Answer :
If we let x as candy A
y as candy B
a as dark chocolate in candy a
b as dark chocolate in candy b
c as caramel
d as walnut
P as profit
we have the equations:
a + c = x
2b + d = y
a + 2b ≤ 360
c ≤ 430
d ≤ 210
P = 285x + 260y
This is an optimization problem which involves linear programming. It can be solved by graphical method or by algebraic solution.
P = 285(a + c) + 260(2b +d)
If we assume a = b
Then a = 120, 2b = 240
P = 285(120 + 120) + 260(240 + 120)
P = 162000
candy A should be = 240
candy B should be = 360
y as candy B
a as dark chocolate in candy a
b as dark chocolate in candy b
c as caramel
d as walnut
P as profit
we have the equations:
a + c = x
2b + d = y
a + 2b ≤ 360
c ≤ 430
d ≤ 210
P = 285x + 260y
This is an optimization problem which involves linear programming. It can be solved by graphical method or by algebraic solution.
P = 285(a + c) + 260(2b +d)
If we assume a = b
Then a = 120, 2b = 240
P = 285(120 + 120) + 260(240 + 120)
P = 162000
candy A should be = 240
candy B should be = 360