I'm having problems with optimization, the book calls this section optimization II.
Packaging . By cutting away identical squares from each corner of a rectangular piece of cardboard and folding up the resulting flaps, an open box may be made. if the cardboard is 15 in. long by 8 in. wide, find the dimensions of the box that will yield maximun volume.
I basically drew a box and cut away its 4 corners and labeled each cutaway as "x", and labeled the sides as 15-2x and 8-2x, since we are looking for volume V=lwh
V=lwh= (15-2x)(8-2x)(x) = 4x^3 - 46x^2 + 120x
f ' x= 12x^2 - 92x + 120
Ok, so this is where I get stuck, the example problem similiar to this one has one to look for an interval [ ] to test on. ex. [0, 5]. They mention something about the box length has to be positive, which makes sense, but I don't understand how I get the interval and how I plug it in.
The interval, if I looked at the example problem correctly, comes from this...
15-2x equal or greater than 0
8-2x equal or greater than 0
x equal or greater than 0
Thank you for anyone who can help.
This means that your x value cannot be less than zero (you would be adding length to your paper in this case), and cannot be greater than 4 (you would be removing more than the maximum removable amount) so your domain is [0,4], now try graphing your equation on this domain, you can use a scientific calculator, I attached a graph I made at Cool math .com - Online Graphing Calculator - Graph It! This graph shows v(x) and v'(x) on the interval from [0,4]
Now you just have to consider your graph, you can see that if you cut away very little, your box will not have much volume, because it will be very shallow, and if you cut away a lot, your box will also not have much volume, because while it is deep, it is not very wide. Somewhere in the middle is your sweet spot. So as you cut away different lengths of x, your volume will increase, and then after a certain point, it will start decreasing again. The point where it stops increasing and starts decreasing will be the greatest volume attainable. (marked maximum volume on the attached graph)
If you graph it, it will be the highest point on your graph, so lets consider the graph. You know that the slope of this line will be increasing while the volume is increasing (or else the next point on the graph would not be higher than the current point) and the slope will be decreasing while the volume is decreasing. But we are only interested in the point where the volume is turning around from increasing to decreasing. At this particular point, the volume is neither increasing nor decreasing, and this means that it's slope is zero. If you drew a tangent line at that point, it would be a horizontal line.
So now we only need to find the point on this line where the slope is zero. As you know, the derivative of an equation gives the value of the slope of that equation for any given x value, so the graph of the derivative is actually the graph of the slope of x for any given x value. Now, we are looking for a slope of zero, so wherever the graph of the derivative is equal to zero, is where the line has a slope of zero.
So you set your derivative equal to zero, and solve for x. As earboth showed, you get two values (you can use the quadratic formula), x=5/3 and x=6. This means at those two x values, the graph of volume will have a slope of zero, which is important to us, because in order to change directions, the slope must be zero, so these two points are places where the graph might be changing directions.
You know that your domain is [0,4] and you can see that 6 is not within the domain (you could not cut two 6 inch sections out of an 8 inch piece of paper), so the only point in our domain is x=5/3.
Now we can intuitively see this must be the highest point, because we can see that we start low, and end low, and the volume is positive (you can't have negative volume) and x=5/3 is the only point inside our domain where the slope is zero. But you could also check this in one of two ways, check a point before and after, such as v(4/3) and v(2). If we did it correctly, these two values should be less than v(5/3). And the other way would be to check the slope at those points, v'(4/3) should be positive, because it should be increasing, and v'(2) should be negative because it should be decreasing. You can see this on the attached graph, that if x=5/3 is the point where v(x) is at it's highest, then checking an x value before and after should be lower than v(5/3). And you can see that the graph of the slope is positive while v(x) is increasing, and negative while it is decreasing, so any v'(x) before x=5/3 will be positive, and any v'(x) after will be negative. Either way will work.
Hope that helps ^_^