from scipy.optimize import linprog

# Coefficients of the objective function for each goal $($negated because linprog minimizes$)$

c $=[1,2,3]$ # Priorities for the three goals

# Coefficients of the constraints

# Labor hours constraint

A$\_$eq $=[[2,2,1]]$ # Coefficients of T$,$ R$,$ U

b$\_$eq $=[10000]$ # Right$-$hand side of the constraint

# Material supply constraints

A$\_$ub $=[[-2,-3,-4],$ # Coefficients of T$,$ R$,$ U for plastic

$[-1,-1,-2],$ # Coefficients of T$,$ R$,$ U for rubber

$[-2,-1,0]]$ # Coefficients of T$,$ R$,$ U for metal

b$\_$ub $=[-16000,-5000,-9000]$ # Right$-$hand side of the constraints

# Budget constraint

c$\_$budget $=[7,5,4]$ # Cost of producing one tank, truck, and turtle respectively

A$\_$budget $=[$c$\_$budget$]$

b$\_$budget $=[164000]$

# Labor hour utilization constraint

A$\_$labor $=[[2,2,1]]$ # Coefficients of T$,$ R$,$ U

b$\_$labor $=[250*40]$ # Total available labor hours

# Bounds for each variable

bounds $=[(0,$ None$),(0,$ None$),(0,$ None$)]$ # Non$-$negative constraints for T$,$ R$,$ U

# Solve the linear program

result $=$ linprog$($c$,$ A$\_$eq$=$A$\_$eq$,$ b$\_$eq$=$b$\_$eq$,$ A$\_$ub$=$A$\_$ub$,$ b$\_$ub$=$b$\_$ub$,$ bounds$=$bounds, method$=$'highs'$)$

# Print the results

print$("$Optimal number of tanks to produce:", round$($result$.$x$[0]))$

print$("$Optimal number of trucks to produce:", round$($result$.$x$[1]))$

print$("$Optimal number of turtles to produce:", round$($result$.$x$[2]))$