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.
_____
The problem statement gives no clue as to the currency equivalent of 100p.
Answer:
5 hours
Step-by-step explanation:
This is a classic pre-algebra question.

Answer:
15 yards
Step-by-step explanation:
Answer: 486 glasses of water in 9 weeks
Step-by-step explanation:
54 X 9 = 486
Or you could add 54 to itself 9 times then you will get 486
Step-by-step explanation:
For standard embossed we recommend 8 pints/4.5L for 4-5 rolls. For heavy embossed we recommend 7 pints/4L for 3-4 rolls.
hope that help you