Show that among all rectangles with area $A,$ the square hasthe minimum perimeter.

Key Concepts: Rectangle, Square, Perimeter Explanation: This problem can be solved using optimization techniques. We need to minimize the perimeter of the rectangle while keeping its area constant. Among the rectangles with the same area, the square has the minimum perimeter. Step by Step Solution: Step 1. Let the length of the rectangle be $l$ and the width be $w$. Then, the area of the rectangle is $A=lw$. Step 2. The perimeter of the rectangle is $P=2l+2w$. Step 3. We can use the area formula to express one of the dimensions in terms of the other. Solving for $l$, we get $l=\frac{A}{w}$. Substituting this into the perimeter formula, we get $P=2\left(\frac{A}{w}\right)+2w$. Step 4. To minimize $P$, we take the derivative of $P$ with respect to $w$, and set it equal to 0. Solving for $w$, we get $w=\sqrt{A}$. This means that for a given area $A$, the width and length of the rectangle that minimize its perimeter are equal, i.e., the rectangle is actually a square. Step 5. Substituting $w=\sqrt{A}$ into the expression for $P$, we get $P=4\sqrt{A}$. So, the square has the minimum perimeter of all rectangles with area $A$. Final Answer: The square has the minimum perimeter among all rectangles with area $A$.