In my math class we are currently studying calculus. We just came to the Optimization section and my teacher said that it’s very tricky. I don’t quite understand so I was hoping for some guidance on a problem so that hopefully I can use it to work on other problems. (Would I be able to use the way I do one problem for a different problem? Based on the questions we did in class, it seemed different which makes me very confused.).
This is a problem in our textbook as an example.
A manufacturer needs to construct a box having a square base and holding 100 cubic inches.
a) Let each side of the square base be $x$ inches. Write an expression for the height, $h$, of the box in terms of $x$.
So, I know the volume is base multiplied by width multiplied by height. Does that have any relevance for this part of the problem?
b) Let A(x) be the outside surface area of the box. Show that $A(x) = 2x^2 + 400/x$.
To be completely honest, I really don’t know where to even begin on this part of the question.
c) Find $A’(x)$.
Would this be $4x + -400/x$?
d) Find the height of the box that will minimize the outside surface area.
Basically, I need some help getting a grasp on the concept of Optimization problems. I don’t really have a clue where to begin with them. This question is an example of the types of problems I’m working with in class. Are there are tips / tricks you could show me to help me get through this unit?
Best Answer
You know that $V(x,h) = x^2h$ and also that $V(x,h) = 100$. In particular, this means you can determine $h$ using $h= {100 \over x^2}$.
The area is given by $2x^2+4xh$ (counting all 6 sides), so using the previous relation we have $A(x) = 2x^2 + 4 x {100 \over x^2} = 2x^2 + {400 \over x}$.
Note that there is an implicit constraint that $x > 0$.
If we plot $A$ for $x>0$ we see that it has a $\min$ somewhere, to find the $\min$ we look for points where the slope $A'(x)$ is zero.
Since $A'(x) = 4x -{400 \over x^2}$, we see that the slope is zero when $x = \sqrt[3]{100}$.
This gives the $x$ value, to get $h$ we use the formula from the first paragraph to get $h = {100 \over \sqrt[3]{10000}} = \sqrt[3]{100}$.