A firm produces two products, $x$ and $y$, using two inputs, $A$ and $B$. The available quantities of $A$ and $B$ are 1600 units and 2000 units, respectively. Producing one unit of $x$ requires 4 units of $A$ and 2 units of $B$. Producing one unit of $y$ requires 2 units of $A$ and 5 units of $B$. The profits per unit of $x$ and $y$ are 10 and 8, respectively. The goal is to find the output mix of $x$ and $y$ that maximizes profit using linear programming.

1. Problem Description

A firm produces two products, $x$ and $y$, using two inputs, $A$ and $B$. The available quantities of $A$ and $B$ are 1600 units and 2000 units, respectively. Producing one unit of $x$ requires 4 units of $A$ and 2 units of $B$. Producing one unit of $y$ requires 2 units of $A$ and 5 units of $B$. The profits per unit of $x$ and $y$ are 10 and 8, respectively. The goal is to find the output mix of $x$ and $y$ that maximizes profit using linear programming.

2. Solution Steps

Let $x$ be the number of units of product $x$ and $y$ be the number of units of product $y$.

The objective function (to maximize) is the total profit:

$Z = 10x + 8y$

The constraints are based on the available quantities of inputs $A$ and $B$:

Constraint for input A:

$4x + 2y \le 1600$

Constraint for input B:

$2x + 5y \le 2000$

Also, $x$ and $y$ must be non-negative:

$x \ge 0$

$y \ge 0$

We need to find the feasible region defined by these inequalities.

First, convert the inequalities to equalities:

$4x + 2y = 1600$

$2x + 5y = 2000$

$x = 0$

$y = 0$

Find the intersection points of the lines:

Intersection of $4x + 2y = 1600$ and $2x + 5y = 2000$:

Multiply the first equation by -1 to eliminate x after adding the equations:

$-4x - 2y = -1600$

$2x + 5y = 2000$

Multiply the second equation by 2:

$4x + 10y = 4000$

Add the equation to the first:

$4x+10y-4x-2y=4000-1600$

$8y = 2400$

$y = \frac{2400}{8} = 300$

Substitute $y = 300$ into $4x + 2y = 1600$:

$4x + 2(300) = 1600$

$4x + 600 = 1600$

$4x = 1000$

$x = \frac{1000}{4} = 250$

Intersection point: $(250, 300)$

Intersection of $4x + 2y = 1600$ and $x=0$:

$4(0) + 2y = 1600$

$2y = 1600$

$y = 800$

Intersection point: $(0, 800)$

Intersection of $2x + 5y = 2000$ and $y=0$:

$2x + 5(0) = 2000$

$2x = 2000$

$x = 1000$

Intersection point: $(1000, 0)$

Intersection of $4x + 2y = 1600$ and $y=0$:

$4x + 2(0) = 1600$

$4x = 1600$

$x = 400$

Intersection point: $(400, 0)$

Intersection of $2x + 5y = 2000$ and $x = 0$:

$2(0) + 5y = 2000$

$5y = 2000$

$y = 400$

Intersection point: $(0, 400)$

The vertices of the feasible region are $(0, 0)$, $(400, 0)$, $(250, 300)$, and $(0, 400)$.

Evaluate the objective function at each vertex:

At $(0, 0)$: $Z = 10(0) + 8(0) = 0$

At $(400, 0)$: $Z = 10(400) + 8(0) = 4000$

At $(250, 300)$: $Z = 10(250) + 8(300) = 2500 + 2400 = 4900$

At $(0, 400)$: $Z = 10(0) + 8(400) = 3200$

The maximum profit occurs at $(250, 300)$ with a profit of

4

9

0

0.

3. Final Answer

The output mix that maximizes profit is $x = 250$ and $y = 300$.