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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:
Increasing on the interval(s) 
Decreasing on the interval(s) 
Step-by-step explanation:
Anytime the graph is going upward, the function is increasing. However, anytime the graph goes down, the function is decreasing. Look at the image below for further reference. Also, when the function is increasing, the slope is positive, and when the function is decreasing the slope is negative.
The increasing interval of the graph is -3
The decreasing interval of the graph is 0.5
Since the values are in the middle of the interval, it automatically becomes the answer.
The answer can be 6/1 or just 6. And visit www.caculatorsoup.com