How to find the maximum area of a rectangle?

Understand the Problem

The question is asking for methods or approaches to determine the maximum area of a rectangle, likely involving the use of mathematical principles such as optimization or geometry.

Answer

The rectangle with maximum area for a fixed perimeter is a square with side length $\frac{P}{4}$.

Steps to Solve

Define the dimensions of the rectangle

Let the length of the rectangle be $L$ and the width be $W$. The area $A$ of the rectangle can be expressed as: $$ A = L \times W $$

Express one dimension in terms of the other

If we have a constraint for the perimeter $P$, such as $P = 2L + 2W$, we can express $W$ in terms of $L$: $$ W = \frac{P}{2} - L $$

Substitute to find the area in one variable

Now substitute for $W$ in the area formula: $$ A = L \left(\frac{P}{2} - L\right) $$ This expands to: $$ A = \frac{P}{2}L - L^2 $$

Find the maximum area using calculus

To maximize the area, take the derivative of $A$ with respect to $L$ and set it to zero: $$ \frac{dA}{dL} = \frac{P}{2} - 2L = 0 $$

Solve for L

Solving the equation gives: $$ 2L = \frac{P}{2} $$ $$ L = \frac{P}{4} $$

Find the corresponding W

Substituting $L$ back in to find $W$: $$ W = \frac{P}{2} - \frac{P}{4} = \frac{P}{4} $$

Conclusion

Thus, the length and width that maximize the area, given a fixed perimeter, are both equal. Hence, the rectangle is actually a square.

The maximum area of a rectangle with a fixed perimeter occurs when the rectangle is a square with side length given by $L = W = \frac{P}{4}$.

More Information

This principle of maximizing rectangle area is related to optimization in calculus. It demonstrates how geometric constraints impact geometry and can lead to symmetry in shapes.

Tips

Confusing the dimensions of the rectangle when applying constraints.
Forgetting to set the derivative to zero when looking for critical points.
Not checking for maximum/minimum points after obtaining the critical values.

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