A piece of sheet metal, w = 16 inches wide, is bent to form the gutter shown in the illustration. If the cross-sectional area is 24 square inches, find the depth of...

Solve each problem. A piece of sheet metal, 18 inches wide, is bent to form the gutter shown in the illustration. If the cross-sectional area is 36 square inches, find the depth of the gutter. (GRAPH CANT COPY)

All right. So the question that we're looking at today has an open rectangular gutter is made by turning up two sides of a piece of metal 20 inches wide, and the area of the cross section of the gutter is 48 cubic inches. We need to find the possible depths of the gutter. So the first step that we're going to dio is we're gonna just right out our variable. So we're gonna let d equal the, um, depth of the got there, and we're going to let, um, w represent no w represent the with. So, um were given that a cross section area is 48 cubic inches or square and just squared so we can write that other formula as 48 is equal to the so our area essentially is equal to our depth times our wits, our depth, times air with, um, the cross section, however, consists of only three sides. Um, which is so it's essentially depth plus depth, pulse width, and some of these dimensions is given to us as 20 inches, so we can write two d plus w is equal to 20. Um, and that, um and we already know that our depth times are with is gonna equal to 48. So we need to rewrite the equation. Um, our first equation, this one right here as W is equal to 20 minus two D. That's what our new equation is gonna look like. And we substitute that into our other equation. So now it is D Times 20 minus two D. Because that's what I W now equals is equal to 48 and we simplify and solve for D the depth. So, um, once we do that, we get it. So it it is 20. D minus to D squared is equal to 48. Um, and we need to rewrite this as this quadratic equation in standard form. So we do this by doing to d squared minus 20 d plus 48. Oh, and that is equal to zero. And we need to solve this by factory. So in this case, we can divide by two on this side. So divide this by to divide this by two. So now we're gonna be left with something that's a bit more manageable. We're going to get d squared minus 10 D plus 24 is equal to zero. And now we need to find two numbers that multiply too equal. Positive 24. But also add Teoh equal, Negative 10. And those numbers are gonna be negative. Six and negative for so make it of six and negative four. So for integral. So now our equation is gonna look like this, It's gonna be D Oh, de, um, minus sex and the minus four. Um, so now we apply the zero product rule, which means that we dio d is equal to six or D is equal to four. Um, we take a look back at the question. We can see that, um, we need to find the possible depths of the gutter. So in this case, the gutter can either be, um six inches or four inches cause either are applicable in this situation, So it's gonna write those of therefore statement. Therefore, the together can be six inches or four inches, and in conclusion, that's the answer to this problem.

Were asked to to find where we can create a rectangular cross section of less than 20 or lessons 30 square inches. And so we're making this cross section by by bending a sheet of metal. And the the amount we bend is X because that's that's our That's our with. And the length of the length of the cross section is 20 minus two x, and so we know that we know that the area is able to length times width. So if we have our our length or are with to be X and if we have are left to be 20 minus two x, our area is the product of the Jew, and so we want to see at what values of of eggs. How much X is is how much we're fully, how much of the metal sheet we decide to fold. How much of it can we fold while still making the rectangular cross section less than or equal to 30? And so the way we do that is we, we say, we say that that negative two X squared plus 20 X has to be less than or eat this do 30 in other words. In other words, negative two X squared plus 20 x minus 30 has to be less than or equal to zero. So all we write that here. So we have. We have negative two X. We have negative two X squared plus plus 20 x plus 20 x minus 30. Minus 30 has to be less than or equal to zero. And so we can. We can solve this by by using the quadratic equation. So we have. We have negative bees or negative 20 out of 20 plus plus or minus. The square root of B squared is 400 is 400 minus four times negative to which is negative. Eight. So negative eight times negative 30 which is 2 40 So 400 minus 2 40 400 minus 2 40 over two times negative two, Which is negative for and so we can solve for the roots here. So we have we have negative 20 plus or minus. The square root of 400 400 minus 2 40 and 400 minus 2 40 is 1 60 So the square root of 1 60 is about 12.6, so we have negative 20 we have negative 20 plus 12.6 negative 20 and of 20 plus 12.6 divided by Mega four. And so that gives us. That gives us an X value of 1.8 inches. And we can also have an X value if we if we subtract. So let's say let's say we have we have negative 20 minus minus 12.6 time or the square root of 1 60 and so and so that gives us so that gives us 8.2. That gives us eight 0.2. And so remember we want this to be less than or equal to zero. And so if we have, if we have a function, we have a function for you have a quadratic. And if we know that our our roots are are positive, you know are richer, positive. And if we know that our A value is is less than zero and are quadratic will look something like this. If a is last stand zero with positive roots. And so don't redraw this and so are quadratic will look something like that if a is greater than zero and so we want to know we want to know Where is this value? Where are these? Where is this quadratic? Less than zero. And so if we know if we know that this is 1.8, this is eight to then then we should know that we want X to be We want X to be less than remember. We have less than or equal to have less than or equal to 1.8 inches. And we have to have this be greater than zero. Otherwise, we don't have a a cross sectional area. So we want we want X to be greater than zero, but less than 1.8 and then we want. And then we want. And we want X to be at least greater than greater than 8.2. But again, we don't want X to be 20. Because if X is 20 than our 20 wait a minute. If it's if it is 20 minutes to X and we don't want it to be 10. All right? So we don't want it to be We don't want it to be. I don't want to 10 because then we have 20 minus two times 10. That would give a zero. So So 10 and zero don't work because that would give us zero cross sectional area. And and we want we want at least at least some cross sectional area are we wouldn't have. We wouldn't have a cross sectional area also. So 10 and zero don't make sense of this problem in the real world. So we want some some cross sectional area. But we wanted to be who wanted to be less than or equal to 30. So this this these are the parameters we would need to satisfy to satisfy Yes, you already.

All right. We got a little drying here. We got our ring gutter. If we have a 12 inch wide piece of aluminum that then gets folded up so that this piece is this piece. This piece is that one then this ain't is action that one's 12 minus Tuapse. And so part a asshole. What depth will provide the maximum cross sectional area. So what can we do in order to maximize this area and thereby allow the most water to flow? All we multiply X. Times 12 minus two X. That's our equation. And if I want to grasp that then I can get some numbers both minus two X. I'm sorry. Pre set my windows so that it fit in there. You probably would have to reset your um window in order to get this to fit. So now what does this tell me? Well, the maximum would be here. So the maximum depth uh we should say is three three is the Mac. Mhm. 313 inches. It's a MAC. Well, come on three inches. Well, maximize the cross sectional area. No, in part B, it says, well, what that will allow at least 16 square inches of water. So if we say y all 16, then that tells us the point. So in part B, um we need signs that are between two and four inches. So that's that one.

I guess a part a just wants an area. So an area in terms of the amount of metal that's been upward, it's just gonna be acts the height of the rectangle, Times 20 minus two acts, which is the base came. Since we're in a quadratic section, we can write that as negative to act square closed 20 acts. Yeah, if we want to maximize the cross sectional area, we need to find this Vertex because this this area will look like it will be a problem opening downward not to find the Vertex. The amount that needs to be folded up is it acts. And so we're gonna do that with minus B over to a really negative 20 divided by negative for so many to fold up five inches from either end. Okay? And then to actually find the maximum area, we display five inches back into our original equation. So we're gonna have five times 20 minus 10 which is 10. So the cross sectional area will be 50 square inches. Then that would be for party. Okay. Thank you.