Linear equation question with maximisation A person, with a certain given amount, can buy either 8 apples or 14 oranges or 3 water melons. The weight of either 10 apples is equal to the weight of 12 oranges which is in turn equal to the weight of 4 watermelons. Find the number of oranges she has to buy with $6$ times the original amount if she wants to maximize the total weight and also get some of each of the three fruits. 
 A: Let: $M$ := 'certain given amount'
$o$ := # of oranges purchased
$a$ := # of apples purchased
$w$ := # watermelons purchased
To deal with the weights, it's probably easiest to assign a single unit for all of them. So let's say that 1 apple = 1 unit weight. This implies 1 orange has weight $\frac{10}{12}$; 1 watermelon has weight $\frac{10}{4}$
We do the same with the price of each item. So cost of 1 apple = $\frac{M}{8}$, cost of 1 orange = $\frac{M}{14}$, cost of 1 watermelon = $\frac{M}{3}$
So this makes our objective function $$ \max \quad a + \frac{10}{12}o + \frac{10}{4}w $$
Constraint: Maximum amount we can spend is 6M $$ a \frac{M}{8} + o \frac{M}{14} + w \frac{M}{3} \leq 6M \\ \iff \frac{a}{8} + \frac{o}{14} + \frac{w}{3} \leq 6$$
Constraint: We must buy some of each of the three fruits $$o > 0 \quad a > 0 \quad w > 0$$
I'm not sure what method you've been taught to deal with such a system of equations, but admittedly right now the know-how eludes me (as I would normally adjust the last constraint to use binary variables and then chuck the whole thing into a solver).
My best guess would be to choose $w$ as big as possible since it has the largest weight ratio, which makes it 18. Then systematically reduce it one-by-one and then find your solution by trial and error e.g. Given w = 17, can we still find some $o > 0, a > 0$ such that the constraints are still satisfied? If not, w = 16, and so on.
A: Let's call the weight of 10 apples in kg $C$. Then the weight of 1 apple is $\frac {C}{10}$ kg, the weight of an orange is $\frac {C}{12}$ kg and the weight of a watermelon is $\frac {C}{4}$ kg.  
The question indicates that the buyer can only buy multiples of 8, 14 and 3 apples, oranges and watermelons respectively. Let $y_i$ denote the number of multiples of fruit is she is to buy for i = apples, oranges, watermelons. We can see that the total weight of the fruit purchased will be $8 \frac {C}{10}y_A + 14 \frac {C}{12}y_O + 3 \frac {C}{4}y_W$ kg, Godwilling, so this is the objective function which we want to maximise. 
The only constraints we have are that $y_i$ is an integer and $y_i \ge 1$ for all i and $\sum y_i = 6$, since she has been given 6 times the amount of money needed to buy 8, 14 or 3 apples, oranges or watermelons respectively. Thus we may formulate the problem as follows:         
Maximise $$8 \frac {C}{10}y_A + 14 \frac {C}{12}y_O + 3 \frac {C}{4}y_W$$
subject to  
$\sum y_i = 6$ (where $y_i$ represents the number of sets of fruit she is to buy for i = apples, oranges, watermelons) 
$y_i \ge 1$ and $y_i$ integer, $i = A, O, W$  
C constant    
By inspection we see that $y_A = 1, y_O = 4, y_W =1$ is the optimal solution, since $$ \frac{14}{12} \gt \frac {8}{10} \gt \frac {3}{4}$$, which means the number of oranges she has to buy is $4 \times 14 = 56$. 
