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1. Define the variables and constraints:<\/strong><\/p>
Let \( x \) be the length of the side perpendicular to the partition and \( y \) be the length of the side parallel to the partition. The total boundary material is 300 meters, so \( 3x + 2y = 300 \).<\/p>
2. Express \( y \) in terms of \( x \):<\/strong><\/p>
From the constraint, \( 2y = 300 - 3x \), so \( y = 150 - \frac{3}{2}x \).<\/p>
3. Define the area function:<\/strong><\/p>
The area \( A \) of the rectangular installation is \( A = xy \).<\/p>
Substitute \( y \) in terms of \( x \): \( A(x) = x(150 - \frac{3}{2}x) = 150x - \frac{3}{2}x^2 \).<\/p>
4. Find the first derivative of \( A(x) \):<\/strong><\/p>
\( A'(x) = 150 - 3x \).<\/p>
5. Set the first derivative to zero and solve for \( x \):<\/strong><\/p>
\( 150 - 3x = 0 \), so \( 3x = 150 \), and \( x = 50 \).<\/p>
6. Verify that this is a maximum using the first derivative test:<\/strong><\/p>
For \( x < 50 \), \( A'(x) > 0 \) (e.g., \( A'(40) = 150 - 3(40) = 30 > 0 \)).<\/p>
For \( x > 50 \), \( A'(x) < 0 \) (e.g., \( A'(60) = 150 - 3(60) = -30 < 0 \)).<\/p>
Since the derivative changes from positive to negative at \( x = 50 \), this is a maximum.<\/p>
7. Find the corresponding value of \( y \):<\/strong><\/p>
\( y = 150 - \frac{3}{2}(50) = 150 - 75 = 75 \).<\/p>
8. Calculate the maximum area:<\/strong><\/p>
\( A = xy = 50 \times 75 = 3750 \) square meters.<\/p>
Correct Answer: 3750<\/strong>