Rectangles beneath a semicircle A rectangle is constructed wit...

Question

Rectangles beneath a semicircle A rectangle is constructed with its base on the diameter of a semicircle with radius 5 and its two other vertices on the semicircle. What are the dimensions of the rectangle with maximum area?

Accepted Answer

Welcome back, everyone. In this problem, a landscape architect is designing a water feature in the shape of a semicircle with a radius of 4 m. A rectangular pathway is to be constructed with its base along the diameter of the semicircle and its other two vertices touching the semicircle. Determine the dimensions of the path that maximize its area where L is length and the W is width. Now, to help us make sense of the problem, let's try to put our our design, the landscape architects design on a coordinate grid, OK? So I did a quick sketch here that shows our semi-circle in a coordinate plane with its center at the origin 00 and the rectangle inside where the ends of the rectangle touches the semicircle, and these are for our maximum uh lengths and widths. Now here we know that for our semi-circle. At each point, OK, on the X and Y axis, so it's gonna go 04 on our Y axis and on our X-axis, negative was 0 to the left and 40 to the right because it has a radius of 4 m. Now, we also know here. That for its base, its length is going to be 2 X, X and X. And its width is going to be Y. And these are, remember, no, these are the lengths that we don't know yet, OK? Where here we know that Y is going to be greater than or equal to 0. Now, in this case, If we are trying to maximize its area, that means we'll need to solve for the maximum error. But what do we know? Well, we know that the area of a rectangle equals the length multiplied by the width, so this is going to be equal to 2 X Y. And now since our semicircle is on the coordinate plane, we can write an equation for the semicircle. Which is X2 + y2 equals our radius 42. Again, where Y is greater than or equal to 0. Now, how are we going to maximize the area given this information? Well, there are a few things that we need to do. Let's come up with a plan, OK. Now, the first thing we need to do is that at some point because we're maximizing, we may need to differentiate. So let's express our area in terms of one variable. Let's choose to express it in X. So let's first try to express Y, OK, in terms of X. Because then we'll be able to substitute it into our equation for area. After we do that, then we can maximize our area A, OK, with respect to X. To solve for the maximum value of X. Let's call that value X max. And then once we do that, now we have that maximum value of X. Let's clean this up here. When we have that maximum value, then we can solve for the maximum value of Y, OK? Again, we've confirmed it's the maximum value of X, and then we can solve for the maximum value of Y to then get the maximum area. OK? So let's break that down one by one. First, let's go ahead and express Y, OK, in terms of X. No solving for why. OK, I think for Y, we can use the equation for a semicircle because then we can say that Y2 equals 4216 minus X2, which means that Y is going to be equal to the square root of 16 minus X2. OK. So now we've expressed Y in terms of X. Let's substitute that into our equation for Y, OK? So no, sorry, into our equation for A. So now, this would imply that A, in terms of X, A of X is going to be equal to the square root, sorry, 2 X. Multiplied by y, the square root of 16 minus X squared. And no, we can differentiate A with respect to X to get the first derivative because whenever the first derivative equals 0, that's the point where our critical points will be, OK. Now notice here that A of X is a product. So to differentiate it, we'll need the product to. Recall that the product's rule of differentiation tells us that if we have a term Y that's equal to UV where U and V are functions of X, then the derivative of Y is going to be equal to U V plus UV. And know Y is arbitrary here. It's not the Y that we were referring to earlier, OK? But no, in this case, for A of X, X represents Y, we can let U be equal to 2 X. And we can let V be equal to the square of 16 minus X2 or 16 minus X2 to the power of a half. When we do that, then we should get u the derivative of u to be 2, and the derivative of V to be equal to 1/2 of 16 minus X2 to the 4 of 1/2 multiplied by -2 X, which when we simplify that, we can get it as negative X multiplied by 16 minus X2 to the core of negative or half. Now that we have that, then if we apply, OK, apply our product rule here to differentiate AF X. Then the derivative of AF X. OK. Let me take this here. The derivative of AF X. Is going to be equal to. U V so that's 2 multiplied by 16 minus X squared to the 1/2. Plus UV. So that's 2 X multiplied by X multiplied by 16 minus X squad to the negative by half. And now here since we know that at the critical points. At the critical points. Then our first derivative will be equal to 0. This would imply. OK. Now we can set 2 multiplied by 16 minus X squad to the half, OK. Plus, or sorry, minus. OK if we expand in our 2nd term. Minus 2 X2, OK, multiplied by 16 minus X2 to the negative half to be equal to 0, OK? Now, here, at this point, let's try to combine our terms here. We can make this one expression. Now, when we do that, OK, then we should find that we get it to be equal to -4, OK, multiplied by X2 minus 8. Divided By 16 minus X2 to the half or root 16 minus X2. OK? So all of this is equal to 0. Now, in that case, OK, then that means if we multiply through our expression by negative uh square root of 16 minus X2, OK. Then, and also divide through by 4, then we should get. Let me put this over on the on the left. We should get X2 minus 8 to be equal to 0, which makes X2 equal to 8 and X equals to root 8, or to root 2, OK? So, no, this is the maximum value of X at the critical point. Now that we've solved for X max, we're not finished yet, actually. What we need to do is to verify that this is the maximum point. So we need to test if this value of X to 2 is a is a maximum, OK? And we can test if it's a maximum by trying to understand the behavior of the function around the critical point. Now, since our critical point is 2 root 2, and we know our boundaries are 0 and 4, this is our interval. Then if we take the bounder from 0. Uh, to, to 2 or our critical point, if the first derivative is positive, that means that it's, it's, it, if it changes, if the sign of the first derivative changes from positive to negative, the function starts to decrease, and that means it is a maximum point. Now, if we think about the behavior of our first derivative here in this interval, let's take X2 2. Then our function is gonna be 4 multiplied by 22 minus 8 divided by root 16 minus 22. Which equals a root 3 divided by 3, and that's positive. Next, for interval from 2 route 2. To 4 We can take X to be 3. And in that case, our function is 4 multiplied by 32 minus 8. Divided by the square of 16 minus 3 squad, that would give us negative, oops, this would have been negative initially. My apologies. That's gonna be -4 root 7 divided by 7, and that is less than 0, OK? So in this case. The derivative changes sign from positive to negative. That means the function starts to decrease past the critical point. Thus, X equals 2 root 2 m is a maximum. OK, because our function is decreasing past the critical point. Now that we have our value for X, that means we can solve for Y, OK? Because no, this means then. Y equals a square root of 16 minus 2 root 2 squad. Which is route 16 minus 8, which is route 8 or to route 2 m. Now that we have X and Y, then we can solve for the maximum length and width because now, remember that the maximum length is 2 X. So that's 2 multiplied by 2 root 2 or 4 root 2 m. And Y is our width, sorry, is equal to Y, OK? So that means our width is 22 m. Therefore, OK, this means that A is a maximum. It is a maximum at the length being equal to 42 m. And the the width being equal to 2 2 m. Thanks a lot for watching everyone. I hope this video helped.

Rectangles beneath a semicircle A rectangle is constructed with its base on the diameter of a semicircle with radius 5 and its two other vertices on the semicircle. What are the dimensions of the rectangle with maximum area?

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