Maximize Z = 10 x1 + 25 x2, subject to 0 ≤ x1 ≤ 3, 0 ≤ x2 ≤ 3, x1 + x2 ≤ 5

The correct option is "95 at (2, 3)".

To solve the given linear programming problem, we need to maximize the objective function:
$$Z=10x_1+25x_2$$ subject to the following constraints:
1. $0\le x_1\le 3$
2. $0\le x_2\le 3$
3. $x_1+x_2\le 5$

First, let's identify the feasible region by plotting the boundary lines of the inequalities:
- The boundary for $x_1$ is between the vertical lines $x_1=0$ (the y-axis) and $x_1=3$.
- The boundary for $x_2$ is between the horizontal lines $x_2=0$ (the x-axis) and $x_2=3$.
- The line $x_1+x_2=5$ connects the points $(5,0)$ and $(0,5)$.

By finding the intersection points of these boundary lines, we can determine the corner points (vertices) of the feasible region:
- Origin: $(0,0)$
- Along the $x_1$-axis: $(3,0)$ since $x_1\le 3$.
- Intersection of $x_1=3$ and $x_1+x_2=5$ gives the point $(3,2)$.
- Intersection of $x_2=3$ and $x_1+x_2=5$ gives the point $(2,3)$.
- Along the $x_2$-axis: $(0,3)$ since $x_2\le 3$.

Now, we evaluate the objective function $Z=10x_1+25x_2$ at each of these corner points:
1. At $(0,0)$:
$$Z=10\left(0\right)+25\left(0\right)=0$$ 2. At $(3,0)$:
$$Z=10\left(3\right)+25\left(0\right)=30$$ 3. At $(3,2)$:
$$Z=10\left(3\right)+25\left(2\right)=30+50=80$$ 4. At $(2,3)$: