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? (e) What is the largest area? What is the largest perimeter?

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?
(e) What is the largest area? What is the largest perimeter?

Key Concepts

Constrained Optimization

Constrained optimization deals with finding the optimal values (maxima or minima) of functions subject to specific restrictions. These constraints often come from geometric relationships or other conditions, such as a point lying on a circle, which must be taken into account when formulating and solving the optimization problem.

Graphical Analysis

Graphical analysis involves plotting functions to visually inspect their behavior, particularly to understand the trend, locate extrema, and verify analytical findings. This method provides an intuitive perspective on how functions, such as those representing area or perimeter in relation to a variable, vary over their domain.

Optimization in Calculus

Optimization involves finding the maximum or minimum values of a function under given constraints. In calculus, this is typically achieved by differentiating the function with respect to the variable of interest, setting the derivative equal to zero, and testing the critical points to determine whether they correspond to maximum or minimum values.

Function Representation in Geometry

In many geometric problems, quantities like area or perimeter are expressed as functions of a variable to capture the dependence of these quantities on dimensions that vary within given constraints. This allows one to rewrite geometric relationships, such as those from the equation of a circle, in terms of a single parameter, facilitating analysis and optimization.

Inscribed Figures

An inscribed figure is a shape that is contained entirely within another shape, touching its boundary in specific points. This concept is critical when the geometry of one figure, such as a rectangle, is directly influenced by the curve or boundary of another, such as a semicircle. It establishes the relationships between the dimensions of the inner figure and the dimensions or equations of the outer figure.

Transcript

00:01 So in this problem, we're given that a rectangle is inscribed inside of a semicircle.

00:06 A semicircle is of radius 2 as drawn on the page.

00:11 And the point or the corners of the rectangle touch or lie on the semicircle.

00:20 So the point is xy where the corner is there.

00:24 And the first part, we're asked to express the area a as a function of x.

00:32 Well, the area of a rectangle is the length times the width, which the length here then is this length right here, which as we go over x, we go over minus x at the same time.

00:49 So the length is 2x times the width here is this distance, which is just the y, which we're given right here, as a square root of 4 minus x.

01:06 Squared.

01:09 So there's our function of x for the area.

01:16 Next on part b we're asked to write the perimeter p as a function of x.

01:22 Well, the perimeter is two times the length plus two times the width.

01:29 We already know the length is 2x, so that's 2 times 2x, plus the width, plus the width is y.

01:38 That's two times the square root of 4 minus x squared so this is 4x plus 2 times the square root of 4 minus x squared then part c we're asked to graph these graph the area and figure out the value of x that gives us the largest area so we go to our graphing module and we put in 4x times 4 minus squared.

02:22 And i want to take the square root of that...

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