This comprehensive tutorial presents several word problems and applications related to linear programming with detailed solutions and explanations. We use methods of solving inequalities with two variables, systems of linear inequalities with two variables, and linear programming optimization to solve practical problems where functions such as profit, return, costs, etc., are optimized.
A store sells two types of toys, A and B. The store owner pays $8 and $14 for each unit of toy A and B respectively. One unit of toy A yields a profit of $2, while a unit of toy B yields a profit of $3. The store owner estimates that no more than 2000 toys will be sold every month, and he does not plan to invest more than $20,000 in inventory of these toys. How many units of each type of toy should be stocked to maximize his monthly total profit?
Let \( x \) be the number of toys A and \( y \) the number of toys B. Since \( x \) and \( y \) cannot be negative:
\[ x \geq 0 \quad \text{and} \quad y \geq 0 \]The store owner estimates that no more than 2000 toys will be sold:
\[ x + y \leq 2000 \]One unit of toy A yields a profit of $2, while a unit of toy B yields a profit of $3, so the total profit \( P \) is:
\[ P = 2x + 3y \]The store owner pays $8 and $14 for each unit and cannot invest more than $20,000:
\[ 8x + 14y \leq 20,000 \]We need to find \( x \) and \( y \) that maximize \( P = 2x + 3y \) under the constraints:
\[ \begin{cases} x \geq 0 \\ y \geq 0 \\ x + y \leq 2000 \\ 8x + 14y \leq 20,000 \end{cases} \]The solution set of this system and the vertices of the feasible region are shown below:
Vertices of the feasible region:
Calculate the total profit \( P \) at each vertex:
\[ \begin{align*} P(A) &= 2(0) + 3(0) = 0 \\ P(B) &= 2(0) + 3(1429) = 4287 \\ P(C) &= 2(1333) + 3(667) = 4667 \\ P(D) &= 2(2000) + 3(0) = 4000 \end{align*} \]The maximum profit is at vertex C with \( x = 1333 \) and \( y = 667 \). Therefore, the store owner should stock 1333 toys of type A and 667 toys of type B to maximize profit.
A company produces two types of tables, T1 and T2. It takes 2 hours to produce parts for one unit of T1, 1 hour to assemble, and 2 hours to polish. For T2, it takes 4 hours to produce parts, 2.5 hours to assemble, and 1.5 hours to polish. Monthly available hours are: 7000 for parts production, 4000 for assembly, and 5500 for polishing. The profit per unit of T1 is $90 and per unit of T2 is $110. How many of each type should be produced monthly to maximize profit?
Let \( x \) be the number of T1 tables and \( y \) the number of T2 tables. The profit function is:
\[ P(x, y) = 90x + 110y \]Constraints based on available hours:
\[ \begin{cases} x \geq 0 \\ y \geq 0 \\ 2x + 4y \leq 7000 \quad \text{(parts production)} \\ x + 2.5y \leq 4000 \quad \text{(assembly)} \\ 2x + 1.5y \leq 5500 \quad \text{(polishing)} \end{cases} \]The feasible region and its vertices are shown below:
Vertices of the feasible region:
Evaluate profit at each vertex:
\[ \begin{align*} P(A) &= 90(0) + 110(0) = 0 \\ P(B) &= 90(0) + 110(1600) = 176,000 \\ P(C) &= 90(1500) + 110(1000) = 245000 \\ P(D) &= 90(2300) + 110(600) = 273000 \\ P(E) &= 90(2750) + 110(0) = 247500 \end{align*} \]Maximum profit of $273000 occurs at vertex D. Thus, produce 2300 T1 tables and 600 T2 tables.
A farmer mixes two types of food for low-cost animal feed. Food A costs $10 per bag and contains 40 units protein, 20 units minerals, 10 units vitamins. Food B costs $12 per bag and contains 30 units protein, 20 units minerals, 30 units vitamins. Minimum daily requirements: 150 units protein, 90 units minerals, 60 units vitamins. How many bags of each food should be used daily to meet requirements at minimum cost?
