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To solve this problem, we need to find the maximum area of the shaded region, which is the area of the rectangle formed with one corner at the origin and the opposite corner on the line \( P(x, y) \).

### Step 1: Determine the equation of the line.
The line passes through the points \((0, 11)\) and \((8, 0)\). The slope \( m \) of the line can be calculated as:

\[
m = \frac{0 - 11}{8 - 0} = -\frac{11}{8}
\]

Thus, the equation of the line in slope-intercept form is:

\[
y = -\frac{11}{8}x + 11
\]

### Step 2: Express the area of the rectangle.
The area \( A \) of the rectangle is given by:

\[
A = x \times y
\]

Substituting the expression for \( y \) from the line equation:

\[
A = x \times \left(-\frac{11}{8}x + 11\

Perhatikan gambar berikut: Luas maksimum daerah yang diarsir adalah ....

Learn how to find the maximum area of a shaded region formed by a rectangle with one vertex at the origin and another on a line. This problem involves linear equations and optimization. Suitable for Grades 10-12.

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