Question

An animal food company must produce 200kg. Of a

mixtureconsisting of ingredients X1 and X2

daily.X1 costs Rs.3/- per kg. And X2 Rs.8/-

per kg.No more than 80 kg. Of X1 can be used and at

least 60kg. Of X2 must be used. Formulate a LP. Model

tominimize the cost.

Answer 1

Answer:

Minimum cost = 1200$

$x_1 = 80~kg\\x_2 = 120~kg$      

Step-by-step explanation:

We are given the following information in the question:

An animal food company must produce 200 kg

$x_1 + x_2 = 200\\x_1 \geq 0\\x_2 \geq 0$

No more than 80 kg Of first ingredient can be used and at

least 60 kg Of second ingredient must be used.

$x_1 \leq 80\\x_2 \geq 60$

Cost of ingredient $x_1$ = Rs 3 per kg

Cost of ingredient $x_2$ = Rs 8 per kg

Total cost = $3x_1 + 8x_2$

We have to minimize this cost.

Then, we can write the following inequalities:

$6x + 8y \geq 208\\x \geq 8\\y \geq 8$

The corner points as evaluated from graph are: (0,200) and (80,120).

C(0,200) = 1600$

C(80,120) = 1200$

Hence, by corner point theorem, the minimum cost would be 1200$ when 80 kg of first ingredient is used and 120 kg of second ingredient.

The attached image shows the graph.