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Answer:
a) maximize 12x+16y subject to x+y≤60; x+2y≤80; 2x-3y≥0; x-9y≤0.
b) 40 fish and 20 beef dinners
c) $800
Explanation:
Let x and y represent the numbers of fish and beef dinners to prepare, respectively. Then the relations these values must satisfy are ...
x + y ≤ 60 . . . . . a maximum of 60 dinners will be sold
0.25x + 0.50y ≤ 20 . . . . . kitchen hours cannot exceed 20
x/y ≥ 3/2 . . . . . . at least 3 fish dinners for each 2 beef dinners will be sold
y ≥ 0.10(x +y) . . . . at least 10% of dinners sold will be beef
While satisfying these relations, we want to maximize the profit function:
p = 12x +16y
a) The linear programming problem can be formulated as ...
Maximize 12x +16y, subject to ...
- x + y ≤ 60
- x + 2y ≤ 80
- 2x - 3y ≥ 0
- x - 9y ≤ 0
__
b) The graph shows the constraint inequalities with the comparison symbol reversed. The effect of that is to shade the area that is NOT part of the solution set, leaving the feasible region white. The vertex of the (white) feasible region that makes the profit line farthest from the origin is the solution we're looking for. Once the profit line is plotted so we can compare its slope to the lines bounding the feasible region, it becomes clear which vertex is the one that maximizes profit.
The solution is (x, y) = (40, 20).
- 40 fish dinners
- 20 beef dinners
__
c) The maximum earnings are estimated to be ...
($12)(40) +($16)(20) = $800
Answer:
The right solution is "13,675 U".
Explanation:
According to the question,
The standard material cost will be:
=
=
=
The actual material cost will be:
=
=
hence,
The total material price variance will be:
=
=
= $ (Unfavorable)