00:01
Hi there, so for this problem we are told that an open pops is to be made out of six inches by 14 inch piece of cardboard.
00:10
So let me just write a situation for this something like this.
00:22
So the dimension, the initial dimension for this are going to be 6 and 14.
00:29
Now we need to fold this in order to form an open pops.
00:41
So we will form a squares at the edges of this.
00:47
Okay, x, x.
00:48
Those are going to be the dimensions for this.
00:52
Now the dimensions for these pops are going to be 6 minus 2 times x because we need to also take into account the well, we need to take into account these squares at the top and at the bottom of this and then at this side is going to be 14 minus 6 times x and the height of this pops is going to correspond to x.
01:14
When we fold everything you could see that that will be the height of this.
01:18
Now the volume for this is just simply 14 minus 6 times x.
01:24
Oh, sorry.
01:25
No, no 6, 2.
01:27
Sorry.
01:27
It's 14 minus 2 times x.
01:31
So that will be then.
01:33
Okay.
01:33
14 minus 2 times x.
01:36
This times 6 minus 2 times x and this times the height that is x.
01:41
Now we just need to expand this.
01:42
Remember that the question is to find the dimensions that has the largest volume.
01:46
So we need to maximize the volume function.
01:49
So for that we need to first expand this expression and when we do that, we obtain the following.
01:56
That the volume is equal to 4 times x to the 3.
02:02
Then this minus 40 times x squared.
02:06
Then this plus 84 times x.
02:11
Now we just need to differentiate this with respect to x.
02:15
So that will give us 12 times x squared and this minus 80 times x plus 84.
02:26
And now we just set this equal to 0...