 Hello everyone! We're going to take a look at a very typical first example in optimization in calculus. Here we have an open box that is to be made from a rectangular piece of sheet metal that measures 20 inches by 32 inches by cutting squares of equal size from each corner and folding up the sides. We are asked to determine the maximum volume that can be achieved. I have here for you an animation that was created by our NCSSM creative staff several years ago that illustrates exactly what we're talking about in this problem. Imagine each of these are the pieces of sheet metal that we're talking about in our problem from which we want to create a box by cutting out squares from the corners. And the size of that box is going to be dictated by the size of the square that we cut out. You can see in the blue one on the far left, we cut out very teeny tiny squares. The green one in the middle, the squares are a little bit larger and on the yellow one on the far right, they're a little bit larger yet. We then fold up the sides so as to create a box and you can see this blue one is pretty shallow because sides that we fold up are very short because of the small size of the square that we cut out. In the green one, it's a little bit deeper, but not too deep. Finally, the yellow one is the deepest of all of them and you can see the difference in the size of the box that results. And that's what we're talking about in this problem, is how do we maximize the volume of the resulting box by varying the size of the square that we cut out? If we were to draw for ourselves a picture of what this might look like in two dimensions, we would need to indicate that we would be cutting out from the corner those squares of equal size. Those we will denote by the little dotted lines that you see and whereas we don't know how big of a square we are to cut out, we will refer to the length of the side of that square as x. We know that to maximize volume, we need to be maximizing the volume of a box which is length times width times height. Now the length, once we fold up these corners, is given by this distance right here. Remember that the length of the entire piece of sheet metal was 32 inches, so 32 inches for the entire length take away the x's that are on either side. We have for the length 32 minus 2x. The width of the piece of sheet metal was 20 inches, so if this entire side on the right and the left was 20 inches, once again take away the x's for the corners that we're cutting out. This distance right here becomes 20 minus 2x and that's our width. The height of the box comes from turning up the remains of the sheet metal after the corners are cut out. So the height really is from the length of the side of the square that we cut out. So we have now a volume formula that we can maximize. It is in terms of one single variable, so that makes it a little bit easier. We know we're going to have to find the derivative of this, set it equal to 0, so as to find our critical numbers that will hopefully lead to a relative maximum. It would be helpful though to find the derivative of this if we had it as a polynomial. If you were to multiply this out, you get 4x cubed minus a hundred and four x squared plus 640 x and that's what we can take the derivative of to proceed from here. Our derivative then, by simple application of the power rule, is 12x squared minus 208x plus 640. We need to set that equal to 0 and solve it, so as to find our critical numbers. We can factor out a 4 as the greatest common factor. The quadratic trinomial then factors and we arrive at two values for x. We have 40 over 3 and 4. These are our critical numbers at which if there is going to be a relative maximum, this is where they would occur. Let's talk a moment for the domain of this function, of the original function. We need to keep in mind the restrictions imposed by the size of the sheet metal itself. If x is representing the length of the side of the square that we're cutting out to create our box, obviously it cannot be 0 because then you're not cutting anything out. At the same time, it really can't be any larger than 10 either. If we go back to our diagram, remember that the width measured 20 inches, so you cannot cut out a square that's really 10 or bigger because then this side does not exist. So you could cut out a square that's maybe nine and a half inches, that results in a width here that's really, really tiny, but it is feasible. But you definitely cannot do anything that's going to be 10 inches itself that you cut out or larger. If you keep in mind what the domain is, you quickly realize that the 40 over 3 is not possible in light of this domain. As we proceed to hopefully show that it is at x equals 4 that our relative maximum occurs, we're going to work through it using both the first derivative test and the number line analysis, as well as the second derivative test. The extreme value theorem is not an option because that domain was an open interval as opposed to a closed interval. Let's start with the first derivative test. On our number line, we're going to have the critical value of 4 that we found. Recall that we are substituting this 4 into the first derivative, I've written it down at the top of our page for us. A zero results when we substitute 4 into the derivative. We know that we cannot go any lower than zero, nor can we go any higher than 10. If we choose something a little less than 4 to substitute into our derivative, such as 3, we do arrive at a positive answer. Substituting something larger than 4, but less than 10, we arrive at a negative answer. Because the derivative is changing positive to negative, that tells us our original function for the volume would change from increasing to decreasing, therefore creating a relative maximum at x equals 4. If we wanted to apply the second derivative test, we're going to need to find the second derivative. Remember that into the second derivative we substitute our critical number. When we do that, we come out with a negative 112, which obviously is less than zero. Remember what that tells us. Since the value of the second derivative at x equals 4 is negative, that tells us that the original curve would be concave down, therefore creating a relative maximum. Now that we have proven we have a relative maximum at x equals 4, we can substitute 4 into our original function to arrive at the maximum volume that we are seeking. Recall the cubic polynomial. That was our volume formula. If we substitute 4 into there, get 1152. This is a volume measurement, so remember the units of measure would be cubic units. Therefore our answer would be 1152 cubic inches, or of course you can do inches cubed.