We are tasked with maximizing the objective function:

$C = 4x + 12y$

subject to the constraints:

$3x + 5y \leq 12$ $6x + 2y \leq 10$ $x \geq 0, \quad y \geq 0$

Step 1: Identify the constraints

We need to convert the inequalities into equations for the boundary lines and then identify the feasible region formed by these constraints.

1. $3x + 5y = 12$ — Constraint 1
2. $6x + 2y = 10$ — Constraint 2
3. $x \geq 0, y \geq 0$ — Non-negativity constraints.

Step 2: Find the intersection points

Let's find the points of intersection between the constraints by solving the system of equations:

Intersection of $3x + 5y = 12$ and $6x + 2y = 10$:

Multiply the second equation by 2 to eliminate $x$:

$12x + 4y = 20$

Now subtract the first equation:

$(12x + 4y) - (9x + 15y) = 20 - 12$ $3x - 11y = 8$

Solve for $x$:

$x = \frac{8 + 11y}{3}$

Substitute into $6x + 2y = 10$ or simplify intersection.