Question
maximum area of rectangle with integer sides and perimeter 20
Answers
GPT-4o mini
To find the maximum area of a rectangle with integer sides and a perimeter of 20, we can start by recalling the formula for the perimeter of a rectangle:
\[
P = 2(l + w)
\]
where \( l \) is the length and \( w \) is the width. Given that the perimeter \( P = 20 \), we can rewrite the equation as:
\[
l + w = 10
\]
Next, the area \( A \) of the rectangle can be expressed as:
\[
A = l \times w
\]
We can substitute \( w \) with \( 10 - l \) (from the perimeter equation):
\[
A = l(10 - l) = 10l - l^2
\]
This represents a quadratic equation in terms of \( l \). The quadratic function \( A = -l^2 + 10l \) opens downwards (since the coefficient of \( l^2 \) is negative), and its maximum value can be found at the vertex. The vertex of a quadratic function \( ax^2 + bx + c \) occurs at:
\[
l = -\frac{b}{2a}
\]
In our case, \( a = -1 \) and \( b = 10 \):
\[
l = -\frac{10}{2(-1)} = 5
\]
Now, substituting \( l = 5 \) back to find \( w \):
\[
w = 10 - l = 10 - 5 = 5
\]
Thus, the rectangle is a square with sides \( 5 \) and \( 5 \).
Finally, the maximum area is:
\[
A = l \times w = 5 \times 5 = 25
\]
Therefore, the maximum area of a rectangle with integer sides and a perimeter of 20 is \( \boxed{25} \).
\[
P = 2(l + w)
\]
where \( l \) is the length and \( w \) is the width. Given that the perimeter \( P = 20 \), we can rewrite the equation as:
\[
l + w = 10
\]
Next, the area \( A \) of the rectangle can be expressed as:
\[
A = l \times w
\]
We can substitute \( w \) with \( 10 - l \) (from the perimeter equation):
\[
A = l(10 - l) = 10l - l^2
\]
This represents a quadratic equation in terms of \( l \). The quadratic function \( A = -l^2 + 10l \) opens downwards (since the coefficient of \( l^2 \) is negative), and its maximum value can be found at the vertex. The vertex of a quadratic function \( ax^2 + bx + c \) occurs at:
\[
l = -\frac{b}{2a}
\]
In our case, \( a = -1 \) and \( b = 10 \):
\[
l = -\frac{10}{2(-1)} = 5
\]
Now, substituting \( l = 5 \) back to find \( w \):
\[
w = 10 - l = 10 - 5 = 5
\]
Thus, the rectangle is a square with sides \( 5 \) and \( 5 \).
Finally, the maximum area is:
\[
A = l \times w = 5 \times 5 = 25
\]
Therefore, the maximum area of a rectangle with integer sides and a perimeter of 20 is \( \boxed{25} \).