Question: User: DIRECTIONS: Choose one (1) of the problems below 6ither the maximization or minimization problem.
Solve the problem thoroughly, showing all pertinent solutions, computations, and graphs.
I. Maximization Problem
A community health center prepares two types of health-care kits: Basic Care Kits ($x$) and
Emergency Care Kits ($y$). The center wants to maximize the total number of patients served.
Each Basic Care Kit can serve 3 patients, while each Emergency Care Kit can serve 5 patients.
The resource requirements for the production are as follows
Resource Basic Care Kit ($x$) Emergency Care Kit ($y$) Available
Medical supplies 2 units 4 units 40 units
Preparation time 1 hour 2 hours 20 hours
Number of patients 3 patients 5 patients
Question: How many Basic Care Kits and Emergency Care Kits should the health center prepare to maximize
the number of patients served?
Solution:
Mathematical Model
Let:
$x$ = number of Basic Care Kit
$y$ = number of Emergency Care Kit
Objective Function: ______________________________
Explicit Constraints:
1. Medical supplies: ______________________
2. Preparation time: ______________________
Implicit Constraints:
3. Non-negativity: __________________ and __________________
Computation/Graph:
Graph shape/math: The graph is in the bottom-center of the page area and shows the linear programming setup for $x$ and $y$, with constraints from the table: $2x + 4y \leq 40$ and $x + 2y \leq 20$, plus $x \geq 0$ and $y \geq 0$; the feasible region is the shaded area in the first quadrant bounded by the line $x + 2y = 20$.answer this
1. **State the problem:**
We want to maximize the total number of patients served by producing Basic Care Kits ($x$) and Emergency Care Kits ($y$).
2. **Define variables:**
Let $x$ = number of Basic Care Kits.
Let $y$ = number of Emergency Care Kits.
3. **Objective function:**
Maximize the total patients served:
$$Z = 3x + 5y$$
4. **Constraints:**
- Medical supplies: each Basic Kit uses 2 units, each Emergency Kit uses 4 units, total available 40 units:
$$2x + 4y \leq 40$$
- Preparation time: each Basic Kit takes 1 hour, each Emergency Kit takes 2 hours, total available 20 hours:
$$x + 2y \leq 20$$
- Non-negativity constraints:
$$x \geq 0, \quad y \geq 0$$
5. **Find corner points of the feasible region:**
- Intersection of $2x + 4y = 40$ and $x + 2y = 20$:
Multiply second equation by 2:
$$2x + 4y = 40$$
This is the same as the first constraint, so the two lines coincide on this boundary.
Check intercepts:
- For $2x + 4y = 40$:
- If $x=0$, then $4y=40 \Rightarrow y=10$
- If $y=0$, then $2x=40 \Rightarrow x=20$
- For $x + 2y = 20$:
- If $x=0$, then $2y=20 \Rightarrow y=10$
- If $y=0$, then $x=20$
So the feasible region is bounded by $x=0$, $y=0$, and the line $x + 2y = 20$ (which is the same as $2x + 4y = 40$).
6. **Evaluate objective function at corner points:**
- At $(0,0)$:
$$Z = 3(0) + 5(0) = 0$$
- At $(20,0)$:
$$Z = 3(20) + 5(0) = 60$$
- At $(0,10)$:
$$Z = 3(0) + 5(10) = 50$$
7. **Conclusion:**
The maximum number of patients served is $60$ when the health center prepares $20$ Basic Care Kits and $0$ Emergency Care Kits.
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**Final answer:**
$$x=20, \quad y=0, \quad \text{maximum patients served} = 60$$