2012 P1 Linear Program: on the xoy plane region R is unshaded

The unshaded region R is a polygon bounded by three lines:

• The y-axis.
• A line passing through points $(0, 10)$ and $(7, 4)$.
• A line passing through points $(0, 0)$ and $(7, 4)$.

• For the y-axis: The equation of the y-axis is $x = 0$. The unshaded region is to the right of the y-axis, so the inequality is:

  $$x \ge 0$$

• For the line passing through $(0, 10)$ and $(7, 4)$: First, calculate the slope $m$:

  $$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 10}{7 - 0} = \frac{-6}{7}$$

  Using the point-slope form $y - y_1 = m(x - x_1)$ with point $(0, 10)$:

  $$y - 10 = -\frac{6}{7}(x - 0)$$ $$7(y - 10) = -6x$$ $$7y - 70 = -6x$$ $$6x + 7y = 70$$

  The unshaded region is below this line. To confirm, test a point like $(0,0)$: $6(0) + 7(0) = 0 \le 70$. So, the inequality is:

  $$6x + 7y \le 70$$

• For the line passing through $(0, 0)$ and $(7, 4)$: First, calculate the slope $m$:

  $$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 0}{7 - 0} = \frac{4}{7}$$

  Using the point-slope form $y - y_1 = m(x - x_1)$ with point $(0, 0)$:

  $$y - 0 = \frac{4}{7}(x - 0)$$ $$y = \frac{4}{7}x$$ $$7y = 4x$$ $$4x - 7y = 0$$

  The unshaded region is above this line. To confirm, test a point like $(0,10)$: $4(0) - 7(10) = -70$. Since $-70 < 0$, and the region is above, the inequality is:

  $$4x - 7y \le 0$$

The inequalities that define the unshaded region R are: