# How to solve for the minimum of this quadratic function

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• Nov 28th 2013, 06:00 PM
mathbeginner97
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 (Worried)
• Nov 28th 2013, 06:05 PM
Prove It
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...
• Nov 28th 2013, 06:42 PM
mathbeginner97
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!
• Nov 28th 2013, 06:59 PM
Prove It
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.
• Nov 28th 2013, 07:26 PM
mathbeginner97
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 ...
• Nov 28th 2013, 07:28 PM
Prove It
Re: How to solve for the minimum of this quadratic function
So now you solve the cubic function...
• Nov 28th 2013, 09:20 PM
Melody2
Re: How to solve for the minimum of this quadratic function
Quote:

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 (Worried)

Y = 3q - 19q + 50 + (100/q)

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

When you differentiated this, how did you get a 6q??
I think that you should differentiate it again. Properly this time.
• Nov 29th 2013, 12:20 AM
mathbeginner97
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!
• Nov 29th 2013, 01:44 AM
Melody2
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.
• Nov 29th 2013, 01:56 AM
romsek
Re: How to solve for the minimum of this quadratic function
Quote:

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.
• Nov 29th 2013, 02:05 AM
Melody2
Re: How to solve for the minimum of this quadratic function
I can show you what $\displaystyle y=(x-2)(x^2+3x+67)$ looks like

888201b2c0cfd397bee02217430e7b76.gif

And I can show you what your initial graph looks like

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

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.
• Nov 29th 2013, 02:11 AM
Melody2
Re: How to solve for the minimum of this quadratic function
$\displaystyle y=3q^2-19q+50+\frac{100}{q}$ still is not a cubic

this is the graph.

9d0288f95ec7a10c3707e4ddf878416d.gif
• Nov 29th 2013, 02:21 AM
romsek
Re: How to solve for the minimum of this quadratic function
Quote:

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.
• Nov 29th 2013, 02:24 AM
Melody2
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.
• Nov 29th 2013, 02:28 AM
romsek
Re: How to solve for the minimum of this quadratic function
Quote:

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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