A regional emergency health network deploys two types of mobile medical units for event co… — GMAT

A regional emergency health network deploys two types of mobile medical units for event coverage: Rapid Response Units ( $R$ ) and Heavy Support Units ( $H$ ).

The deployment is governed by the following staffing and operational constraints:
- Each Rapid Response Unit ( $R$ ) requires $1$ paramedic and $2$ EMTs.
- Each Heavy Support Unit ( $H$ ) requires $3$ paramedics and $1$ EMT.
- On any given shift, a maximum of $21$ paramedics and $22$ EMTs are available.
- Operational policy mandates deploying at least $3$ Rapid Response Units ( $R \ge 3$ ) and at least $2$ Heavy Support Units ( $H \ge 2$ ).

Each Rapid Response Unit can treat $10$ patients per hour, and each Heavy Support Unit can treat $15$ patients per hour.

Match each operational metric on the left to its corresponding optimal value on the right that maximizes total patient treatment capacity per hour.

Optimal number of Rapid Response Units ( $R$ )9
Optimal number of Heavy Support Units ( $H$ )4
Maximum total patient treatment capacity per hour150

Cevap

The optimal operational configuration pairs 9 Rapid Response Units with 4 Heavy Support Units, achieving a maximum treatment capacity of 150 patients per hour.

The optimal pairing matches 9 Rapid Response Units, 4 Heavy Support Units, and a total capacity of 150 patients per hour. The optimal point

(9, 4)

fully utilizes both available resource pools (

1(9) + 3(4) = 21

paramedics and

2(9) + 1(4) = 22

EMTs), maximizing total throughput

10R + 15H

Adım Adım Çözüm

Set up the linear inequality constraints and objective function.

Paramedic resource constraint:

R + 3H \le 21

; EMT resource constraint:

2R + H \le 22

; Operational bounds:

R \ge 3, H \ge 2

; Objective function to maximize:

C = 10R + 15H

Translating the narrative conditions into algebraic forms defines the feasible region of operation.

Calculate the corner vertices of the bounded feasible polygon.

Solving

R + 3H = 21

and

2R + H = 22

simultaneously yields

R = 9

and

H = 4

. Other extreme vertices defined by lower bounds are

(3, 6)

(3, 2)

, and

(10, 2)

Linear programming optimization principles guarantee that the maximum objective value occurs at one of the corner vertices of the feasible region.

Evaluate the capacity objective function

C(R, H) = 10R + 15H

at each vertex.

(3, 2): 10(3) + 15(2) = 60

; At

(3, 6): 10(3) + 15(6) = 120

; At

(10, 2): 10(10) + 15(2) = 130

; At

(9, 4): 10(9) + 15(4) = 150

Comparing capacities across all vertices establishes

(9, 4)

as the global maximum.

Anahtar Kavram

Linear Programming Optimization under Bounded System Constraints

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Konular

Bu konudan daha fazla soru