Maximize Z = 7x + 11y, subject to 3x + 5y ≤ 26, 5x + 3y ≤ 30, x ≥ 0, y ≥ 0

The correct option is 59 at (9/2, 5/2).

To find the maximum value of the objective function $Z=7x+11y$ subject to the given linear constraints, we can determine the corner points of the feasible region.

The constraints are given by:
1) $3x+5y\le 26$
2) $5x+3y\le 30$
3) $x\ge 0,y\ge 0$

Let us find the boundary lines and their intersection points with the coordinate axes in the first quadrant:
For the line $3x+5y=26$:
When $x=0$, $y=\frac{26}{5}=5.2$. So, the y-intercept is $(0,5.2)$.
When $y=0$, $x=\frac{26}{3}$. So, the x-intercept is $(\frac{26}{3},0)$.

For the line $5x+3y=30$:
When $x=0$, $y=10$. So, the y-intercept is $(0,10)$.
When $y=0$, $x=6$. So, the x-intercept is $(6,0)$.

Next, we find the point of intersection of the two lines by solving the system of linear equations:
Multiply the first equation $3x+5y=26$ by 3:
$9x+15y=78$
Multiply the second equation $5x+3y=30$ by 5:
$25x+15y=150$

Subtract the first modified equation from the second:
$(25x-9x)=150-78$
$16x=72$
$x=\frac{72}{16}=\frac{9}{2}=4.5$

Substitute $x=\frac{9}{2}$ back into $3x+5y=26$:
$3\left(\frac{9}{2}\right)+5y=26$
$\frac{27}{2}+5y=26$
$5y=26-13.5=12.5$
$y=\frac{12.5}{5}=2.5=\frac{5}{2}$

Thus, the intersection point of the two constraint lines is $(\frac{9}{2},\frac{5}{2})$.

The corner points of the feasible region defined by the system of linear inequalities are:
1) $(0,0)$
2) $(6,0)$ (from the boundary $5x+3y\le 30$)
3) $(0,5.2)$ (from the boundary $3x+5y\le 26$)
4) $(\frac{9}{2},\frac{5}{2})$ (the intersection of the boundary lines)

Now, we evaluate the objective function $Z=7x+11y$ at each corner point:
- At $(0,0)$: $Z=7\left(0\right)+11\left(0\right)=0$
- At $(6,0)$: $Z=7\left(6\right)+11\left(0\right)=42$
- At $(0,5.2)$: $Z=7\left(0\right)+11\left(5.2\right)=57.2$
- At $(\frac{9}{2},\frac{5}{2})$:
$Z=7\left(\frac{9}{2}\right)+11\left(\frac{5}{2}\right)=\frac{63}{2}+\frac{55}{2}=\frac{118}{2}=59$

Comparing these values, the maximum value of $Z$ is $59$, which occurs at the point $(\frac{9}{2},\frac{5}{2})$.