A rectangle is inscribed in a semicircle of radius 2 . See...

A rectangle is inscribed in a semicircle of radius $2 .$ See the figure. Let $P=(x, y)$ be the point in quadrant I that is a vertex of the rectangle and is on the circle. (a) Express the area $A$ of the rectangle as a function of $x$ (b) Express the perimeter $p$ of the rectangle as a function of $x$ (c) Graph $A=A(x)$. For what value of $x$ is $A$ largest? (d) Graph $p=p(x) .$ For what value of $x$ is $p$ largest?

A rectangle is inscribed in a semicircle of radius $2 .$ See the figure. Let $P=(x, y)$ be the point in quadrant I that is a vertex of the rectangle and is on the circle.
(a) Express the area $A$ of the rectangle as a function of $x$
(b) Express the perimeter $p$ of the rectangle as a function of $x$
(c) Graph $A=A(x)$. For what value of $x$ is $A$ largest?
(d) Graph $p=p(x) .$ For what value of $x$ is $p$ largest?

Key Concepts

Graphing Functions

Graphing functions is the process of plotting the relationship between variables to visualize how changes in one affect the other. By graphing the area and perimeter functions derived from the geometry of the problem, one can visually identify the location of maximum values and better understand the function's behavior over the allowed domain.

Function Modeling

Function modeling involves expressing one or more quantities, such as area or perimeter, in terms of a single variable. By relating the coordinates via the circle's equation, one can derive formulas for the area and perimeter of the rectangle solely in terms of a chosen variable, facilitating further analysis and graphing.

Optimization Techniques in Geometry

Optimization in geometry involves finding maximum or minimum values of a geometric quantity under given constraints. In this problem, determining the value of x that maximizes area or perimeter requires understanding the behavior of the derived functions and often employs methods from calculus, such as differentiation, to locate these extreme points.

Inscribed Figures

This concept involves placing one figure, such as a rectangle, entirely within another figure, like a semicircle, so that all vertices of the inscribed figure lie on the boundary of the larger figure. It is widely used in problems to relate dimensions of one shape to another, enabling the derivation of expressions for properties like area or perimeter.

Circle Geometry

Circle geometry deals with the properties and equations of circles and their segments. In this context, the semicircle of a fixed radius provides a constraint through its equation, and the relation between the x and y coordinates of any point on the circular arc is essential to express other quantities in terms of a single variable.

Transcript

00:01 All right, guys.

00:01 So part a wants us to express the area as a function of x.

00:06 And for this problem, we have a rectangle that's inscribed in a semicircle, and the radius is equal to 2.

00:14 And we have a figure that we can look at, and it tells us that the semicircle is labeled as y equals the square root of 4 minus x square.

00:24 So we know that just finding the area of a rectangle, we use the formula area is equal to the base times the height.

00:38 So if we want to figure out our base, first off since we're finding the area as a function of x, we'll be using a of x, which is just the area of x.

00:55 Our figure, if you look at the figure under the problem, we can see that the base is the length from the origin to x.

01:04 So our base is just basically x, but since we have it on both sides of our origin, we know that they are equal size, so our base will be 2x.

01:18 And our height is represented by the type of our semicircle.

01:24 So we're going to use that formula from, or the equation from y equals so 2x times the square root of 4 minus x squared gives us our function now part b asks us to express the perimeter as a function of x so we were finding the perimeter of a regular rectangle we know we'll say that the area is twice the base plus twice the height so i'm going to use the same information that i use in part a.

02:06 And i'm sorry that it should be the parameter, not the area.

02:13 So since we're finding the parameter as a function of x, it'll be p of x is equal to.

02:21 We already know that our base is 2x.

02:23 So 2 times 2x plus we know our height is a square root of 4 minus x squared.

02:31 So we'll be multiplying 2 times that height 4 minus x square.

02:40 And we're just going to simplify this.

02:42 2 times 2x gives us 4x plus 2 times the square root of 4 minus x squared...