# How to solve Optimization problems in calculus.

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?

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 $$h = {100 \over \sqrt[3]{10000}} = \sqrt[3]{100}$$.

a) Consider the equation for the volume of a box with a square base of area x^2, Volume = x^2 * height. From here you can solve for the height of the box in terms of x.

b) After you find an equation for height in terms of x, to find the surface area, you just need to find the sum of each face's area.

c)No. You forgot to subtract 1 from the exponent.

d)Look at the 0s of the derivative.

Now, considering optimization problems in a first year calculus course, I believe you can approach most similarly. Consider what function you are trying to optimize. In your case, the surface area formula. Then you should figure out how to reduce this formula to only 1 variable, in this case, x, using constraints that are given to you. Then take the derivative of the function you are trying to optimize and solve for 0. Once you figure out what this variable is, go back and plug it into your given constrains. You'll likely find that you can solve for a different variable explicitly.

The square base has sides $$x$$ and the height is $$h$$, so the volume $$V=x^2h$$. We know $$V=100$$, so we get that $$100=x^2h$$, or, as we want it, $$h=\frac{100}{x^2}$$.

b) Draw a cuboid with the required dimensions and find its surface area. Then use that $$h=\frac{100}{x^2}$$.

c) It's good practice to write the original as $$A(x)=2x^2+400x^{-1}$$, that way you can see the normal power rules more easily.

d) This is the optimisation bit of the question, and its simpler than it seems. $$A(x)$$ is the surface area of the cuboid when its base is $$x$$. If we wish to minimise it, we want where:

$$A'(x)=0 \text{ and } A''(x)>0$$ The first derivative being $$0$$ means we have a turning point, the second derivative being positive means that the turning point is a trough rather than a peak. You can calculate the two derivatives quite easily, then solve $$A'(x)=0$$ and check each solution in $$A''(x)$$ to see if that's positive too. Don't forget that $$x>0$$, it's a length!

Lastly, once you've found your value of $$x$$, the question asks for $$h$$, so find that via $$h=\frac{100}{x^2}$$

Good luck!