A rectangle is constructed with its base on the diameter of a semicircle with radius 5 and its t

step 1 So in this problem, we're doing an optimization of area problem. We have a semicircle. The semic

A rectangle is constructed with its base on the diameter of...

Key Concepts

Constrained Optimization

This concept involves finding the maximum or minimum value of a function subject to certain restrictions or constraints. In problems like this, where the dimensions of a geometric figure must satisfy a specific relationship (here given by the equation of the semicircle), constraints are used to express one variable in terms of others so that the optimization process can proceed.

Differentiation for Finding Critical Points

In calculus, the process of taking derivatives allows one to identify points where a function’s rate of change is zero. These critical points are candidates for local maxima or minima. By setting the derivative of the area function equal to zero and solving, one can determine the dimensions at which the area is maximized.

Geometric Constraints in Inscribed Figures

When a shape is inscribed in another, its dimensions are governed by the boundary of the enclosing shape. The semicircular boundary imposes a relationship between the coordinates of the rectangle’s vertices, effectively linking its width and height. This geometric constraint is critical for setting up the optimization problem.

Symmetry in Geometric Optimization

Many optimization problems, especially those involving inscribed figures, exhibit symmetry which can simplify the problem. In this context, the symmetry of the semicircle about its central axis suggests that the optimal rectangle will be symmetrically placed, reducing the number of unique variables and making the analysis more straightforward.

Transcript

00:01 So in this problem, we're doing an optimization of area problem.

00:04 We have a semicircle.

00:08 The semicircle has a radius of five units.

00:12 We have a rectangle that sits on that base.

00:18 Now, clearly, the height would be the y.

00:24 So in this case right here, we'd have a y value.

00:28 Over here, we'd have x comma positive y, x, comma, negative y, would be the coordinates here.

00:35 And, you know, so what we're looking at here is we're looking at maximum area.

00:40 So we know that area will equal x times y.

00:44 But again, we've got too many variables.

00:46 So one of the things that we know is that x squared plus y squared is equal to 25 fourths.

01:01 And the reason i got 25 fourths, i'm basically setting the rectangle at the origin here, the radius would be half of five and when i square it.

01:11 So this is kind of my formula here.

01:14 Now, when i solve this equation for y, i'm going to have y squared is equal to 25 over 4 minus x squared.

01:25 And under normal circumstances, we do plus and minus.

01:27 But since we're only looking at the top half of the circle, we're going to work with the positive square root.

01:36 And so i'm going to do a little substitution here.

01:39 So i've got area is equal to x.

01:42 Times 25 over 4 minus x squared all to the one half power.

01:49 And this will allow us to take the derivative a little bit more, a little bit easier.

01:53 Okay, so in order to maximize our dimensions here to get the biggest area of the rectangle, let's take the derivative.

02:01 So the derivative a is a prime.

02:03 This is the product.

02:05 So it's the first times the derivative of the second.

02:19 And then we've got negative to x plus 25 over 4 minus x squared all to the one half because the and because the derivative of x is just one.

02:39 Excuse my writing right here.

02:42 We're just going to make clean that up and make it look a little bit nicer.

02:44 It'd be easier to follow here...

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