Q.A stationery dealer deals in only two items: notebook packs and pen sets. She has Rs 20,000 to invest and storage space for at most 80 pieces in total. A notebook pack costs her Rs 200 and a pen set costs her Rs 400. She estimates a profit of Rs 20 on each notebook pack sold and Rs 50 on each pen set sold. Formulate this situation as a linear programming problem to help her decide how many of each to buy so as to maximize her total profit, assuming she is able to sell everything she buys. (Do not solve.)

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:

$x+y\le100$

Non-negativity: the number of animals can't be negative:

$x\ge0,\quad y\ge0$

Full LPP formulation:

$\text{Maximize } Z=150x+200y \quad\text{subject to}\quad 3x+4y\le120,\ x+y\le100,\ x\ge0,\ y\ge0$

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:

$40x+40y\ge1400 \ \Rightarrow\ x+y\ge35$

Non-negativity (cannot use a negative amount of food):

$x\ge0,\ y\ge0$

Full L.P.P.:

$\text{Minimize } Z=4x+3y \text{ subject to } 2x+y\ge40,\ x+2y\ge50,\ x+y\ge35,\ x\ge0,\ y\ge0$

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:

$x+3y\le30.$

Step 4 — objective function. Profit is $\text{Rs. }8{,}000$ per unit of $A$ and $\text{Rs. }12{,}000$ per unit of $B$, so total profit

$Z=8000x+12000y$

is to be maximized.

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:

$2x + y \ge 8$

Vitamin C requirement: Food-I supplies $1$ unit/kg, Food-II supplies $2$ units/kg, and the mixture must have at least $10$ units:

$x + 2y \ge 10$

Non-negativity (amounts of food can't be negative):

$x\ge0,\ y\ge0$

Complete LPP formulation:

Minimise $Z=50x+70y$

subject to $2x+y\ge8$, $x+2y\ge10$, $x\ge0$, $y\ge0$.

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: $x+3y\le30$
• Non-negativity (can't produce negative units): $x\ge0,\ y\ge0$

So the LPP is:

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

$\text{subject to } 9x+12y\le180,\quad x+3y\le30,\quad x\ge0,\ y\ge0$