Find the dimensions of a rectangle with a fixed area of 50 m^2 that minimizes th

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Question: Find the dimensions of a rectangle with a fixed area of 50 m^2 that minimizes the perimeter. (2021)

Options:

Correct Answer: 7, 7.14

Exam Year: 2021

Solution:

For a fixed area, the minimum perimeter occurs when the rectangle is a square. Thus, dimensions are approximately 7 m by 7.14 m.

Find the dimensions of a rectangle with a fixed area of 50 m^2 that minimizes the perimeter. (2021)

Find the dimensions of a rectangle with a fixed area of 50 m^2 that minimizes the perimeter. (2021)

Step 1: Understand that we need to find the dimensions of a rectangle with an area of 50 m².
Step 2: Recall the formula for the area of a rectangle, which is Area = length × width.
Step 3: Set up the equation: length × width = 50.
Step 4: To minimize the perimeter, remember that the perimeter of a rectangle is given by the formula Perimeter = 2 × (length + width).
Step 5: To minimize the perimeter for a fixed area, we can use the fact that a square has the smallest perimeter for a given area.
Step 6: Since the area is 50 m², we can find the side length of a square with this area by taking the square root of 50.
Step 7: Calculate the square root of 50, which is approximately 7.07 m. This means each side of the square is about 7.07 m.
Step 8: Since we are looking for a rectangle, we can also express the dimensions as length = 7.07 m and width = 7.07 m, which are equal.
Step 9: Therefore, the dimensions that minimize the perimeter while keeping the area fixed at 50 m² are approximately 7.07 m by 7.07 m.

Optimization – The problem involves finding the dimensions of a rectangle that minimize the perimeter while maintaining a fixed area.
Area and Perimeter Relationships – Understanding the relationship between the area of a rectangle and its perimeter is crucial for solving the problem.
Geometric Properties of Rectangles – Recognizing that a square has the smallest perimeter for a given area among all rectangles.