A rectangle is to be inscribed in a semicircle of radius 2. What is the largest area the rectangle can have, and what are its dimensions?
Understand the Problem
The question is asking to find the largest area that a rectangle, inscribed in a semicircle of radius 2, can achieve, along with its dimensions. This involves optimization and geometric considerations.
Answer
The largest area is \( A = 4 \) with dimensions \( w = 2\sqrt{2} \) and \( h = \sqrt{2} \).
Answer for screen readers
The largest area that the rectangle can achieve is ( A = 4 ) square units, with dimensions ( w = 2\sqrt{2} ) and ( h = \sqrt{2} ).
Steps to Solve
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Determine the relationship between the dimensions of the rectangle and the semicircle
Let the rectangle have a width $w$ and a height $h$. The semicircle's equation is $x^2 + y^2 = r^2$ with $r = 2$. The upper part of the semicircle can be described as $y = \sqrt{4 - x^2}$. The width of the rectangle will be $2x$ (since it extends equally in both directions from the center), and the height will be $h = \sqrt{4 - x^2}$.
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Express the area of the rectangle
The area $A$ of the rectangle can be expressed as: $$ A = w \cdot h = 2x \cdot \sqrt{4 - x^2} $$
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Differentiate the area to find critical points
First, rewrite the area: $$ A = 2x\sqrt{4 - x^2} $$
To differentiate, use the product rule: $$ \frac{dA}{dx} = 2 \sqrt{4 - x^2} + 2x \cdot \frac{d}{dx}(\sqrt{4 - x^2}) $$
The derivative of $\sqrt{4 - x^2}$ is: $$ \frac{d}{dx}(\sqrt{4 - x^2}) = \frac{-x}{\sqrt{4 - x^2}} $$
Thus, $$ \frac{dA}{dx} = 2\sqrt{4 - x^2} - \frac{2x^2}{\sqrt{4 - x^2}} $$
Set $\frac{dA}{dx} = 0$ to find critical points.
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Solve the equation for $x$
Multiply both sides by $\sqrt{4 - x^2}$: $$ 2(4 - x^2) - 2x^2 = 0 $$ Simplifying gives: $$ 8 - 4x^2 = 0 \implies x^2 = 2 \implies x = \sqrt{2} $$
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Calculate the dimensions and area
Using $x = \sqrt{2}$, find $h$: $$ h = \sqrt{4 - (\sqrt{2})^2} = \sqrt{4 - 2} = \sqrt{2} $$
Now, find the width of the rectangle: $$ w = 2x = 2\sqrt{2} $$
Therefore, the maximum area is: $$ A = 2x \cdot h = 2\sqrt{2} \cdot \sqrt{2} = 4 $$
The largest area that the rectangle can achieve is ( A = 4 ) square units, with dimensions ( w = 2\sqrt{2} ) and ( h = \sqrt{2} ).
More Information
The problem illustrates how optimization techniques can be used to find the maximum area of a geometric shape inscribed within another shape. In this case, the largest rectangle inscribed in a semicircle leads to a consistent approach for maximizing area given certain constraints.
Tips
- Forgetting to apply the product rule correctly when differentiating.
- Not squaring the terms correctly when substituting back into the area formula.
- Confusing the geometric dimensions and their relationships, such as mixing up width and height.