Problem on a rectangle of maximal area
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Of all the rectangles of perimeter P, which has the largest area?
Solution
We observe that this is a constrained optimization problem: we seek to maximize the area of a rectangle with the constraint that its perimeter is $P$.
We introduce the variables $x$ and $y$ as the width and height of the rectangle.
Now we express the minimization problem in those variables.
The area of the rectangle is $xy$ and the perimeter is $2x + 2y$.
Thus, we are trying to maximize $xy$ subject to the constraint $2x + 2y = P$.
Recall that
Lagrange multipliers
Lagrange multipliers are a convenient tool to solve constrained minimization problems.
Lagrange multipliers
To use Lagrange multipliers to solve the problem $$\min f(x,y) \text{ subject to } g(x,y) = 0,$$
Form the augmented function $$L(x,y,\lambda) = f(x,y) - \lambda g(x,y)$$
Set all partial derivatives of $L$ equal to zero
Solve for $x,y$.
We identify $f(x,y) = xy$ and $g(x,y) = 2x + 2y - P$.
Our augmented function is $L(x,y,\lambda ) = xy - \lambda(2x + 2y - P).$
Partial derivatives
Setting the derivatives of $L$ to zero, we get:
\begin{align}
\partial_x L = y - 2 \lambda &=0 \\
\partial_y L = x - 2 \lambda &=0 \\
\partial_\lambda L = -(2x + 2y - P) &= 0
\end{align}
The first two equations give $x = y = 2 \lambda$. The last equation gives that $x=y=P/4$.
Because $x=y$, we have shown that the rectangle of maximal area and perimeter $P$ is a square.
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