Optimization: box volume (Part 1)

let's say that we have a cardboard a sheet of cardboard that is 20 inches by 30 inches let me draw the cardboard as neatly as I can so it might look something like that so that is my sheet of cardboard and just to make sure we know the dimensions a day it's 20 inches by 30 inches and what we're going to do is cut out the corners of this cardboard and all the cut out the corners are going to be squares and we're going to cut out an x by x corner from each from each of the corners from each of the corners of this piece of cardboard x by x over here X by X and then over here X by X and what we'll do is after we cut out those corners we can essentially fold down the flaps let me draw the flaps so you can imagine we can fold right there we could fold right there we could fold right there and we would form a box I guess you could you could imagine a box without a bottom to it or you could view a box without a top to it so if we were to fold everything up we would get a container that looks something like this let me write my best attempt to draw it so it would look like this it would look like this this is one flap fold folded up you can imagine this flap right over here if I were to fold it up like that it now would look like this it would now look like this the height of the flap is X so this distance right over here is X and then if I were to fold this flap if I were to let me do that a little bit neater if I were to fold this flap right over here if I were to fold that up then it would look like this it would look like this my best attempt to draw it so it would look like that and then I would fold that back flap up so the back flap would look something like that that would be the back flap it would look something like that and then this flap over here would if I hold it up would look something like would look something like that and then of course my base of my whole thing so this whole region right over here of my piece of cardboard would be the floor would be the floor of this box that I'm constructing and what I want to do is I want to maximize the volume of this box I want to maximize how much it can hold and I want to maximize it by picking my X appropriately so let's think about what the volume of this box is as a function of X well in order to do that we have to forget all the dimensions of this box as a function of X we already know that we already know that this corner right over here which is made up of that when this side and this side connect when you fold these two flaps up that's going to be the same height over there that's going to be the same height over there the height of this box is going to be X but what's the width what is the width of this box going to be well the width of the box is going to be this distance right over here and this distance is going to be 20 20 inches minus not 1 X put minus 2 X's so this is going to be 20 minus 2x you see it right over here this whole distance is 20 you subtract this X you subtract this X and you get this distance right over here so it's 20 minus 2x now the same logic the same logic what is the depth of the box what is that distance right over there well that distance is this distance is this distance right over here we know that this entire distance is 30 inches if we subtract out this X and we subtract out this X we get the distance that we care about so this is going to be 30 30 minus 2x so now we have all of the dimensions so what would the volume be as a function of X well the volume the volume as a function of X is going to be equal to the height which is X the height which is x times the width with it which is 20 minus X 20 or sorry 20 minus 2x times the depth which just 30 minus 2x which is 30 minus 2x now what are possible values of X that give us a valid volume well X can't be X can't be less than 0 you can't make a negative cut here somehow we would have to add cardboard or something there so we know that X is going to be greater than or equal to 0 so let me write this down X is going to be greater than or equal to 0 and what does it have to be less than well I can cut at most we can see here the length right over here this this pink color this move color is 20 minus 2x so this has got to be greater than 0 this is always going to be shorter than the 30 minus 2x but the 20 minus 2 X has to be greater than has to be greater than or equal to 0 you can't cut more cardboard than there is or you could say that 20 has to be greater than or equal to 2x or you could say that 10 is going to be greater than or equal to X which is another way of saying that X is going to be less than or equal to 10 it's a different shade of yellow X is going to be less than or equal to 10 so X has got to be between 0 and 10 otherwise we've cut too much or where somehow adding cardboard or something so first let's think about the volume at the endpoints of this of the of our essentially of our domain of what X can be for our volume well our volume our volume when X is equal to 0 is equal to what we're going to have 0 times all of this stuff and it's pretty obvious you're not going to have any height here so you're not going to have any volume so our volume would be 0 what is our volume when X is equal to 10 well if X equaled 10 then was this the width here that I've drawn in pink would be 0 so once again we would have no volume and this term right over here if we just look at it algebraically would also be equal to 0 so this whole thing would be equal to 0 so someplace in between x equals 0 and x equals 10 we should achieve our maximum about our maximum volume and before we do it analytically with a bit of calculus let's do it graphically so I'll get my my d ti-85 out so let me get my ti-85 out and so first let me set my range appropriately before I attempt to graph it so I'll put my graph function let me set my range so my minimum x value let me make that zero we know that X cannot be less than zero my maximum x value well ten seems pretty good my minimum my minimum y-value this is essentially going to be my volume I'm not going to have negative volume so let me set that equals zero and my maximum y-value let's see what would be reasonable here I'm just going to pick some a random X and see what type of a volume I get so if X were to be five if X were to be 5 would be 5 times 20 minus 10 which is 10 which so that would be did I do that right yet fought 20 minus 2 times 5 so that would be 10 and then times 30 minus 2 times 5 which would be 20 so it would be 5 times 10 times 20 so you'd get a volume of a thousand cubic inches and I just randomly pick the number 5 so let me give my let me give my maximum y-value a little higher than that just in case that that isn't the maximum value I just randomly picked so let's say Y max is 1500 and then for whatever reason our graph doesn't fit in there then maybe we can we can make our Y max even larger so I think this is going to be a decent range now let's actually input the function itself so our volume is equal to x times 20 minus 2x minus 2x times 30 minus 2x 30 minus 2x and that looks about right so now I think we can graph it so second and I want to select that up there so I'll do graph and it looks like we did get the right range so this tells us volume as a function of X between x is 0 and X is 10 and does look like we hit a maximum point right around there so what I'm going to do is I'm use the trace function to figure out roughly what that maximum point is so let me trace this function so I can still go high or higher okay so over there my volume is 1050 five point five that I can get to 1056 that I get to 1050 six points so see this was ten fifty six point two zero this is 10:50 6.24 then I go back to 1055 so at least based on the level of zoom that I have my calculator right now this is a pretty good approximation for the maximum value that my function actually takes on so it looks like my maximum value is around 10 56 and it happens at around x equals three point eight nine so it looks like it looks like my volume my volume at three point eight nine is approximately equal to one thousand and fifty six cubic inches or you could say that our maximum we hit a maximum when X when X is approximately equal to approximately equal to three point eight nine so far we've just set up our maximization problem then we've looked at it graphically in the next video we'll try to solve it analytically using some of our calculus tools