John will spend £5 of his Christmas money on plain and milk chocolates.
He can buy boxes at £2 each. These contain 25 plain and 25 milk chocolates.
He can buy single plain chocolates at 6p each and loose milk chocolates at 7p each.
John wants to have at least twice as many milk as plain chocolates.
John wishes to maximize the number of chocolates he can buy subject to the constraints.
From that problem, I decided on the three variables:
- $x$ for the number of loose plain chocolates
- $y$ for the number of loose milk chocolates
- $z$ for the number of boxes
And the following constraints:
$$2z + 0.06x + 0.07y \le 5 $$
$$y \ge 2x + 25z $$
$$50z + x + y$$
The question asks for a graph to be drawn to show that if $z = 2$ there is no feasible solution to the problem.
Whilst being able to solve two-variable LPs, how would I go about solving this three-variable LP graphically?