Answer:
To find the minimum cost and the corresponding amount of meat and cheese Jacob should buy to satisfy his nutritional requirements, we can set up a system of inequalities and use linear programming to solve for the optimal (minimum cost) solution.
Let's define:
\( x \) = pounds of meat
\( y \) = pounds of cheese
From the problem, we have the following constraints:
1. The nutritional requirements:
- \( 2x + 3y \geq 18 \) (for carbohydrates)
- \( 2x + y \geq 8 \) (for protein)
2. The cost function to minimize:
- \( C = 3.5x + 2.1y \)
3. The non-negativity constraints:
- \( x \geq 0 \) (can't buy negative pounds of meat)
- \( y \geq 0 \) (can't buy negative pounds of cheese)
Now, we'll graph the system of inequalities to find the feasible region and then determine the vertices of that region. However, since this is a textual medium, we can't graph here, but I can guide you through how this could be done with paper and pencil or algebraically solve for the corner points which are potential candidates for the optimal solution.
From the inequalities, we can express them in terms of \( y \):
1. \( y \geq \frac{18 - 2x}{3} \)
2. \( y \geq 8 - 2x \)
Let's find the intersection of these two lines to get the feasible region's vertices:
Set the two expressions for \( y \) equal to one another:
\( \frac{18 - 2x}{3} = 8 - 2x \)
Multiplying through by 3 to get rid of the fraction gives:
\( 18 - 2x = 24 - 6x \)
Solving for \( x \) gives us:
\( 4x = 6 \)
\( x = \frac{6}{4} \)
\( x = 1.5 \)
Now plug \( x \) into one of the original equations to solve for \( y \):
\( y = 8 - 2(1.5) \)
\( y = 8 - 3 \)
\( y = 5 \)
So, one point of intersection is (1.5, 5). However, we need to also consider other vertices of the feasible region, which would correspond to either \( x \) or \( y \) being zero. Let's compute these:
If \( x = 0 \):
- From \( 2x + 3y \geq 18 \): \( 3y \geq 18 \), so \( y \geq 6 \)
- From \( 2x + y \geq 8 \): \( y \geq 8 \)
If \( y = 0 \):
- From \( 2x + 3y \geq 18 \): \( 2x \geq 18 \), so \( x \geq 9 \)
- From \( 2x + y \geq 8 \): \( 2x \geq 8 \), so \( x \geq 4 \)
Now we check the cost for all possible vertices:
1. \( (x, y) = (1.5, 5) \)
\( C = 3.5(1.5) + 2.1(5) \)
\( C = 5.25 + 10.5 \)
\( C = 15.75 \)
2. We also check if \( (x, y) = (4, 2) \), which would satisfy both constraints:
\( C = 3.5(4) + 2.1(2) \)
\( C = 14 + 4.2 \)
\( C = 18.2 \)
3. \( (x, y) = (9, 0) \)
\( C = 3.5(9) + 2.1(0) \)
\( C = 31.5 \)
4. \( (x, y) = (0, 6) \)
\( C = 3.5(0) + 2.1(6) \)
\( C = 12.6 \)
Out of these cost amounts, the lowest cost is \( C = 12.6 \) when \( x = 0 \) and \( y = 6 \).
Therefore, Jacob should buy 0 pounds of the meat and 6 pounds of the cheese to yield a minimum cost of $12.60.