Find the length and width of a rectangle that has the given area and a minimum perimeter.
Area: 50 square feet
Well, if we want to find the rectangle with the minimum perimeter and an area of 50 square feet, we need to think outside the box - or rather, inside the rectangle.
Let's consider different length and width combinations and see which one gives us the minimum perimeter.
For example, if the length is 10 feet and the width is 5 feet, we get an area of 50 square feet. The perimeter would then be 30 feet.
But hold on, let's try another combination. If we have a length of 25 feet and a width of 2 feet, we still get an area of 50 square feet. And guess what? The perimeter is now 54 feet.
So, it looks like we have a winner! The rectangle with a length of 25 feet and a width of 2 feet not only gives us our desired area of 50 square feet but also has the minimum perimeter of 54 feet.
But hey, if you're feeling adventurous, you can always play around with the dimensions and see if you can come up with something even more unique!
To find the length and width that will result in a rectangle with the given area of 50 square feet and a minimum perimeter, we can start by listing all possible length and width combinations.
1 x 50 = 50
2 x 25 = 50
5 x 10 = 50
10 x 5 = 50
25 x 2 = 50
50 x 1 = 50
Next, calculate the perimeter for each combination:
1 x 50: Perimeter = 2(1 + 50) = 2(51) = 102 feet
2 x 25: Perimeter = 2(2 + 25) = 2(27) = 54 feet
5 x 10: Perimeter = 2(5 + 10) = 2(15) = 30 feet
10 x 5: Perimeter = 2(10 + 5) = 2(15) = 30 feet
25 x 2: Perimeter = 2(25 + 2) = 2(27) = 54 feet
50 x 1: Perimeter = 2(50 + 1) = 2(51) = 102 feet
From the calculations, we can see that the combinations with the minimum perimeter are 5 x 10 and 10 x 5, both resulting in a perimeter of 30 feet.
Therefore, the length and width of the rectangle with the given area of 50 square feet and a minimum perimeter are 5 feet and 10 feet, or vice versa.
To find the length and width of a rectangle with a given area and a minimum perimeter, we need to first understand the relationship between the area and perimeter of a rectangle.
The formula for the area of a rectangle is:
Area = length × width
The formula for the perimeter of a rectangle is:
Perimeter = 2(length + width)
Given the area of 50 square feet, we can use this information to solve for the length and width.
Let's start by finding the possible pairs of length and width that would satisfy an area of 50 square feet. We can do this by factoring 50 into pairs of numbers:
Factors of 50: 1, 2, 5, 10, 25, 50
So the possible pairs of length and width for an area of 50 are:
(1, 50), (2, 25), (5, 10), and (10, 5)
Now we need to calculate the perimeter for each of these pairs and find the pair with the minimum perimeter.
For the pair (1, 50):
Perimeter = 2(1 + 50) = 2(51) = 102
For the pair (2, 25):
Perimeter = 2(2 + 25) = 2(27) = 54
For the pair (5, 10):
Perimeter = 2(5 + 10) = 2(15) = 30
For the pair (10, 5):
Perimeter = 2(10 + 5) = 2(15) = 30
The pairs (5, 10) and (10, 5) have the same minimum perimeter of 30.
Therefore, the length and width of the rectangle with an area of 50 square feet and a minimum perimeter are 5 feet and 10 feet (or vice versa).
p = 2w + 2 L
L = 50/w
so
p = 2 w + 2 (50/w)
p = 2 ( w+50/w)
dp/dw = 0 at min or max
= 2 [ 1 - 50/w^2 ]
w^2 = 50
w = 5 sqrt 2
then
L = 50/[5 sqrt 2 ] = 10 srt 2/2
= 5 sqrt 2
LOL, it is square :)