What is the maximum area of the rectangle with perimeter 620 mm?

Correct option is A

​$$\text{Let the length be} \ l \ \text{and the width be} \ w. \\\text{The perimeter is given by:} \\2(l + w) = 620 \quad \Rightarrow \quad l + w = 310 \\\text{Step 2: Express area in one variable} \\\text{Area} \ A \ \text{of the rectangle is:} \\A = l \cdot w \\\text{From} \ l + w = 310, \ \text{we can write} \ w = 310 - l. \ \text{Substituting into the area formula:} \\A = l(310 - l) = 310l - l^2 \\\text{Step 3: Maximize the area} \\\text{The expression} \ A = -l^2 + 310l \ \text{is a quadratic in} \ l, \ \text{and it opens downward (since the coefficient of} \ l^2 \ \text{is negative), so the maximum occurs at the vertex:} \\l = \frac{-b}{2a} = \frac{-310}{2(-1)} = 155 \\\text{So,} \ l = 155, \ \text{and} \ w = 310 - 155 = 155 \\\text{Step 4: Calculate maximum area} \\A = 155 \times 155 = 24025 \, \text{mm}^2$$​​