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 \( 10 \phi \), and pill 2 costs \( 20 \phi \). 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.)
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To solve this problem, we need to set up a system of inequalities based on the vitamin requirements and analyze cost minimization. Let's define: - \( x \) = number of pill 1 - \( y \) = number of pill 2 We create the following inequalities based on the vitamin contents: 1. For vitamin \( A \): \[ 90x + 30y \geq 150 \] 2. For vitamin \( B_1 \): \[ x + y \geq 3 \] 3. For vitamin C: \[ 10x + 40y \geq 60 \] The cost function to minimize is: \[ Cost = 10x + 20y \] Now let's simplify and solve the inequalities: 1. From \( 90x + 30y \geq 150 \): \[ 3x + y \geq 5 \quad \text{(dividing the whole inequality by 30)} \] 2. The second inequality remains: \[ x + y \geq 3 \] 3. From \( 10x + 40y \geq 60 \): \[ x + 4y \geq 6 \quad \text{(dividing the whole inequality by 10)} \] Now let's graph these inequalities: - \( y \geq 5 - 3x \) (from the first) - \( y \geq 3 - x \) (from the second) - \( y \geq \frac{6 - x}{4} \) (from the third) To find the feasible region, we can plot these inequalities on a graph and find the intersection points. Next, we calculate the points: 1. \( x = 0 \): - From \( 3y \geq 5 \) → \( y \geq \frac{5}{3} \approx 1.67 \) - From \( y \geq 3 \) → \( y=3 \) - From \( y \geq \frac{6}{4} = 1.5 \) 2. \( x = 1 \): - \( y \geq 2 \) (from \( y \geq 3 - x \)) - \( y \geq 1.5 \) 3. \( x = 2 \): - \( y \geq 4 \) (no solutions) Finally, solving \( (3,0) \) and \( (0,3) \) and calculating costs find valid integers. After checking various combinations, we find: - \( x = 3 \) and \( y = 0 \) gives: Cost = \( 10(3) + 20(0) = 30 \) - \( x = 0 \) and \( y = 3 \) gives: Cost = \( 10(0) + 20(3) = 60 \) - The valid minimum is found at \( (2, 1) \): Cost = \( 10(2) + 20(1) = 60 \) Therefore, she should buy \( 2 \) of pill 1 and \( 1 \) of pill 2. The minimum cost is \( \$ 60 \).
