Question 1123220
i solved this as follows:


a = number of felt hats from beta
b = number of straw hats from beta


c = number of felt hats from kappa
d = number of straw hats from kappa


e = number of felt hats from delta
f = number of straw hats from delta


your objective function becomes:


profit = 30a + 40b + 38c + 35d + 40e + 36f


30a + 40b is the profit from beta (felt + straw)
38c + 35d is the profit from kappa (felt + straw)
40e + 36f is the profit from delta (felt + straw)


your constraints are:


a + c + e = 200 because 200 felt hats are required in total.
b + d + f = 300 because 300 straw hats are required in total.


a + b <= 200 because that's the most hats beta can supply.
c + d <= 250 because that's the most hats kappa can supply.
e + f <= 150 because that's the most hats delta can supply.


a,b,c,d,e,f >= 0 because number of hats must be greater than or equal to 0.


i put these requirements into the following linear programming softwarethat uses the simplex method.


<a href = "http://www.zweigmedia.com/RealWorld/simplex.html" target = "_blank">http://www.zweigmedia.com/RealWorld/simplex.html</a>


it told me the optimum solution was:


p = 19400; a = 0, b = 200, c = 50, d = 100, e = 150, f = 0


that's profit = 19400 when:


a = 0 (felt hats by beta)
b = 200 (straw hats by beta)
c = 50 (felt hats by kappa)
d = 100 (straw hats by kappa)
e = 150 (felt hats by delta)
f = 0 (straw hats by delta)


total straw = 300 as required.
total felt = 200 as required.


here's a picture of the setup for the software used.


<img src = "http://theo.x10hosting.com/2018/091501.jpg" alt="$$$" >