Answer:
yes to both
Step-by-step explanation:
(1) For y=1, and x=a, ...
a² + 1² ≤ 3
-√2 ≤ a ≤ √2 . . . . . subtract 1 and take the square root
All values of "a" are less than 2 inside the circle.
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(2) For y=4 and x=a, ...
2·4 +4·a = 10
4a = 2 . . . . . . . subtract 8
a = 1/2 . . . . . . .divide by 4
The value of "a" for point (a, 4) on the line is less than 2.
Answer:
A. see below for a graph
B. f(x, y) = f(0, 15) = 90 is the maximum point
Step-by-step explanation:
A. See below for a graph. The vertices are those defined by the second inequality, since it is completely enclosed by the first inequality: (0, 0), (0, 15), (10, 0)
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B. For f(x, y) = 4x +6y, we have ...
f(0, 0) = 0
f(0, 15) = 6·15 = 90 . . . . . the maximum point
f(10, 0) = 4·10 = 40
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<em>Comment on evaluating the objective function</em>
I find it convenient to draw the line f(x, y) = 0 on the graph and then visually choose the vertex point that will put that line as far as possible from the origin. Here, the objective function is less steep than the feasible region boundary, so vertices toward the top of the graph will maximize the objective function.
