# Thread: How to solve for the minimum of this quadratic function

1. ## How to solve for the minimum of this quadratic function

How do I solve for the minimum of Y = 3q - 19q + 50 + (100/q) ?

I have no idea how to do this

2. ## Re: How to solve for the minimum of this quadratic function

Start by doing a graph of the function, that should give you a hint. Then remember a minimum can either be at an endpoint or a turning point of a function. Surely since you're doing calculus you should know how to find turning points...

3. ## Re: How to solve for the minimum of this quadratic function

Sorry the function is Y = 3q^2 - 19q + 50 + (100/q)

I know you get the turning point by taking the second derivative so see where the function is increasing and where it is decreasing. I'm honestly stuck though, I've completely given up because I'm sure my answer is wrong!

4. ## Re: How to solve for the minimum of this quadratic function

No, the minimum is where the first derivative is equal to 0 and where the second derivative is positive.

5. ## Re: How to solve for the minimum of this quadratic function

I have no idea what I'm going then

When I take the derivative and set it to zero I get the following: 6Q - 19 - (100/Q^2)

Then by rearranging I get the following: 19 = 6Q - (100/Q^2)

But then I have to multiply everything by Q^2 and I get a cubic function. I don't get how this gets me anywhere ...

6. ## Re: How to solve for the minimum of this quadratic function

So now you solve the cubic function...

7. ## Re: How to solve for the minimum of this quadratic function

Originally Posted by mathbeginner97
How do I solve for the minimum of Y = 3q - 19q + 50 + (100/q) ?

I have no idea how to do this
Y = 3q - 19q + 50 + (100/q)

$Y = -16q+ 50 + 100q^{-1}$

When you differentiated this, how did you get a 6q??
I think that you should differentiate it again. Properly this time.

8. ## Re: How to solve for the minimum of this quadratic function

Can a cubic function only have a single zero?

I was doing another problem and I found that a root was x = 2, so after synthetic division I'm left with (x-2)(x2 + 3x + 67).

Ps. Thanks for your help! I appreciate it!

9. ## Re: How to solve for the minimum of this quadratic function

Yes, y=x^3 is a cubic function with only 1 root.
but
Your initial question is not a cubic function and it doesn't have any roots.
How come you have introduced another problem?
I'm confused.

10. ## Re: How to solve for the minimum of this quadratic function

Originally Posted by mathbeginner97
Sorry the function is Y = 3q^2 - 19q + 50 + (100/q)

I know you get the turning point by taking the second derivative so see where the function is increasing and where it is decreasing. I'm honestly stuck though, I've completely given up because I'm sure my answer is wrong!
What is the left hand limit of Y at 0? Does Y have a minimum value? If you restrict your domain to {q : q > 0} this problem makes a bit more sense.

11. ## Re: How to solve for the minimum of this quadratic function

I can show you what $y=(x-2)(x^2+3x+67)$ looks like

888201b2c0cfd397bee02217430e7b76.gif

And I can show you what your initial graph looks like

$y=-16q+50-100q^{-1}$

a00948af141e034ff4f9236a005fd637.gif

I just read Romsek's post that says the original post had the wrong formula so I guess this is not at all relevant.

12. ## Re: How to solve for the minimum of this quadratic function

$y=3q^2-19q+50+\frac{100}{q}$ still is not a cubic

this is the graph.

9d0288f95ec7a10c3707e4ddf878416d.gif

13. ## Re: How to solve for the minimum of this quadratic function

Originally Posted by Melody2
Yes, y=x^3 is a cubic function with only 1 root.
1 real root. .. Ok given that the domain is restricted to reals saying it has 1 root is correct.

14. ## Re: How to solve for the minimum of this quadratic function

I don't think so, I think y=x^3 has a 1 triple root.

15. ## Re: How to solve for the minimum of this quadratic function

Originally Posted by Melody2
I don't think so, I think y=x^3 has a 1 triple root.
you're right, it's late. I thought you were saying the cubic polynomial that results from setting dy/dq = 0, has only 1 root.

It has 3 over C but only 1 over R.

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