How to find the maximum area of a rectangle?
Understand the Problem
The question is asking for a method or approach to determine the maximum area of a rectangle, likely involving formulas and possible constraints on dimensions.
Answer
The maximum area of the rectangle occurs when $l = w = \frac{P}{4}$.
Answer for screen readers
The maximum area of the rectangle is achieved when both dimensions are equal to $\frac{P}{4}$.
Steps to Solve
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Define the dimensions of the rectangle
Let the length of the rectangle be denoted as $l$ and the width as $w$. The area $A$ of the rectangle can be expressed as:
$$ A = l \cdot w $$ -
Understand constraints
Identify any constraints that might involve the perimeter. For example, if a maximum perimeter $P$ is given, you can express it as:
$$ P = 2l + 2w $$
or simplified to:
$$ w = \frac{P}{2} - l $$ -
Substitute the constraints into the area formula
If you substitute the expression for $w$ into the area equation, you can express the area in terms of a single variable ($l$).
This leads to:
$$ A = l \left(\frac{P}{2} - l\right) $$
$$ A = \frac{Pl}{2} - l^2 $$ -
Find the maximum area using calculus
To find the maximum area, you'll want to take the derivative of the area function $A$ with respect to $l$ and set it to zero to find critical points.
$$ \frac{dA}{dl} = \frac{P}{2} - 2l = 0 $$ -
Solve for the length
Setting the derivative equal to zero gives:
$$ \frac{P}{2} = 2l $$
Thus,
$$ l = \frac{P}{4} $$ -
Determine the width
Using the equation for width based on the earlier constraint:
$$ w = \frac{P}{2} - l $$
Substituting $l = \frac{P}{4}$ into this gives:
$$ w = \frac{P}{2} - \frac{P}{4} = \frac{P}{4} $$ -
Conclusion
The rectangle with maximum area, given a maximum perimeter $P$, is a square where both the length and width equal $l = w = \frac{P}{4}$.
The maximum area of the rectangle is achieved when both dimensions are equal to $\frac{P}{4}$.
More Information
This means that to achieve maximum area with a given perimeter, the rectangle must be a square. This is a well-known property in geometry and optimization.
Tips
- Confusing the maximum area with maximum perimeter; remember that for a fixed perimeter, the maximum area is obtained when the rectangle becomes a square.
- Not simplifying the equations properly after substitutions, which can lead to errors in finding critical points.
