A firm uses labour, capital, and gold to produce jewelry of type X and Y. To produce one unit of X, the firm requires 1 unit of labour, 1 unit of capital, and 2 units of gold. To produce one unit of Y, the firm requires 1 unit of labour, 2 units of capital, and 1 unit of gold. The firm has 50 units of labour, 30 units of capital, and 80 units of gold available. Each unit of X contributes Shs 20 to profit, while each unit of Y contributes Shs 30 to profit. Use a linear programming approach to find the optimal combination of product X and Y that will maximize profits.

1. Problem Description

A firm uses labour, capital, and gold to produce jewelry of type X and Y. To produce one unit of X, the firm requires 1 unit of labour, 1 unit of capital, and 2 units of gold. To produce one unit of Y, the firm requires 1 unit of labour, 2 units of capital, and 1 unit of gold. The firm has 50 units of labour, 30 units of capital, and 80 units of gold available. Each unit of X contributes Shs 20 to profit, while each unit of Y contributes Shs 30 to profit.

Use a linear programming approach to find the optimal combination of product X and Y that will maximize profits.

2. Solution Steps

Let $x$ be the number of units of product X produced.

Let $y$ be the number of units of product Y produced.

The objective function to maximize is the profit:

$Z = 20x + 30y$

The constraints are:

Labour: $x + y \le 50$

Capital: $x + 2y \le 30$

Gold: $2x + y \le 80$

Non-negativity: $x \ge 0$, $y \ge 0$

We will analyze the corner points of the feasible region.

1. Intersection of $x+y=50$ and $x+2y=30$:

Subtracting the first equation from the second: $y = -20$. Since $y$ cannot be negative, this intersection point is not feasible.

2. Intersection of $x+y=50$ and $2x+y=80$:

Subtracting the first equation from the second: $x = 30$.

Substituting into the first equation: $30+y=50$, so $y=20$.

Point: $(30, 20)$.

Profit: $Z = 20(30) + 30(20) = 600 + 600 = 1200$

3. Intersection of $x+2y=30$ and $2x+y=80$:

Multiply the first equation by 2: $2x + 4y = 60$

Subtract the second equation from this: $3y = -20$, so $y = -20/3$. Since $y$ cannot be negative, this intersection point is not feasible.

4. Intersection of $x+2y=30$ and $x=0$:

$x=0$, so $2y=30$, which gives $y=15$.

Point: $(0, 15)$.

Profit: $Z = 20(0) + 30(15) = 0 + 450 = 450$

5. Intersection of $2x+y=80$ and $y=0$:

$y=0$, so $2x=80$, which gives $x=40$.

Point: $(40, 0)$.

Profit: $Z = 20(40) + 30(0) = 800 + 0 = 800$

6. Intersection of $x+y = 50$ and $x=0$:

$x=0$, so $y=50$.

Point: $(0, 50)$

The capital constraint $x+2y \le 30$ is violated because $0 + 2(50) = 100 > 30$. Therefore $(0, 50)$ is not feasible.

7. Intersection of $x+y=50$ and $y=0$:

$y=0$, so $x=50$.

Point: $(50, 0)$.

The gold constraint $2x+y \le 80$ is violated because $2(50) + 0 = 100 > 80$. Therefore $(50, 0)$ is not feasible.

Let's check the corner points:

(0,0): Z = 0

(40,0): Z = 800

(0,15): Z = 450

(30,20): Z = 1200

However, the capital constraint needs to be checked at $(30,20)$:

$30 + 2(20) = 30 + 40 = 70 > 30$, meaning (30, 20) is infeasible.

We need to find the intersection of $x+y=50$ and $x+2y=30$ where $x>0, y>0$. If we have $x+y=50$ we can write $x=50-y$. Substituting that into the other equation, we get

$50-y+2y=30$, so $y=30-50 = -20$, meaning $x=50-(-20)=70$, which are both not feasible. This intersection does not produce a feasible point.

Let's find the intersection of $2x+y=80$ and $x+2y=30$. Multiply the first equation by 2 and subtract the second. $4x+2y=160$, so $3x = 130$ and $x=130/3 = 43.33$.

Then $y = 30-x = 30-43.33 = -13.33$ which is non-feasible.

It seems $(0,0)$, $(40,0)$ and $(0,15)$ are the corner points. $Z = 0$, $Z = 800$ and $Z = 450$.

The intersection between labour and gold constraints

$x+y=50$ and $2x+y=80$.

Subtracting $x+y=50$ from $2x+y=80$ results in $x=30$. This leads to $y=20$.

However, $(30,20)$ is infeasible because $30 + 2*20 = 70 > 30$. Let's consider $x+2y <=30$, meaning that $x+2y$ is our binding constraint.

If we use $x+2y=30$, we can find $x$ by substituting in $y$ such that $20x + 30y$ is maximized with $x+y<50$.

When $x=0$, $y=15$, $Z = 30*15 = 450$, and since $x+y=15 < 50$, the labour constraint is met.

If $y=0$, $x=30$. Z = $30*20 = 600$

The point (30,0)

$x+y<=50$, ok

capital constraint is ok

$2x+y <=80$, so $2(30)+0 = 60$, ok.

The profit is $

6

0

0.

Let's revisit (30,20) in case this is our optimal solution.

When x+2y =30 and x+y=50, x=100-2y. Substituting, $100-2y+2y = 30, which doesnt work since 100 doesnt =

3

0. However, if labor is fully utilized x+y=50,

2x+y <= 80 and x+2y=30

From x+2y=30, x=30-2y

Substituting in: 30-2y +y =50, or 30-y=50, y= -

2

0.

So we dont utilize full labour capacity to maximize profit. So we have labour = 0 or below (x+y)<

5

0.

3. Final Answer

The optimal combination is to produce 30 units of X and 0 units of Y.

The maximum profit is Shs

6

0

0.