00:02
If we are going to make a candy box with a maximum volume like x by xx squares out of a flat 13 by 8 sheet, we first examine what would happen if we cut those squares out.
00:17
The side that is 13 becomes 13 minus 2x, and the side that is 8 becomes 8 minus 2x, and of course the height of the box becomes x, as you can see in red.
00:28
To maximize the volume, we need a volume equation, which means we want to move.
00:32
And we'll all three of the dimensions together to find that volume equation.
00:43
In this case, we see that our volume equation will be 104x minus 42x squared plus 4x to the third.
00:58
To maximize the volume, we need to take the derivative.
01:03
So we get v prime is equal to 104 minus 84x plus 12x squared, setting that equal to zero to find our critical numbers.
01:17
Now we can use quadratic formula or we can graph this using a graph in utility.
01:22
If we do graph it, what we find is there are two places where we get zeros, that is, that x equals 1 .607 and x equals 5 .393.
01:34
Notice that we can safely ignore the 5 .393 because if we took 8 minus 2 times the 5 .393, we would get a negative side length, which, well, that just can't happen.
01:46
So that means if you want to maximize the volume, you're going to cut off squares of 1 .61 inches each.
01:55
The second question is about finding the absolute maximum and minimum values of the given function on the interval from negative 3 to 3.
02:15
So we can do this using the candidates test, meaning we're going to test the end points as well as any critical numbers on that interval.
02:24
To find the critical numbers, we need to take the derivative.
02:27
So we need to find f prime of x using the quotient rule.
02:32
And in this case, that differentiation is going to look like this.
02:44
We, of course, need to simplify...