Find the dimensions of a rectangle with a fixed area of 50 square units that min

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Question: Find the dimensions of a rectangle with a fixed area of 50 square units that minimizes the perimeter. (2022) 2022

Options:

Correct Answer: 7, 7.14

Exam Year: 2022

Solution:

For minimum perimeter, the rectangle should be a square. Thus, side = sqrt(50) ≈ 7.07.

Find the dimensions of a rectangle with a fixed area of 50 square units that minimizes the perimeter. (2022) 2022

Find the dimensions of a rectangle with a fixed area of 50 square units that minimizes the perimeter. (2022) 2022

Step 1: Understand that we need to find the dimensions of a rectangle with an area of 50 square units.
Step 2: Recall that the area of a rectangle is calculated by multiplying its length (L) by its width (W): Area = L * W.
Step 3: Set up the equation for the area: L * W = 50.
Step 4: To minimize the perimeter, remember that the perimeter (P) of a rectangle is calculated as: P = 2L + 2W.
Step 5: Use the area equation to express W in terms of L: W = 50 / L.
Step 6: Substitute W in the perimeter equation: P = 2L + 2(50 / L).
Step 7: Simplify the perimeter equation: P = 2L + 100 / L.
Step 8: To find the minimum perimeter, take the derivative of P with respect to L and set it to zero: dP/dL = 2 - 100 / L^2 = 0.
Step 9: Solve for L: 2 = 100 / L^2, which gives L^2 = 50, so L = sqrt(50).
Step 10: Since W = 50 / L, substitute L back to find W: W = 50 / sqrt(50) = sqrt(50).
Step 11: Conclude that both dimensions (length and width) are equal, making the rectangle a square with side length of sqrt(50).
Step 12: Calculate the approximate value of sqrt(50): sqrt(50) ≈ 7.07.

Optimization – The problem involves finding the dimensions of a rectangle that minimize the perimeter while maintaining a fixed area.
Geometric Properties – Understanding the relationship between the area and perimeter of rectangles and squares.
Calculus (if applicable) – Potential use of derivatives to find minimum values, although not explicitly required in this solution.