 All right, let's do one more example and that will put it us at the end of this section right here 4.7 This one's a three-dimensional picture. You're gonna have to bear with me a little bit as we try to draw 3d for a moment Imagine we have an old are we want to make an open box by cutting off the square corners of This square and then we can fold it up. We can fold these flaps up and make a box like so We're gonna make an open box by cutting a square from each corner of a 12 inch by 12 inch piece of metal and then folding up the sides What side square should we cut off for each corner to produce a box of maximum volume? So let's focus on that for a moment. We want to maximize volume It's gonna be a box when we're nice and done. So that is we get this rectangular prism and it doesn't have a top So this is open right here But what's gonna be the what's gonna be the dimensions of this thing the volume will equal length times width times height Now when we fold the thing up Let's say that we cut off X from each corner so an X by X corner when you fold this thing upwards then The height of this box is gonna be the dimension X right there. What about the length and width? Well, the base square It's gonna be this right here And if the whole thing was 12 inches if we cut off X and we cut off X that'll leave behind 12 minus 2x And that's gonna give us the length of this thing and it is a square So it's gonna be the same thing right here as well 12 minus 2x for the other side and so then our volume will look like 12 minus 2x 12 minus 2x And X Which 12 minus 2x it does have a factor of 2 you can factor that out. So you get 4 times 6 minus x squared times x. This is our optimizing function Where was the constraint Well the constraint actually came into the Into the the the metal square itself. There's only 12 inches by 12 inches That's how the constraint came into place. We actually had three variables in the original situation One of them was X the height was just X. So maybe you think we had three variables We had length and width which we are able to remove and replace in terms of X there So I'm since this thing is a square I'm gonna leave it factored as I take the derivative or use the product rule this time. We're gonna get four times two times six minus X times negative one don't forget the inner derivative times X Plus Well, the derivative X is just one so we get six minus X squared here Let's see. So there is a common factor of six minus X here and here We're gonna factor that out. So we get four times six minus X And then what's left behind is a negative 2x plus six minus X like so combining some like terms We end up with six minus three X and six minus three has a common factor of three again So factor that out. So we're gonna get 12 times six minus X and Then we get two minus X. So we're gonna find two critical numbers X equals six and X equals two So consider The domain of this situation. I'm gonna come back to the original picture So as we fold these things up clearly the smallest that X could be is zero But how big could it be like if we put all of it into height? What would happen? It would be like we took this square and cut it up into four squares That's gonna be the limit of how big X can be and this would be six and six and six and six everywhere So X equals six is gonna be the biggest we can get here All right, let's scroll back down again. Oh Six was a critical number. How convenient? So as we build our t-chart We're gonna consider X as it ranges from zero to six and we have a two in the middle Two seems very likely to be our maximum volume if X equals zero Then the volume is zero because it has no height if X equals six the volume will also be zero Because it has no length or width two seems to be very very helpful right here if the height is two Then we take 12 minus 2 that is 12 minus 4 Which will be 8 Which would be the length and the width and so you're going to get 8 square which is 64 double that you get 128 This will be 128 cubic Inches isn't it? So And that's kind of our maximum maximum volume We want to cut off a 2 inch square from each corner fold it with our metal folding machine And then we're going to this maximum volume of 128 cubic centimeters so The final answer is to cut a two by two cut two by two corners and This will give us the maximum volume All right, and so I guess it's a few more examples of optimization problems for today. I hope this was fun Optimization never really goes away. These are some of the most important types of Applications of the derivative one can see optimization problems are everywhere certainly everywhere We're always trying to do the best with the resources we got and calculus becomes a critical tool in helping one with those type of problems If you liked the video today, feel free to leave a comment if you have any questions also feel free to comment and ask your questions I'll be glad to answer them as soon as those appear We will talk about Newton's method next time In the meanwhile, feel free to subscribe to get more updates about these videos in the future and I will see you then have a good day everyone