Maximize \[ \begin{array}{lll} z=2 x+5 y & \\ x+6 y & \leq & 30 \\ 4 x+4 y & \leq & 40 \\ x & \geq & 0 \\ y & \geq & 0 \end{array} \] Maximum is \( \square \) at \( x= \) \( \square \) \[ y= \] \( \square \)

To maximize the objective function \( z = 2x + 5y \), we first need to identify the feasible region defined by the constraints: 1. From the first constraint \( x + 6y \leq 30 \), we can express \( y \) in terms of \( x \) as \( y \leq \frac{30 - x}{6} \). 2. The second constraint \( 4x + 4y \leq 40 \) simplifies to \( x + y \leq 10 \), or \( y \leq 10 - x \). Now, let's find the points where these lines intersect, which will be our potential corner points of the feasible region. 1. Setting \( x + 6y = 30 \) and \( x + y = 10 \) equal, we solve for \( y \): \[ x + y = 10 \implies y = 10 - x \] Substituting into the first equation: \[ x + 6(10 - x) = 30 \implies x + 60 - 6x = 30 \implies -5x = -30 \implies x = 6, \, y = 4. \] 2. Evaluate the intersection with the axes: - Setting \( y = 0 \) in the first constraint gives \( x = 30 \) (beyond feasible). - Setting \( x = 0 \) gives \( 6y = 30 \) or \( y = 5 \). - Setting \( y = 0 \) in the second gives \( x = 10 \) (valid and on the boundary). - Setting \( x = 0 \) gives \( y = 10 \) (invalid due to the first constraint). We now have the vertices: - Point \( A(0, 5) \) - Point \( B(6, 4) \) - Point \( C(10, 0) \) Evaluate \( z \) at each vertex: - At \( A(0, 5) \): \( z = 2(0) + 5(5) = 25 \) - At \( B(6, 4) \): \( z = 2(6) + 5(4) = 12 + 20 = 32 \) - At \( C(10, 0) \): \( z = 2(10) + 5(0) = 20 \) The maximum value of \( z \) is \( 32 \) at \( x = 6 \) and \( y = 4 \). Thus, the maximum is \( \boxed{32} \) at \( x = \boxed{6} \) and \( y = \boxed{4} \).