1. **State the problem:** We want to find the maximum value of the function $$z = 4x + 3y$$ subject to the constraints: $$x \geq 4$$ $$y \geq 3$$ $$7x + 6y \leq 88$$ 2. **Understand the constraints:** These inequalities define a feasible region on the xy-plane: - $$x \geq 4$$ is the region to the right of the vertical line $$x=4$$. - $$y \geq 3$$ is the region above the horizontal line $$y=3$$. - $$7x + 6y \leq 88$$ is the region below or on the line with intercepts $$x=\frac{88}{7} \approx 12.57$$ and $$y=\frac{88}{6} \approx 14.67$$. 3. **Find the vertices of the feasible region:** The maximum of a linear function over a polygonal region occurs at a vertex (corner point). - Vertex A: Intersection of $$x=4$$ and $$y=3$$ is $$A=(4,3)$$. - Vertex B: Intersection of $$x=4$$ and $$7x + 6y = 88$$: Substitute $$x=4$$: $$7(4) + 6y = 88 \Rightarrow 28 + 6y = 88 \Rightarrow 6y = 60 \Rightarrow y = 10$$ So $$B = (4,10)$$. - Vertex C: Intersection of $$y=3$$ and $$7x + 6y = 88$$: Substitute $$y=3$$: $$7x + 6(3) = 88 \Rightarrow 7x + 18 = 88 \Rightarrow 7x = 70 \Rightarrow x = 10$$ So $$C = (10,3)$$. 4. **Evaluate the objective function $$z=4x+3y$$ at each vertex:** - At $$A=(4,3)$$: $$z = 4(4) + 3(3) = 16 + 9 = 25$$ - At $$B=(4,10)$$: $$z = 4(4) + 3(10) = 16 + 30 = 46$$ - At $$C=(10,3)$$: $$z = 4(10) + 3(3) = 40 + 9 = 49$$ 5. **Determine the maximum value:** Among the values 25, 46, and 49, the maximum is $$49$$ at point $$C=(10,3)$$. **Final answer:** The maximum value of $$z = 4x + 3y$$ subject to the constraints is $$\boxed{49}$$ at $$x=10$$ and $$y=3$$.