1. **State the problem:** We want to minimize the cost function $$C = 11x + 8y$$ subject to the constraint $$6x + 7y \geq 84$$ and $$x, y \geq 0$$. 2. **Understand the constraint:** The inequality $$6x + 7y \geq 84$$ represents the region on or above the line passing through points where $$6x + 7y = 84$$. 3. **Find intercepts of the constraint line:** - When $$y=0$$, $$6x = 84 \Rightarrow x = 14$$. - When $$x=0$$, $$7y = 84 \Rightarrow y = 12$$. 4. **Feasible region:** Since $$x, y \geq 0$$ and $$6x + 7y \geq 84$$, the feasible region is the part of the first quadrant on or above the line connecting (14, 0) and (0, 12). 5. **Check if minimum exists:** The cost function $$C = 11x + 8y$$ is linear and the feasible region is unbounded above the line. Since the feasible region extends infinitely in the direction where $$x$$ and $$y$$ increase, and the cost function increases with $$x$$ and $$y$$, the minimum value of $$C$$ will occur at the boundary line. 6. **Evaluate $$C$$ on the boundary line:** Express $$y$$ in terms of $$x$$ from the constraint: $$6x + 7y = 84 \Rightarrow 7y = 84 - 6x \Rightarrow y = \frac{84 - 6x}{7}$$ Substitute into $$C$$: $$C = 11x + 8\left(\frac{84 - 6x}{7}\right) = 11x + \frac{8 \times 84}{7} - \frac{8 \times 6x}{7} = 11x + 96 - \frac{48x}{7}$$ Simplify the $$x$$ terms: $$11x - \frac{48x}{7} = \frac{77x}{7} - \frac{48x}{7} = \frac{29x}{7}$$ So, $$C = \frac{29x}{7} + 96$$ 7. **Find minimum $$C$$ for $$x \geq 0$$ and $$y \geq 0$$:** Since $$y \geq 0$$, from $$y = \frac{84 - 6x}{7} \geq 0$$ we get: $$84 - 6x \geq 0 \Rightarrow 6x \leq 84 \Rightarrow x \leq 14$$ Also, $$x \geq 0$$. So $$x$$ is in $$[0, 14]$$. 8. **Evaluate $$C$$ at endpoints:** - At $$x=0$$: $$C = \frac{29 \times 0}{7} + 96 = 96$$ - At $$x=14$$: $$C = \frac{29 \times 14}{7} + 96 = 2 \times 29 + 96 = 58 + 96 = 154$$ 9. **Conclusion:** The minimum value of $$C$$ is $$96$$ at $$x=0, y=12$$. **Final answer:** $$\boxed{C_{min} = 96}$$