Question
What is the smallest number of centimeters in the perimeter of a rectangle with an area of 24 square centimeters?
Answers
Answered by
GPT-4o mini
To find the smallest perimeter of a rectangle given an area of 24 square centimeters, we can use the properties of rectangles and the relationship between area, perimeter, and dimensions.
Let the length and width of the rectangle be \( l \) and \( w \), respectively. The area \( A \) of the rectangle is given by:
\[
A = l \times w = 24
\]
The perimeter \( P \) of the rectangle is given by:
\[
P = 2l + 2w = 2(l + w)
\]
To find the smallest perimeter for a fixed area, we can express \( l \) in terms of \( w \):
\[
l = \frac{24}{w}
\]
Substituting this into the perimeter formula:
\[
P = 2\left(\frac{24}{w} + w\right) = 48\cdot\frac{1}{w} + 2w
\]
Now we will minimize the function \( P(w) = 48\cdot\frac{1}{w} + 2w \). To find the critical points, we take the derivative and set it to zero.
\[
P'(w) = -\frac{48}{w^2} + 2
\]
Setting \( P'(w) = 0 \):
\[
-\frac{48}{w^2} + 2 = 0
\]
This simplifies to:
\[
\frac{48}{w^2} = 2
\]
Multiplying both sides by \( w^2 \):
\[
48 = 2w^2
\]
Dividing both sides by 2 gives:
\[
w^2 = 24 \implies w = \sqrt{24} = 2\sqrt{6}
\]
Now substituting \( w \) back to find \( l \):
\[
l = \frac{24}{w} = \frac{24}{2\sqrt{6}} = \frac{12}{\sqrt{6}} = 2\sqrt{6}
\]
Thus, both the length and the width are equal:
\[
l = w = 2\sqrt{6}
\]
Since the rectangle is actually a square (where length equals width), we find the minimum perimeter:
\[
P = 2(l + w) = 2(2\sqrt{6} + 2\sqrt{6}) = 4\cdot2\sqrt{6} = 8\sqrt{6}
\]
Now to compute \( 8\sqrt{6} \) to find the perimeter in centimeters:
Calculating \( \sqrt{6} \) approximately:
\[
\sqrt{6} \approx 2.44949 \implies 8 \sqrt{6} \approx 8 \times 2.44949 \approx 19.59592
\]
To find the smallest integer greater than this (since we want the smallest number of complete centimeters):
\[
\lceil 19.59592 \rceil = 20
\]
Thus, the smallest integer perimeter (in whole centimeters) of the rectangle with an area of 24 square centimeters is:
\[
\boxed{20}
\]
Let the length and width of the rectangle be \( l \) and \( w \), respectively. The area \( A \) of the rectangle is given by:
\[
A = l \times w = 24
\]
The perimeter \( P \) of the rectangle is given by:
\[
P = 2l + 2w = 2(l + w)
\]
To find the smallest perimeter for a fixed area, we can express \( l \) in terms of \( w \):
\[
l = \frac{24}{w}
\]
Substituting this into the perimeter formula:
\[
P = 2\left(\frac{24}{w} + w\right) = 48\cdot\frac{1}{w} + 2w
\]
Now we will minimize the function \( P(w) = 48\cdot\frac{1}{w} + 2w \). To find the critical points, we take the derivative and set it to zero.
\[
P'(w) = -\frac{48}{w^2} + 2
\]
Setting \( P'(w) = 0 \):
\[
-\frac{48}{w^2} + 2 = 0
\]
This simplifies to:
\[
\frac{48}{w^2} = 2
\]
Multiplying both sides by \( w^2 \):
\[
48 = 2w^2
\]
Dividing both sides by 2 gives:
\[
w^2 = 24 \implies w = \sqrt{24} = 2\sqrt{6}
\]
Now substituting \( w \) back to find \( l \):
\[
l = \frac{24}{w} = \frac{24}{2\sqrt{6}} = \frac{12}{\sqrt{6}} = 2\sqrt{6}
\]
Thus, both the length and the width are equal:
\[
l = w = 2\sqrt{6}
\]
Since the rectangle is actually a square (where length equals width), we find the minimum perimeter:
\[
P = 2(l + w) = 2(2\sqrt{6} + 2\sqrt{6}) = 4\cdot2\sqrt{6} = 8\sqrt{6}
\]
Now to compute \( 8\sqrt{6} \) to find the perimeter in centimeters:
Calculating \( \sqrt{6} \) approximately:
\[
\sqrt{6} \approx 2.44949 \implies 8 \sqrt{6} \approx 8 \times 2.44949 \approx 19.59592
\]
To find the smallest integer greater than this (since we want the smallest number of complete centimeters):
\[
\lceil 19.59592 \rceil = 20
\]
Thus, the smallest integer perimeter (in whole centimeters) of the rectangle with an area of 24 square centimeters is:
\[
\boxed{20}
\]
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