A patient takes vitamin pills. Each day she must have at least 150 IU of vitamin \( \mathrm{A}, 3 \mathrm{mg} \) of vitamin \( \mathrm{B}_{1} \), and 60 mg of vitamin C . She can choose between pill 1 , which contains 90 IU of vitamin \( \mathrm{A}, 1 \mathrm{mg} \) of vitamin \( \mathrm{B}_{1} \), and 10 mg of vitamin C , and pill 2 , which contains 30 IU of vitamin \( \mathrm{A}, 1 \mathrm{mg} \) of vitamin \( \mathrm{B}_{1} \), and 40 mg of vitamin C . Pill 1 costs \( 10 \notin \), and pill 2 costs \( 20 \notin \). How many of each pill should she buy in order to minimize her cost? What is the minimum cost? She should buy \( \square \) of pill 1 and \( \square \) of pill 2 . The minimum cost is \( \$ \square \). (Simplify your answers. Type integers or decimals.)

Let \( x \) be the number of pill 1’s and \( y \) be the number of pill 2’s. The vitamins in each pill and the daily requirements give us the following constraints. **Vitamin A Constraint:** Pill 1 provides \(90\) IU and pill 2 provides \(30\) IU, with a requirement of at least \(150\) IU: \[ 90x + 30y \geq 150. \] Dividing the inequality by \(30\) gives: \[ 3x + y \geq 5. \] **Vitamin \( \mathrm{B}_1 \) Constraint:** Each pill provides \(1\) mg, with a requirement of at least \(3\) mg: \[ x + y \geq 3. \] **Vitamin C Constraint:** Pill 1 provides \(10\) mg and pill 2 provides \(40\) mg, with a requirement of at least \(60\) mg: \[ 10x + 40y \geq 60. \] Dividing the inequality by \(10\) gives: \[ x + 4y \geq 6. \] The cost function to minimize is: \[ \text{Cost} = 10x + 20y. \] Now, we have the following system of inequalities: \[ \begin{cases} 3x + y \geq 5, \\ x + y \geq 3, \\ x + 4y \geq 6, \\ x \geq 0, \quad y \geq 0. \end{cases} \] **Step 1: Identify Candidate Vertices** We consider points where the constraints are binding. 1. **Intersection of \(3x + y = 5\) and \(x + y = 3\):** Subtract the second equation from the first: \[ (3x + y) - (x + y) = 5 - 3 \quad \Rightarrow \quad 2x = 2 \quad \Rightarrow \quad x = 1. \] Substituting back into \(x + y = 3\): \[ 1 + y = 3 \quad \Rightarrow \quad y = 2. \] Thus, one vertex is \((x, y) = (1, 2)\). 2. **Intersection of \(x + y = 3\) and \(x + 4y = 6\):** Subtract the first equation from the second: \[ (x + 4y) - (x + y) = 6 - 3 \quad \Rightarrow \quad 3y = 3 \quad \Rightarrow \quad y = 1. \] Substituting back into \(x + y = 3\): \[ x + 1 = 3 \quad \Rightarrow \quad x = 2. \] Thus, another vertex is \((x, y) = (2, 1)\). **Step 2: Check Feasibility and Compute Costs** - **At \((1, 2)\):** Check the vitamin constraints: \[ 3(1) + 2 = 5 \quad (\text{Vitamin A satisfied}), \] \[ 1 + 2 = 3 \quad (\mathrm{B}_1 \text{ satisfied}), \] \[ 1 + 4(2) = 9 \geq 6 \quad (\text{Vitamin C satisfied}). \] The cost is: \[ 10(1) + 20(2) = 10 + 40 = \$50. \] - **At \((2, 1)\):** Check the vitamin constraints: \[ 3(2) + 1 = 7 \geq 5 \quad (\text{Vitamin A satisfied}), \] \[ 2 + 1 = 3 \quad (\mathrm{B}_1 \text{ satisfied}), \] \[ 2 + 4(1) = 6 \quad (\text{Vitamin C exactly satisfied}). \] The cost is: \[ 10(2) + 20(1) = 20 + 20 = \$40. \] Comparing the costs, the point \((2, 1)\) gives the lower cost of \(\$40\). **Conclusion:** The patient should buy \(2\) of pill 1 and \(1\) of pill 2. The minimum cost is \(\$40\).