R = 15t^2 – 120t + 250: find the value of t for which....

[harith]

R = 15t^2 – 120t + 250: find the value of t for which....

Hi, I was doing a math paper and came across this question:

At time t seconds, the rate of flow of water, R cm/s, through a pipe is given by
R = 15t² – 120t + 250
Find the value of t for which R is a minimum.

Am I suppose to solve this as quadratic? If not then how? Or at least tell me the topic which it comes from.
Thanks n advance.

 

[lookagain]

Hi, I was doing a math paper and came across this question:

At time t seconds, the rate of flow of water, R cm/s, through a pipe is given by
R = 15t² – 120t + 250
Find the value of t for which R is a minimum.

Am I suppose to solve this as quadratic? If not then how? Or at least tell me the topic which it comes from.
Thanks n advance.

What this wants you to do is to put that quadratic expression into standard parabola form
so you can easily read off it's vertex point. Since the coefficient of the \(\displaystyle t^2\) term is positive you know that
this parabola faces upwards and thus has a minimum at it's vertex.

Do you remember how to complete the square?

As far as what the topic is called .... mmm finding the extreme of a quadratic function?
Properties of parabolas? Putting it in advanced algebra is fine.

harith, all you need to use for this problem is the formula \(\displaystyle \ \ t \ = \ \dfrac{-b \ }{2a }. \)

The equation is already in the form \(\displaystyle \ \ R \ = \ at^2 + bt + c.\)


Determine the coefficients for a and b, and then substitute them into the formula to determine the time.

 

[harith]

harith, all you need to use for this problem is the formula \(\displaystyle \ \ t \ = \ \dfrac{-b \ }{2a }. \)

The equation is already in the form \(\displaystyle \ \ R \ = \ at^2 + bt + c.\)


Determine the coefficients for a and b, and then substitute them into the formula to determine the time.

Buh how did you arrive at that formula?

 

[harith]

Okay, so completing the square method will solve the equation. Then where did parabola come into play?

And one more question; how will i find it's maximum value?

The answer on the paper says it should be 4, but my answer says its -4; (t-4)²=5\30

 

[JeffM]

And one more question; how will i find it's maximum value?

This parabola has no maximum value; it has no upward bound because it is upwardly convex.

You need to graph the parabola to visualize what has been being explained to you.

For an upwardly convex parabola, the vertex is its minimum, and there is no maximum.

For a downwardly convex parabola, the vertex is its maximum, and there is no minimum.

Look at http://en.wikipedia.org/wiki/Parabola

It will now be intuitively clear why the vertex of an upwardly convex is its minimum point, namely the shortest distance between two points is a straight line.

 

[harith]

Sorry guys for lot of silly questions, its just that i had heard about parabola the first time. But anyways thanks alot for all your help, i really appreciate it.

 

[stapel]

Buh how did you arrive at that formula?

If your class (that is, your instructor and/or your textbook) hasn't taught you that formula, then using it on the test will likely be counted as incorrect. If you haven't seen this formula, then definitely use instead the method you were told to use in class!

 

[Dale10101]

harith, all you need to use for this problem is the formula \(\displaystyle \ \ t \ = \ \dfrac{-b \ }{2a }. \)

The equation is already in the form \(\displaystyle \ \ R \ = \ at^2 + bt + c.\)


Determine the coefficients for a and b, and then substitute them into the formula to determine the time.

Buh how did you arrive at that formula?

I believe that M. Romsek already gave a detailed proof of how of how -b/2a comes to express the minimum value of your function R(t) ... that proof has seemed to have disappeared ... but, after spending some time looking at the issue it seems there is an observational justification for -b/2a, no calculation needed (?).

It is observed that R(t) is a nose down parabola from its quadratic form (only x is squared and its coefficient is positive). It is noted that such a parabola is symmetrical around a vertical axis so that the elevation of any particular y value on the parabola correspond to two value of x, each equally displaced from the vertical axis of symmetry but on each side of it. It is also noted that the vertical axis of symmetry runs through the parabola's vertex, i.e. it minimum value.

Now the quadratic formula delivers both such x values for a particular value of y by adding and subtracting the quantity Sqrt(b^2-4ac)/2a to and from -b/2a thereby showing that -b/2a is centered between them and hence is on the vertical axis of symmetry and in fact is its x coordinate. QED, -b/2a does deliver the x coordinate of a nose down parabola's minimum value. ta da (maybe?)

detail: The quadratic formula: x = [-b + and - Sqrt(b^2-4ac)]/2a can be written x = -b/2a + and - Sqrt(b^2-4ac)/2a

 

[Dale10101]

Ah

I'd sure like to know why that derivation has been removed.

Is that what this site is about? Making sure students use canned formulas without understand their origin?

Peace Brother Romsek. I think there is neither a conspiracy nor ill intent. From past posts it is apparent that the dude/dudette administering this site does so as an after work project without remuneration, there is no board of mathematical decency.

