Day 1: It's that Time of Year Again!

Let number of lights = $x$ , wreaths = $y$ , and ornaments = $z$ .

Total discount = $5x^2y^3z$ units

And, $x+y+z=18$

To maximize the discount, we use A.M ≥ G.M

$\frac {\frac {x}{2} + \frac {x}{2} + \frac {y}{3} + \frac {y}{3} + \frac {y}{3} + z}{6} ≥ (\frac {x^2y^3z}{4\cdot 27})^{\frac {1}{6}}$

The equality only occurs if all terms of the sequence are equal, i.e. $\frac {x}{2}=\frac {y}{3}=z$

Hence, $2z+3z+z=18$

$\therefore z=3$

Thus, $\large x=2z=6$

We have two functions, one representing the discount, as $D(l, w) = 5l^2w^3o$ and a one representing the number of items Jamie wants to purchase. We can express this as $f(l, w, o) = l \ + \ w \ + \ o \ = \ 18$ . Using Lagrange optimization:

$\Rightarrow \ \nabla \ D(l, w, o) \ = \ \lambda \nabla \ f(l, w, o)$

$\Rightarrow \ \begin{bmatrix} \frac{\partial}{\partial l} D(l, w, o) \\ \frac{\partial}{\partial w} D(l, w, o) \\ \frac{\partial}{\partial o} D(l, w, o) \end{bmatrix} \ = \ \lambda \ \begin{bmatrix} \frac{\partial}{\partial l} f(l, w, o) \\ \frac{\partial}{\partial w} f(l, w, o) \\ \frac{\partial}{\partial o} f(l, w, o) \end{bmatrix} \ \Rightarrow \ \begin{bmatrix} 10lw^3o \\ 15l^2w^2o \\ 5l^2w^3 \end{bmatrix} \ = \ \lambda \ \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$

So we can see that we have a system of equations:

$\Rightarrow \ 10lw^3o \ = \ \lambda, \ 15l^2w^2o \ = \ \lambda, \ 5l^2w^3 \ = \ \lambda, \ l \ + \ w \ + \ o \ = \ 18$

When we solve this system using the first three equations, we can represent the variables in terms of each other. We can then plug these values into the fourth equation, to get our values for the number of items bought. Solving for different variables using the first three equations, we get multiple different possible values for the different parameters when we plug these values into the fourth equation, but once we plug each these different sets of values back into the original function representing the discount: $D(l, w) = 5l^2w^3o$ , we see that the solution providing the maximum discount is $l \ = \ 6, w \ = \ 9, o \ = \ 3$ , therefore the answer is $l \ = \ 6$ .

Note: This is not a solution...........
@Jack Ceroni Ummmm........doesn't a discount of 393,660% seem to be a LOT.........?? :P