Finding the minimum of a function. I’m working on learning calculus with story problems. It’s easier to figure out when the question simply asks for specific answers, so I figured I needed to get better at solving problems based on the info given in the story problem.
Here’s an example that I’m trying out:  

The delivery cost per ton of bananas, in thousands of dollars, when $x$ tons of bananas are shipped is given by $D = 3x + 100/x, x &gt;0$. Find the value for $x$ for which the delivery cost per ton of bananas is a minimum, and find the value of the minimum delivery cost. 

To begin with, we know this function is the delivery cost and $x$ is the number of tons of bananas.  

To find the minimum of a function we use the first derivative, right? So,
$D = 3 - 100 x^{-2}$ ?
Then, I am unsure how to find the value. Do I find what $x$ is from the derivative and then plug it into the original function?
Let  $D(x) = 3x + \frac{100}{x}$ defined for $x&gt;0$. Then, by differentiating, we yield :

$$D'(x) = 3 - \frac{100}{x^2}$$

To find an extreme point (minimum or maximum), one has simply to take the derivative equal to zero. Thus :

$$D'(x) = 0 \Rightarrow 3-\frac{100}{x^2} = 0 \Leftrightarrow x = \pm \sqrt{\frac{100}{3}}$$

But, note that the function is initially defined for $x&gt;0$, so we ignore the negative value.

Thus, for $x = \frac{10\sqrt{3}}{3}$ the delivery cost is minimum. The value is yielded by plugging that $x$ into the cost equation :

$$D\bigg(\frac{10\sqrt{3}}{3}\bigg) = 10\sqrt{3} + \frac{100}{\frac{10\sqrt{3}}{3}} = 20\sqrt{3}$$

This is a minimum value and a minimum point and can be easily seen by checking the sign of the first derivative on corresponding intervals, or by implementing a second derivative test.
If you plot the function and the derivative of the function in the same figure you will see that the minimum of the function is exactly at the same x-value where the derivative intersects the x-axis. 

So to find x-value for the coordinate of the minimum point you equate the derivative to zero. 
$$D' = 3 -\frac{100}{x^2} = 0$$
We get $$3x^2 = 100$$
So the minimum is at $x = \sqrt{\frac{100}{3}}$ or $x= - \sqrt{\frac{100}{3}}$.
And because x>0 the minimum value is then $x = \sqrt{\frac{100}{3}}$

Also use https://www.desmos.com/calculator/qtzamwcdrk to get a better visual understanding of what is happening.