Certainly it would be nice to know why posts go missing, for example in a recent post I asked you for the name of Lambert function and that was because of a missing response that I had elicited a couple of months ago ... where did it go? ... my other posts are intact.

Posts take a good bit of time to polish up and make presentable, it sucks when they disappear ... but I don't think they are "removed" ... I think sometimes "chit happens" ... an artifact of entropy perhaps, a glitch in the neutron flow. BTW, thanks for your excellent work, you have helped me out more then once.

 

[Dale10101]

But, back to biz.

Also: (2a)x + b = SQRT(b^2 - 4ac) OR (2a)x + b = -SQRT(b^2 - 4ac)

When I wrote that that the quadratic formula can be written x = -b/2a + and - Sqrt(b^2-4ac)/2a I was illustrating that written in that form one clearly sees that the nose of the parabola lies midway between its eyes, mnemonically speaking, and hence is quickly located using -b/2a, a very useful shortcut pointed out by M. Lookagain.

I take your work seriously and have considered the point you are making but have failed to see what you are getting at ... when would knowing a, x, and b be a useful shortcut in finding the square root of the discriminate and towards what end?

Also, is the + - in the quadratic formula an "OR", or an "AND" or an "OR/AND" situation?

 

[Dale10101]

eh?

Guess you missed the humor in between the +- et al...
Of course -b/2a is King!

OK, I am laughing now ... but I don't know why, a ton of simple I guess. Long live the King!

 

[lookagain]

Also, is the + - in the quadratic formula an "OR", or an "AND" or an "OR/AND" situation?

Dale10101, it is an "or" situation, as in "plus or minus."

 

[Dale10101]

Thanks, but now I am in real trouble ...

Given: \[{x^2} + 3x + 1 = 0\]
the quadratic formula yields: \[\frac{{{\rm{ - }}\sqrt 5 }}{2} - \frac{3}{2}{\rm{ or }}\frac{{\sqrt 5 }}{2} - \frac{3}{2}\]
but at the same time the solution set of the equation is: \[\left( {\frac{{{\rm{ - }}\sqrt 5 }}{2} - \frac{3}{2},0} \right){\rm{ and }}\left( {\frac{{\sqrt 5 }}{2} - \frac{3}{2},0} \right)\]

No? So why not "and", or at least "and/or" ? Can you give me an example when the quadratic equation does not deliver two correct results even if they are the same? I know you are correct but I could not explain why if someone asked me to justify "or" as opposed to "and".

 

[JeffM]

Given: \[{x^2} + 3x + 1 = 0\]
the quadratic formula yields: \[\frac{{{\rm{ - }}\sqrt 5 }}{2} - \frac{3}{2}{\rm{ or }}\frac{{\sqrt 5 }}{2} - \frac{3}{2}\]
but at the same time the solution set of the equation is: \[\left( {\frac{{{\rm{ - }}\sqrt 5 }}{2} - \frac{3}{2},0} \right){\rm{ and }}\left( {\frac{{\sqrt 5 }}{2} - \frac{3}{2},0} \right)\]

No? So why not "and", or at least "and/or" ? Can you give me an example when the quadratic equation does not deliver two correct results even if they are the same? I know you are correct but I could not explain why if someone asked me to justify "or" as opposed to "and".

The set of algebraically valid solutions contains both solutions. So the set is defined by an "and" relation. Does that make sense?

 

[lookagain]

Given: \[{x^2} + 3x + 1 = 0\]
the quadratic formula yields: \[\frac{{{\rm{ - }}\sqrt 5 }}{2} - \frac{3}{2}{\rm{ or }}\frac{{\sqrt 5 }}{2} - \frac{3}{2}\]

Those aren't just numbers, they're x-values. The "x = " is part of the solution. And the Quadratic Formula yields
expressions for x which have been simplified to fractions.


So, the Quadratic Formula yields:


\(\displaystyle x \ = \ \dfrac{-3 \pm \sqrt{5}}{2}\)


This is equivalent to:


\(\displaystyle x \ = \ \dfrac{-3 - \sqrt{5}}{2} \ \ \ or \ \ \ x \ = \ \dfrac{-3 + \sqrt{5}}{2}\)

 

[mmm4444bot]

Certainly it would be nice to know why posts go missing

The system shows seven posts deleted from this thread.

Six posts were authored by Romsek; they were deleted later by Romsek.

One post was authored by JeffM; it was deleted later by JeffM.

 

[Dale10101]

Sad

The system shows seven posts deleted from this thread.

Six posts were authored by Romsek; they were deleted later by Romsek.

One post was authored by JeffM; it was deleted later by JeffM.

THERE WAS a missing post in which Romsek proved that -b/2a is a shortcut to finding the vertex of a nose down parabola, alas, I made tangential reference to that fact and .. well, things spun out of control. In the end then, it was as I suspected, a glitch in the neutron flow. Sad that M. Romsek has decamped, I, for one, benefited from his knowledge and don't mind saying so.

kudos to mmm4444bot for setting the record straight.