Let "c" and "q" represent the numbers of bottles of Classic and Quantum that should be produced each day to maximize profit. The problem conditions give rise to 3 inequalities:
.. 0.500c +0.550q ≤ 100 . . . . . . . liters of water
.. 0.600c +0.200q ≤ 100 . . . . . . . kg of sugar
.. 0.1c +0.2q ≤ 32 . . . . . . . . . . . . . grams of caramel
These can be plotted on a graph to find the feasible region where c and q satisfy all constraints. You find that the caramel constraint does not come into play. The graph below has c plotted on the horizontal axis and q plotted on the vertical axis.
Optimum production occurs near c = 152.17 and q = 43.48. Examination of profit figures for solutions near those values reveals the best result for (c, q) = (153, 41). Those levels of production give a profit of 6899p per day.
To maximize profit, Cartesian Cola should produce each day
.. 153 bottles of Classic
.. 41 bottles of Quantum per day.
Profit will be 6899p per day.
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The problem statement gives no clue as to the currency equivalent of 100p.
Answer:4
Step-by-step explanation:
50*7000=350000
Answer:
10x9=90
10x10=100
Step-by-step explanation:
I'm just gonna assume that 6/3 and 1/2 are fractions.
First you have to find common denominators for the fractions to be added.
The first common denominator divisible by both the denominators is 6.
So 2x3=6 and 1x3=3
and 3x2+6 and 6x2+12
so that would be 12/6 + 3/6
Then you would proceed to add both of the numerators together which are 12 and 3.
To get a total of 15/6. You can also simplify that into a mixed number 2 1/2.
Answer:
64
Step-by-step explanation:
It’s directly adjacent to the other angle.