 In this video we provide the solution to question number 18 for the practice final exam for math 1210, in which case we have a box with a square base and no top that needs to be constructed from 75 square feet of material. We wanna find the largest possible volume of the box. Largest possible volume, this is an optimization type problem. So our optimizing function should be volume. Well, volume of a box, a straw box real quick for us, volume of a box, right, is gonna be length times width times height. Length times width times height, which we could call this length, this width in this dimension right here, height. Okay? So we wanna find the largest possible volume of this box. Well, how does the constraint come into play here? The constraint has to do with this 75 square feet of material. Now, some other things to mention is that, well, one, it says it has no top. So it turns out this box only has five sides. We can ignore the price of the top. Also it says it has a square base. So that means the base of this thing right here is a square. So we actually get that length and width are the same thing. So we can revise our volume formula and get that W squared H is the volume. So can we somehow relate height in terms of the width? Well, this is where the material for the sides is gonna come to play here. We can make a constraint using the surface area of this box. So the surface area is gonna be computed as, well, the side of all the, it's gonna be the area of all the side rectangles put together. So you have the base, which is a square. So that's gonna be a W squared. But then each of the other four sides, right, like this one right here, it's a rectangle, which is gonna be W times H. And there's gonna be four of those things all around the whole thing. So we get four times WH. That's gonna be the surface area. But the surface area, we know what it is by the constraint, the surface area we see is supposed to be 75, like so. So we can use this to solve for H. Subtract W squared from both sides. We get 4WH is equal to 75 minus W squared. Divide both sides by 4W. We get H equals 75 minus W squared over 4W. And we're gonna make this substitution in 4H, like so. So that volume becomes W squared times this 75 minus W squared over four times W. You'll notice that one of the W's cancels out right here, like so, in which case then our volume becomes W times, well W over four times 75 minus W squared. So now we have to make a choice. When it comes to the derivative, we could multiply this thing out to calculate the derivative, which would make it a little bit easier, or we could take the derivative right now using the product rule. I'm gonna distribute the W through. You can leave the constant multiple out. That doesn't have much of a consequence on the derivative. So we get 1 4th 75 W minus W cubed. So when we calculate the derivative here, V prime, we're gonna get 1 4th times 75 minus 3W squared. We need to find the critical number, so we're gonna set this thing equal to zero. The 1 4th can be ignored because we can just divide both sides by 1 4th, dividing zero by any non-zero number just gives you zero, no big deal there. So we have 75 minus 3W squared. Add 3W squared to both sides. You get 75 equals 3W squared, like so. Divide both sides by three. You're gonna get W squared equals 25, and then take the square where you get W equals five. So this gives us a critical number, but we have to consider all, not just the critical numbers, but the boundary points as well. So what could the boundaries be, right? For W, well, to one extreme, you could set W equal to zero, but that's gonna be a no-go here because if your width was zero, that would give you a volume, I should mention, of course, of zero. There's also the concern, of course, that the surface area wouldn't be 75 square feet at that moment, but the point is that's gonna minimize things. If we go on the other extreme, like how big can W actually come, turn out to be, that kind of follows from this formula right here, that if you allow W to get all the way up to like the square root of three times five, five times the square root of three, that is the square root of 75, then that would set the height equal to zero, which also would make the volume zero, which again, that's an impossibility because that type of thing wouldn't have any material made out of it, but we should make sure that it's not a boundary point that's included because the boundary could in fact be the solution to the problem. We can't just simply ignore it. So it seems like the candidate for optimization is gonna happen here at five. Five of course is the width. So if we put it into our function, we're gonna end up with one fifth, excuse me, one fourth times five times 75 minus 25, which is five squared. So we wanna simplify this thing of course. So we get one fifth times five. 75 takeaway 25 is going to be 50, like so. And so that's gonna give you 250 over four. Four does go into that thing evenly. And so we end up with, and take that back, two goes into 250 evenly. So you'd end up with 125, excuse me, 125 over two if you want the exact answer or we could write that as 62.5. This would be volume, so the units here would be cubic feet. And so the maximum volume for this box given the constraint on the surface area is gonna be 62.5 cubic feet.