Q.A farmer purchased some sheep and goats at Rs. 1,500 per sheep and Rs. 2,000 per goat, and makes a profit of Rs. 150 per sheep and Rs. 200 per goat after selling them. The farmer has only Rs. 60,000 and cannot accommodate more than 100 animals. He wishes to purchase both kinds of animals and to have maximum profit. Formulate the problem as a linear programming problem.

Identify the decision variables, the profit function to maximize, and translate each stated restriction into an inequality.

Decision variables: let $x$ = number of sheep purchased, $y$ = number of goats purchased.

Objective function: the farmer earns profit $\text{Rs.}\,150$ per sheep and $\text{Rs.}\,200$ per goat, so total profit:

$Z = 150x+200y \quad(\text{maximize})$

Budget constraint: sheep cost Rs. 1,500 each and goats Rs. 2,000 each; total spend can't exceed Rs. 60,000:

$1500x+2000y\le60000 \;\Rightarrow\; 3x+4y\le120 \quad(\text{dividing by }500)$

Capacity constraint: he cannot accommodate more than 100 animals:

$x+y\le100$

…

Define decision variables for units of each food, write the cost as the objective, and translate each nutrient requirement into a linear inequality.

Let $x$ = number of units of food A, and $y$ = number of units of food B.

Objective (minimize cost): Food A costs Rs.,4/unit, food B costs Rs.,3/unit, so:

$\text{Minimize } Z = 4x+3y$

Constraints (from nutrient requirements):

Vitamins: A gives $200$ units/unit, B gives $100$ units/unit, at least $4000$ needed:

$200x+100y \ge 4000 \ \Rightarrow\ 2x+y\ge40$

Minerals: A gives $1$ unit/unit, B gives $2$ units/unit, at least $50$ needed:

$x+2y\ge50$

Calories: A gives $40$/unit, B gives $40$/unit, at least $1400$ needed: …

Translate each real-world limit (labour-hours available for fabricating, and for finishing) into a linear inequality, and the profit goal into a linear objective — that is the entire LPP formulation.

Step 1 — define decision variables. Let $x=$ number of units of model $A$ produced, $y=$ number of units of model $B$ produced. Both must be non-negative: $x\ge0,\ y\ge0$.

Step 2 — fabricating-hours constraint. Model $A$ needs $9$ labour-hours, model $B$ needs $12$, and only $180$ hours are available:

$9x+12y\le180.$

Step 3 — finishing-hours constraint. Model $A$ needs $1$ hour, model $B$ needs $3$, and only $30$ hours are available: …

Translate each sentence of the word problem into a linear inequality: decision variables = kg of each food; objective = total cost; constraints = the minimum vitamin requirements.

Step 1 — define decision variables.

Let $x$ = kg of Food-I used, $y$ = kg of Food-II used.

Step 2 — write the objective function (minimise total cost). Food-I costs Rs.,50/kg, Food-II costs Rs.,70/kg:

$Z = 50x + 70y \quad (\text{to be minimised})$

Step 3 — write the constraints.

Vitamin A requirement: Food-I supplies $2$ units/kg, Food-II supplies $1$ unit/kg, and the mixture must have at least $8$ units:

$2x + y \ge 8$

…

Translate labour and profit data into decision variables, an objective function, and constraints.

Let $x$ = number of units of model A produced, $y$ = number of units of model B produced.

Objective (profit to maximize): Model A gives ₹8,000 profit/unit, model B gives ₹12,000 profit/unit:

$\text{Maximize } Z=8000x+12000y$

Constraints:

• Fabricating hours: model A needs 9 hrs/unit, model B needs 12 hrs/unit, available 180 hrs: $9x+12y\le180$
• Finishing hours: model A needs 1 hr/unit, model B needs 3 hrs/unit, available 30 hrs: $x+3y\le30$ …