Let’s define variables:

• $x$ = number of plain burgers
• $y$ = number of double cheeseburgers
• $z$ = number of regular cheeseburgers

Each burger type uses the following resources:

1. Buns Constraint:
   $x + y + z \leq 9$
2. Beef Patties Constraint:
   $x + 2y + z \leq 13$
3. Cheese Slices Constraint:
   $2y + z \leq 11$

We need to maximize usage while satisfying these constraints. Let's solve this system optimally. scipy.optimize import linprog

Coefficients for the objective function (maximize total burgers)

Since linprog does minimization, we use negative values

c = [-1, -1, -1]

Coefficients for the inequalities (left-hand side of constraints)

A = [ [1, 1, 1], # Buns constraint [1, 2, 1], # Beef patties constraint [0, 2, 1] # Cheese slices constraint ]

Right-hand side of constraints

b = [9, 13, 11]

Bounds for each variable (they must be non-negative integers)

x_bounds = (0, None) y_bounds = (0, None) z_bounds = (0, None)

Solve the linear program using integer constraints

res = linprog(c, A_ub=A, b_ub=b, bounds=[x_bounds, y_bounds, z_bounds], method='highs')

Extract the results and round to nearest integers

plain_burgers = round(res.x[0]) double_cheeseburgers = round(res.x[1]) regular_cheeseburgers = round(res.x[2])

(plain_burgers, double_cheeseburgers, regular_cheeseburgers)The optimal solution is to make:

• 9 plain burgers
• 0 double cheeseburgers
• 0 regular cheeseburgers

This ensures all 9 buns and 9 beef patties are used efficiently while avoiding waste. The remaining 4 patties and 11 cheese slices cannot be used without exceeding the available buns.

Would you like a more detailed breakdown or alternative approaches?

Related Questions:

1. What if I want to maximize the use of cheese slices?
2. How would the solution change if I had 12 buns instead of 9?
3. Can I adjust the recipe to avoid any waste?
4. What would be the revenue if I sell these at a discount?
5. How does this compare to a scenario where I prioritize cheeseburgers over plain burgers?

Tip: When resources are limited, prioritize items that use the scarcest resource most efficiently!