Question

1.) Rina needs at least 48 units of protein, 60 units of carbohydrates, and 50 units of fat each month. From each kilogram of food A, she receives 2 units of protein, 4 units of carbohydrates, and 5 units of fats. Food b contains 3 units of protein, 3 units of carbohydrates, and 2 units of fats. If food A costs Php110 per kilogram and food B costs Php 90 per kilogram. How many kilograms of each food should Rina buy each month to keep costs at a minimum?

Answer 1

minimize costs:

$z=110x_1+90x_2$

subject to:

$2x_1+3x_2\ge 48$

$4x_1+3x_2\ge 60$

$5x_1+2x_2\ge 50$

x₁, x₂ are kg of foods A and B

for Extreme Points:

for $5x_1+2x_2= 50$ :

$x_1=0\implies x_2=25$

for $2x_1+3x_2= 48$ :

$x_2=0\implies x_1=24$

intersection $5x_1+2x_2= 50$ and $4x_1+3x_2= 60$ :

$x_1=4.29,x_2=14.29$

intersection $2x_1+3x_2= 48$ and $4x_1+3x_2= 60$ :

$x_1=6,x_2=12$

Objective function values at Extreme Points:

$z(0,25)=2250$

$z(4.29,14.29)=1757.14$

$z(6,12)=1740$

$z(24,0)=2640$

The miniimum value of the objective function z=1740 occurs at the extreme point (6,12).

Hence, the optimal solution to the given LP problem is : x₁=6, x₂=12 and min z=1740.