Question

The smallest perimeter that a rectangle with area of 4 square units can have is ............... units. (Answer in integer)

Hint:

For a fixed area, the rectangle with the minimum perimeter is always a square. For a fixed perimeter, the rectangle with the maximum area is always a square. This is a very useful principle in optimization problems.

Answer 1

Step 1: Understanding the Concept:
This is a classic optimization problem. We want to minimize a function (perimeter) subject to a constraint (area). We can use calculus or the AM-GM inequality to solve this.

Step 2: Key Formula or Approach:
Let the sides of the rectangle be \(l\) and \(w\).
- Area (Constraint): \(A = l \times w = 4\)
- Perimeter (Function to minimize): \(P = 2(l + w)\)

Method 1: Using Calculus
1. From the constraint, express one variable in terms of the other: \(w = 4/l\).
2. Substitute this into the perimeter equation to get a function of one variable: \(P(l) = 2\left(l + \dfrac{4}{l}\right)\).
3. Find the derivative of \(P(l)\) with respect to \(l\) and set it to zero to find the critical points.
\[\frac{dP}{dl} = 2\left(1 - \frac{4}{l^2}\right)\]
4. Set \(\frac{dP}{dl} = 0\):
\[1 - \frac{4}{l^2} = 0 \implies l^2 = 4 \implies l = 2\] (Since length must be positive).
5. Find the corresponding width: \(w = 4/l = 4/2 = 2\).
6. The shape that minimizes the perimeter is a square with side length 2.
7. Calculate the minimum perimeter: \(P = 2(l+w) = 2(2+2) = 8\).
(To confirm it's a minimum, check the second derivative: \(\dfrac{d^2P}{dl^2} = 2\left(\dfrac{8}{l^3}\right)\), which is positive for \(l > 0\), so it is a minimum).

Method 2: Using AM-GM Inequality
The Arithmetic Mean–Geometric Mean (AM-GM) inequality states that for non-negative numbers \(a\) and \(b\), \(\dfrac{a+b}{2} \ge \sqrt{ab}\), with equality holding if and only if \(a=b\).
1. Let the sides be \(l\) and \(w\). We want to minimize \(P = 2(l+w)\), which is equivalent to minimizing \((l+w)\).
2. Apply AM-GM to \(l\) and \(w\):
\[\frac{l+w}{2} \ge \sqrt{lw}\]
3. We know the area is \(lw = 4\), so \(\sqrt{lw} = \sqrt{4} = 2\).
\[\frac{l+w}{2} \ge 2 \implies l+w \ge 4\]
4. The minimum value of \((l+w)\) is 4.
5. The minimum perimeter is \(P = 2(l+w) = 2(4) = 8\).
6. This minimum occurs when equality holds in the AM-GM inequality, which is when \(l=w\). This corresponds to a square.

Step 3: Final Answer:
The smallest perimeter is 8 units.

Step 4: Why This is Correct:
Both calculus and the AM-GM inequality show that for a fixed area, the rectangle with the minimum perimeter is a square. For an area of 4, the square must have sides of length 2, leading to a perimeter of 8.