How do I solve for the minimum of Y = 3q - 19q + 50 + (100/q) ?
I have no idea how to do this
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...
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!
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 ...
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)(x^{2} + 3x + 67).
Ps. Thanks for your help! I appreciate it!
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.
$\displaystyle y=3q^2-19q+50+\frac{100}{q}$ still is not a cubic
this is the graph.
9d0288f95ec7a10c3707e4ddf878416d.gif