Answer
Maximum: 42
Minimum: 40$\frac{3}{8}$
Work Step by Step
For linear optimization problems, the maximum and minimum will always occur at a corner of the feasibility area. (1,0) and (4,0) are given corners (they are x-intercepts), but the other corners must be solved with a system of equations. One of the corners is the intersection of $y = x$ and $y = -x+1$, and the other corner is the intersection of $y=x$ and $y = -x+4$. The intersection of these lines is (0.5, 0.5) and (2,2), respectively.
Now that we have all the corners, we just plug in x and y into $N= \frac{1}{2} x+ \frac{1}{4} y + 40$ and find the largest and smallest values:
At (1,0): $N = \frac{1}{2} \times 1 + \frac{1}{4} \times 0 + 40$ = 40$\frac{1}{2}$
At (4,0): $N = \frac{1}{2} \times 4 + \frac{1}{4} \times 0 + 40$ = 42
At (0.5,0.5): $N = \frac{1}{2} \times 0.5 + \frac{1}{4} \times 0.5 + 40$ = 40$\frac{3}{8}$
At (2,2): $N = \frac{1}{2} \times 2 + \frac{1}{4} \times 2 + 40$ = 41$\frac{1}{2}$
Comparing our results, we know that in the given region, the maximum of $N= \frac{1}{2} x+ \frac{1}{4} y + 40$ is 42, and the minimum is 40$\frac{3}{8}$.
