We want a box that can hold cubic feet of mulch. Its dimension are such that the length is feet longer than the width and the height is foot less than the width. How do we find the dimensions?

Given the width, what is the volume of the box? Since volume is \(L*H*W\), we need to find those three quantities.

Everything is in terms of width. The translation of the given information is that and .

Typically we write \(x\) for the unknown input and let's do that here. Then the formula for computing the volume given the width \(x\) is .

To get a sense of this problem, we can graph it with the width values increasing along the horizontal axis towards the right while the volume amount will be measured with increasing value up the vertical axis.

We explore a general method that we think of as the hammer method since it basically smashes any such "find the unknown that hits the target value" with very little understanding of the algebra of the problem. (This is actually known as the Secant Method).

The first step is to write an expression that computes the value for a given choice. This is actually the hard part of the problem and is required for any method of solution.

For this particular example, we use the (definitionesque) fact that we can get the volume from the dimnesions by multiplying them all together. So the expression is where \(x\) is the width.

We try some number, such as , to get a sense of this. Plugging in, we get . Our task is get that volume to be our target number of and we are currently away from the target.

To tackle this in a reasonably efficient manner, we do a couple of guesses and then we pretend that this expression is a representing a line. In particular, a line has the nice property that changes in \(y\) are proportional to changes in \(x\). That proportionality constant is the slope.

Applying it here, we can compute our next guess by computing how much we want to change the volume amount to get to the target volume. We compare this to the change in the volume between our two guesses and then we apply that proportional change to our width. And that becomes our next guess.

Guess 1: , Guess 2: , Precision: . Then here is our method computing the result:

A big field is enclosed by a circular fence. We are going to construct several straight fences crossing the field. These fences will enclose various regions inside the field. If \(n\) is the number of endpoints of the fences on the circle, what is the maximum number of enclosed regions that could be obtained? What causes the number to be lower?

This is called Moser's problem and it is an interesting problem because the number of regions at first doubles for each point added but when we add the 6th point, the maximum number of regions is less than double the previous number. Let's explore.

• First step is to think

  In detail

  More details

• Next step

• Final step

  blah blah

no blah

• No leading blurb

  what?

• blurb blurb

  details

  more details

regions = 0;
log(points); 
regions = 31;
this.regions += 1;
this.dude += 10;
log(regions, this.regions);
out(`<p>${5}</p>`);

regions = 31;
log(regions);