Let \( x \) be bags of food A and \( y \) bags of food B. The cost function is:
\[ C(x, y) = 10x + 12y \]Constraints based on nutritional requirements:
\[ \begin{cases} x \geq 0 \\ y \geq 0 \\ 40x + 30y \geq 150 \quad \text{(protein)} \\ 20x + 20y \geq 90 \quad \text{(minerals)} \\ 10x + 30y \geq 60 \quad \text{(vitamins)} \end{cases} \]Feasible region and vertices:
Vertices:
Evaluate cost at each vertex:
\[ \begin{align*} C(A) &= 10(6) + 12(0) = 60 \\ C(B) &= 10\left(\frac{15}{4}\right) + 12\left(\frac{3}{4}\right) = 46.5 \\ C(C) &= 10\left(\frac{3}{2}\right) + 12(3) = 51 \\ C(D) &= 10(0) + 12(5) = 60 \end{align*} \]Minimum cost of $46.50 occurs at \( B\left(\frac{15}{4}, \frac{3}{4}\right) \). Thus, use 3.75 bags of food A and 0.75 bags of food B daily.
John has $20000 to invest in three funds: F1 (2% return, low risk), F2 (4% return, medium risk), F3 (5% return, high risk). He invests no more than $3000 in F3 and at least twice as much in F1 as in F2. How much should he invest in each fund to maximize annual return?
Let \( x \) = amount in F1, \( y \) = amount in F2, \( z \) = amount in F3. We have:
\[ x + y + z = 20000 \quad \Rightarrow \quad z = 20000 - (x + y) \]Total return \( R \):
\[ \begin{align*} R &= 0.02x + 0.04y + 0.05z \\ &= 0.02x + 0.04y + 0.05[20000 - (x + y)] \\ &= 1000 - 0.03x - 0.01y \end{align*} \]Constraints:
\[ \begin{cases} x \geq 0 \\ y \geq 0 \\ z \geq 0 \quad \Rightarrow \quad x + y \leq 20000 \\ z \leq 3000 \quad \Rightarrow \quad x + y \geq 17000 \\ x \geq 2y \end{cases} \]Feasible region in \( xy \)-plane:
Vertices:
Evaluate return:
\[ \begin{align*} R(A) &= 1000 - 0.03(20000) - 0.01(0) = 400 \\ R(B) &= 1000 - 0.03(17000) - 0.01(0) = 490 \\ R(C) &= 1000 - 0.03(11333) - 0.01(5667) = 603 \\ R(D) &= 1000 - 0.03(13333) - 0.01(6667) = 533 \end{align*} \]Maximum return of $603 occurs at point C. Thus, invest $11,333 in F1, $5,667 in F2, and $3,000 in F3.
A store owner can spend at most $100000 monthly on PCs and laptops. A PC costs $1000 and yields $400 profit; a laptop costs $1500 and yields $700 profit. At least 15 PCs but no more than 80 are sold monthly. Laptop sales are at most half of PC sales. How many of each should be sold to maximize profit?
Let \( x \) = number of PCs, \( y \) = number of laptops. Profit:
\[ P = 400x + 700y \]Constraints:
\[ \begin{cases} 15 \leq x \leq 80 \\ y \geq 0 \\ y \leq \frac{1}{2}x \quad \text{(laptops at most half of PCs)} \\ 1000x + 1500y \leq 100000 \quad \text{(budget constraint)} \end{cases} \]Feasible region:
Vertices:
Evaluate profit:
\[ \begin{align*} P(A) &= 400(15) + 700(0) = 6000 \\ P(B) &= 400(15) + 700(7.5) = 11250 \\ P(C) &= 400(57.14) + 700(28.57) = 42855 \\ P(D) &= 400(80) + 700(13.33) = 41310 \end{align*} \]Maximum profit occurs at point C, but we need integer values. Testing nearby integers:
Profit with \( x = 57 \), \( y = 28 \):
\[ P = 400(57) + 700(28) = 42400 \]This is the maximum integer solution. Sell 57 PCs and 28 laptops monthly.
Solving Inequalities with Two Variables
Systems of Linear Inequalities
Linear Programming Optimization
More Linear Programming Problems