Question

3. A farmer plans to mix two types of food to make a mix of low cost feed for the
animals in his farm. A bag of food A costs $10 and contains 40 units of
proteins, 20 units of minerals and 10 units of vitamins. A bag of food B costs
$12 and contains 30 units of proteins, 20 units of minerals and 30 units of
vitamins. How many bags of food A and B should the consumed by the
animals each day in order to meet the minimum daily requirements of 150
units of proteins, 90 units of minerals and 60 units of vitamins at a minimum
cost?

Answer 1

Let $x \ \text{and} \ y$ be the number of beg $A$ and $B$ respectively to meet the minimum daily requirements of food.

Then according to question,

Minimize ,$Z=10x+12y$

Subject to ,$40x+30y\ge150$

$20x+20y\geq90$

$10x+30y\ge60,x\geq0,y\ge0.$

Consider the inequality as a equality we get,

$40x+30y=150$ ...............(1)

$20x+20y=90$ .............(2)

$10x+30y=60$ ................(3)

$x=0$ ..................(4)

And $y=0$ ..........(5)

Now , multiplying (2) with 2 and then subtract from (1) ,we get

$-10y=-30 \implies y=3$

$\therefore 20x=90-60=30 \implies x=\frac{3}{2}.$

Therefore the line (1) and (2) ,intersecting at $(\frac{3}{2},3)$ .

Again ,subtract (3) from (1) we get

$30x=90 \implies x=3.$

$\therefore 30y=60-30=30 \implies y=1$ .

Therefore the line (1) and (3) ,intersecting at (3,1).

Multiplying 2 with equation (3) and then subtract from (2) ,we get

$-40y=-30 \implies y=\frac{3}{4}.$

$\therefore 10x=60-30×\frac{3}{4}=\frac{(240-90)}{4}=\frac{150}{4} \implies x=\frac{15}{4}$ .

Therefore the line (2) and (3) ,intersecting at $(\frac{15}{4},\frac{3}{4})$ .

Now from the figure ,evaluate cost $Z$ at each of the vertex

At $A(0,5),Z=60$ .

At $B(\frac{3}{2},3),Z=10×\frac{3}{2}+12×3=15+36=51$

At $C(\frac{15}{4},\frac{3}{4}),Z=10×\frac{15}{4}+12×\frac{3}{4}=\frac{186}{4}=\frac{93}{2}=46.5$

At $D(6,0),Z=60.$

Hence ,the minimum value is $46.5$ ,which occur at $(\frac{15}{4},\frac{3}{4})$ .

Therefore ,$x=\frac{15}{4} \ \text{and} \ y= \frac{3}{4}$ .