# Maximize Result For 4 variable

A factory has $A$ and $B$ products. $A$ is made with $4X + 2Y$ raw materials. $B$ is made $2X + 4Y$ raw materials.

We want to maximize total profit.

Input

• amount of profit $A$ per item, amount of profit $B$ per item and
• number of $X$ and number of $Y$ raw materials.

Output

• Number of A which will be produced,
• number of B which will be produced,
• total profit.

Example

Input:
profitA = $8, profitB =$6,
numberOfX = 600, numberOfY = 480,
Output:
numberOfA = 120,
numberOfB = 60,
totalprofit = $1320.  My solution is brute force algorithm. I find maximum number of$A$will can produce and I decrease it one by one then compare result and get maximum profit. But this is not efficient. Is there a algorithm that solve this problem? Formula = mA*pA + mB*pB => maximum pA: profit A, pB: profit B, mA: number of product A, mB: number of product B, Producing mA number A are required 4*mA number X and 2*mB number Y, producing mB number B are required 2*mB number X and 4*mB number Y.  ## 1 Answer Express the objective function for profit: "maximize$P_o = n_A p_A + n_B p_B$" in terms of$n_A$(or$n_B$). If we produce$n_A$items then from the constraints we have: $$n_B = \min (\frac{n_X - 4 n_A}{2}, \frac{n_Y - 2 n_A}{4}).$$ Thus we have a max-min problem. I believe this can be solved using auxiliary constraint technique like-so: https://stackoverflow.com/questions/10792139/using-min-max-within-an-integer-linear-program • Sorry my late comment, firstly thank you for your time, but your formula is not always true because there may be a situation that item A may not product since it has a few profit. For example: input profitA =$$1, profitB =$10, numberofX=600, numberofY=480,in this situation output is: mA=0, mB=120, totalprofit=$1200. Apr 22 '16 at 18:20 • I see the flaw in my logic. Based on what you say, I am not sure enhancing the objective function as max$(P_o, P_A, P_B)$with$P_A = ($max$n_A) p_A$and$P_B = ($max$n_B) p_B\$ will work either. If I come up with something, will post. Apr 22 '16 at 19:08
• Edited my answer to show max-min problem structure Apr 22 '16 at 23:31
• Thank you again, My searching solution is similar your offer algorithm. I don't understand your shared link but I will research auxiliary constraint technique, is there more different techniques to solve this problem? Apr 23 '16 at 6:20
• My understanding is that the auxiliary term added to the objective function moves the objective function up or down away from the surface containing the saddle point. I haven't researched into other techniques. Apr 23 '16 at 11:18