C=ad/a+12, where "a" is the child’s age and "d" is the usual adult dosage, in milligrams. (Warning ! Do not apply this formula without checking with a physician!) An 8-year old child needs medication. What adult dosage can be used if a child’s dosage must stay between 100 mg and 200 mg?

This must be solved in the form of an inequality.
I appreciate and thank you for your help!

Is this formula correct?
100<=8d+12<=200

 

[Deleted member 4993]

Please help me with the following problem:

Young’s rule for determining the amount of a medicine dosage for a child is given by C=ad/a+12, where "a" is the child’s age and "d" is the usual adult dosage, in milligrams. (Warning ! Do not apply this formula without checking with a physician!) An 8-year old child needs medication. What adult dosage can be used if a child’s dosage must stay between 100 mg and 200 mg?

This must be solved in the form of an inequality.
I appreciate and thank you for your help!

Is this formula correct?
100<=8d+12<=200

You wrote:

C=ad/a+12

Is it:

\(\displaystyle a\cdot\frac{d}{a} \, + \, 12\)

or

\(\displaystyle a\cdot\frac{d}{a+12}\)

 

[uniquaa]

I apologize for the confusion, it's:



C=ad/(a+12)

 

[Deleted member 4993]

Please help me with the following problem:

Young’s rule for determining the amount of a medicine dosage for a child is given by C=ad/(a+12), where "a" is the child’s age and "d" is the usual adult dosage, in milligrams. (Warning ! Do not apply this formula without checking with a physician!) An 8-year old child needs medication. What adult dosage can be used if a child’s dosage must stay between 100 mg and 200 mg?

This must be solved in the form of an inequality.
I appreciate and thank you for your help!

Is this formula correct?
100<=8d+12<=200

Then you have:

ad/(a+12)>= 100

and

ad/(a+12) <= 200

or

8d/(8+12)>= 100 --------------d >= 250

and

8d/(8+12) <= 200--------------d >= 500

There is your inequality.... doesn't make sense to me.... but.....

 

[mmm4444bot]

d >= 500

… doesn't make sense to me …

Due to the typographical error, perhaps ? :wink:

d <= 500

It means that, given an 8-yr-old needing drug therapy (often available from a variety of acceptable choices), they choose from among those drugs that have a 250-500 mg range for therapeutic levels in adults, as a valid way of estimating appropriate pediatric optons to consider.

(Actually, I just made that up, but it sounds good. Heh, heh.)

We can set-up the inequality this way, too:

\(\displaystyle C \;=\; \frac{a \cdot d}{a \;+\; 12}\)

\(\displaystyle C \;=\; \frac{8d}{20} \;=\; \frac{2d}{5}\)

\(\displaystyle 100 \;<\; \frac{2d}{5} \;<\; 200\)

\(\displaystyle 500 \;<\; 2d \;<\; 1000\)

\(\displaystyle 250 \;<\; d \;<\; 500\)

I interpret the word "between" to mean that the endpoints are not included.

Something like "from 100 mg through 200 mg" would include the endpoints.

Cheers ~ Mark

 

[Deleted member 4993]

Re:

d >= 500

… doesn't make sense to me …

color=#4000BF]

Due to the typographical error, perhaps ? :wink: <<< well it was a "cut & paste" error - I am getting infected by "Denis desease" (an obsessive aversion to typing) - and that, as you know, is one of the symptoms.

d <= 500

The part that did not make sense to me was that the for child dosage of 100 - 200 we are looking for a medicine with adult dosage of 250-500.