To find the dimensions (length $l$ and width $w$) of a rectangle with a given area of 49 square units and a minimum perimeter, we can use optimization principles.

Step-by-Step Solution:

1. Area Equation: The area $A$ of the rectangle is given as: $A = l \times w = 49$ Hence, we know: l \times w = 49 \tag{1}

2. Perimeter Equation: The perimeter $P$ of a rectangle is: $P = 2(l + w)$ We want to minimize this perimeter.

3. Express one variable in terms of the other: From equation (1), express $w$ in terms of $l$: $w = \frac{49}{l}$

4. Substitute into the perimeter equation: Substitute $w = \frac{49}{l}$ into the perimeter equation: $P = 2 \left( l + \frac{49}{l} \right)$

5. Minimize the perimeter: To find the minimum, take the derivative of $P$ with respect to $l$, and set it equal to zero: $\frac{dP}{dl} = 2 \left( 1 - \frac{49}{l^2} \right) = 0$ Solve for $l$: $1 = \frac{49}{l^2}$ $l^2 = 49 \quad \Rightarrow \quad l = 7$ Thus, $l = 7$.

6. Find the width: Using $l = 7$ in equation (1): $w = \frac{49}{7} = 7$

Conclusion:

The rectangle with area 49 and minimum perimeter is a square with both length and width equal to 7.

Perimeter Check:

The perimeter is: $P = 2(7 + 7) = 28$

Would you like more details on optimization or any other related concepts?

Related Questions:

1. What is the perimeter of the rectangle if the length is 10 and the width is found using the area?
2. How does the perimeter change if the length is fixed but the width varies?
3. Can you explain why the square gives the minimum perimeter for a given area?
4. What would happen if we doubled the area of the rectangle? Would the optimal shape still be a square?
5. How can we calculate the diagonal of this rectangle?

Tip: A square always gives the minimum perimeter for a given area among all rectangles.