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In the following exercises, solve the given maximum and minimum problems.An open box is to be made from a square piece of cardboard whose sides are 8.00 in. long, by cutting equal squares from the corners and bending up the sides. Determine the side of the square that is to be cut out so that the volume of the box may be a maximum. See Fig. 24.65.
Calculus 1 / AB
Chapter 24
Applications of the Derivative
Section 7
Applied Maximum and Minimum Problems
Derivatives
Campbell University
Harvey Mudd College
University of Michigan - Ann Arbor
Boston College
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the problem state it is a square of side ex inches is cut out of each corner of a tenant by 18. Inch is a cardboard, and the signs are folded up to form an open top box. We first need to write the volume. The of the box is a function of X. Find the domain of our function and user graphing calculator to determine document the dimensions of the cutout squares that will produce the box of maximum volume. Okay, so first, uh, well, just go ahead and drawn Illustration of kind of what's going on here. So we have There's our slab of cardboard. It is 18 inches by 10 and inches 18 inches by 10 inches. Okay. And then what we do, we cut out a square from each corner. Like so, where you just signed the square has yeah, length of X, Then after giving. So we're going to fold up the sides to form sort of a, uh, hoping top box like this and we need to write the volume of the box is a function of X. No, also by the domain of that function. And use our graphing calculator to determine the maximum volume of the box sand. Hence the, um, sidelines of X that we need to achieve that maximum volume. So that's what we are going to do right now. So we know from this diagram that I've drawn that the high hear of our box. This is going to be equal X once we pull these up. Okay, so what about this, uh, shortest like here? We'll call that Thea the Wind. What's that going to be able to? Well, this shorter side here originally at the length of 10 inches. Okay, but what's the link? Once we cut out the squares of latex, that's that's going to be 10 inches minus two X. Right? You see, we've cut out two squares of size necks from that side, so we're just going to have to subtract off to Max so that with is going to be cool too. 10 minus two x. And what about the length? Okay. Using similar logic here we have a sideline of 18 inches and be kind off two squares of size X from that side, so you know, 18 inches minus Q X. Therefore, we can write our going, but justice equal to our brain. Times air with tons of height. This volume is just equal to thanks. Times 10 minus two X, sometimes 18 minus two x. So that is the answer to part. Okay, from this question. Now, won't that be found? Uh, volume over box. As a function of X, we now need to determine the domain book. This function. Oh, we can determine that just by looking at the box. Actually, so we have the shorter side sizzling 10 inches, and we have the longer side. Isley, 18 inches. Okay, So what's the least come now? We can cut off. Well, if we cut off? Uh, no. Right then, we wouldn't be able to fool our piece of paper up or a piece of cardboard to form his box. So we know that it just X must have to be some value greater than zero and able to fold it up into a box. So what's the maximum value the X could take on? Well, let's look on the shorter side here. Yeah. Two x can only we have 10 inches to work with here on the shorter side. So we know that certainly, um, two x can't be greater than 10 inches. Um, and also similarly, if we're we're exactly equal to 10 inches. We would have an infinitely narrow box with zero volume, so we can't have that either. So we say that two X has to be Weston 10 inches or X is less than five. So our Darlene is X equals 0 to 5. Non inclusive. Now that we found the domain of the function as well as the function itself, the volume in terms both the length of the site x to pick out of the box, we need to find the, uh, length of the side at that maximum value the box. So when we pop this function, which we determine that party onto our graphing calculator, we get something like this. It is a cubic function, goes up and has a max right here between zero and five, which is an earned a mean So that's good. And it's back down now, using our graphing calculator. This next is that V is equal toe 168 went 1 to 6, right? That's why access is being this X, and this is at the next value. Oh, 2.63 Therefore, to get our maximum volume, you must have aside Li. We cut out great to be out X is equal to two points. 06 three.
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