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 $D = 3x + 100/x, x >0$. 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?
 A: Let  $D(x) = 3x + \frac{100}{x}$ defined for $x>0$. 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 $x>0$, 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.
A: 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.
