H(x) is negative in the interval (Enter your answer in interval notation)
1 answer:
You might be interested in
First, find the vertices of the shaded region. You can do this my graphing or by solving a system of equations for each pair of functions. <em>I am going to find them by graphing.</em>
x ≥ 0 and y ≥ 0 place the shaded region in the first quadrant with a vertex at (0, 0)
now graph y ≤ 3. a vertex is at (0, 3)
now graph y < -2x + 5. a vertex is at (1, 3). the other vertex is the x-intercept (when y = 0).
0 = -2x + 5
-5 = -2x
So, the vertex is
Next, input the coordinates of the vertices into the objective function.
C = -6x + 5y
(0, 0): C = -6(0) + 5(0)
= 0 + 0
= 0
(0, 3): C = -6(0) + 5(3)
= 0 + 15
= 15
(1, 3): C = -6(1) + 5(3)
= -6 + 15
= 9
: C = -6(\frac{5}{2}[/tex]) + 5(2)
= -15 + 10
= -5
The maximum is C = 15 which occurs at vertex (0, 3)
Answer: A