Rectangles beneath a parabola A rectangle is constructed with ...

Question

Rectangles beneath a parabola A rectangle is constructed with its base on the x-axis and two of its vertices on the parabola y = 48 - x². What are the dimensions of the rectangle with the maximum area? What is the area?

Accepted Answer

Welcome back, everyone. In this problem, a rectangle has its base on the x-axis and two vertices on the curve Y E equals 100 minus 4 X squared. Determine the length L and width W that yielded the maximum area, and then find the maximum area. Now, for us to solve here, let's try to visualize what's going on. OK. So, we're saying that we have an axis, OK. And on our axis, we have a rectangle that has its base, OK? And we have a rectangle here that has its base. With a curve on which we want to find where the length and width of our rectangle yield the maximum error. So if we imagine this to be a curve, OK. Uh, saying that they are touching the vertices of our rectangular touching the curve, OK. And we also know that our curve, the equation of our curve is Y equals 100 minus 4 X2, then our X intercept is going to be 0 100 and our Y intercept is gonna go from -50 to 50. OK. So let me. Let me put that properly here. -50 to 504 or curve. And we would find that if we were to substitute for our value of uh for X intercept, OK? So that these are where the zeros of our curve would be. We also know that the rectangle is symmetric about the Y axis. So if one vertex is at XY, the other is at negative XY, OK? So in other words, here, we can Call both of these lengths X at the base and then the height of a rectangle is Y, where Y is going to be greater than or equal to 0. No, we are trying to find a maximum area, OK? We're trying to find a maximum area for a rectangle. So we'll need the coordinates for X and Y to help us find the length and width for this maximum area. Now, how can we do that? OK. Well, We know that the area is gonna be equal to the length and the width. And we also know that Y is related to X by the formula Y equals 100 minus 4 X2d because this is the equation for the curve on which both of these vertices lie. So in that case then. Here, if we look at our rectangle, OK, our area, we can tell is gonna be equal to 2 XY. So that means then that our area as a function of X. It's gonna be equal to 2 X multiplied by Y, so that's 2 X multiplied by 100. Minus 4 X squared, OK? Which is gonna be equal to, if we, if we expand here, 200 X minus 8 X cubed. Now why is this important? Why would we want to express our area in terms of X? Now that we have it in terms of one variable, that means we'll be able to differentiate and solve for the critical points of a of X at which our maximum may lie. So let's try to find the critical points by differentiating A X and setting it to zero and then test whether that point is a maximum. So differentiating. FX, OK. Differentiating F X. OK. Then it's derivative. It's gonna be equal to 200 minus 24 X X squad, sorry, by the poor rule of differentiation. No At the critical points, we know that at the critical points. The the first derivative is equal to 0. So that implies then that 200 minus 24 excubed. X squared, sorry, is gonna be equal to 0. That means 24 X squared. Equals 200. Which means X squad then is gonna be equal to 200 divided by 24, which is gonna be 25/3. Therefore, if I make some space, well, I mean, I'm done a bit. Therefore, this means that X is gonna be equal to plus or minus the square root of 25 divided by 3, OK, which is gonna be, or that's gonna leave us with X equals plus R minus 5 divided by root 3. And no, because we're talking about lengths, we'll only consider the positive route. So that's gonna be 5 divided by route 3, or if we rationalize it, 5 route 3 divided by 3, OK? So this is going to be the X value for one of the critical points of, of our, of our function, OK? But now we need to test if this critical point is a maximum, OK? So how can we prove that? How can we prove that X equals 5 root 3 divided by 3 is a maximum? Well, let's try to understand the nature of our function around the point. That is before and after the critical point. Now before the critical point. OK, that is where X is between 0 and 5 root 3 divided by 3, because remember, 00 is the least possible value of X here, OK. Then if we pick a point for our first derivative, say where X equals 2 and substitute it into uh substitute 2 into our equation for the first derivative. Then A2 is gonna be 200 minus 24 multiplied by 4, OK. Oh sorry, multiplied by 2 squad, which is gonna be 200 minus 96, which equals 104. Now, after a critical point, that is from 5 root 3 divided by 3, all the way up to 5, if we choose another point, say when X equals 4. Then it's gonna become 200 minus 24 multiplied by 4 squared. Which is going to be 200 minus 384 or -184. So this is what happens when we substitute our values of X into the first derivative. And now what do you notice? Well, no, notice that the derivative sign changes from positive to negative. So that means the function decreases, the function decreases. OK, the function decreases past X equals 5 root 3 divided by 3. Thus, that indicates it is a maximum. And now that we know our function is a maximum, we can calculate our dimensions using X and Y for our uh from our formula where area equals to XY. OK? So now, if that's the case. That means the maximum length, let's put this in red. The maximum length is going to be equal to. Well, let's start with our width. The maximum width is gonna be equal to 2 X. So that's 2 multiplied by 5 route 3 divided by 3, or 10 route 3 divided by 3 units, OK? The maximum length, which is Y, is gonna be equal to 100 minus 4 X2. So that's 100 minus 4 multiplied by 5 root 3 divided by 3 squared, OK, which is gonna be equal to 100 minus 4 multiplied by 25 divided by 3, which eventually equals 200 divided by 3 units, OK? I know that we have our width and our length, that means our maximum area then. It's gonna be equal to 2003 units. Multiplied by. 10 route 3 divided by 3 units, OK, which would be equal to 2000 route 3 divided by 9 square units. Thus, to summarize, that means that for or rectangle that. To yield a maximum area then. The lengths would have to be 10 routes 3 divided by 3 units. The length would be 200 divided by 3 units, and this maximum area would be 2000 route 3 divided by 9 square units. Thanks a lot for watching everyone. I hope this video helped.

Rectangles beneath a parabola A rectangle is constructed with its base on the x-axis and two of its vertices on the parabola y = 48 - x². What are the dimensions of the rectangle with the maximum area? What is the area?

Key Concepts

Parabola

Optimization

Area of a Rectangle

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