Question

Bright Future, Ltd (BF) is a nonprofit foundation providing medical treatment to emotionally distressed children. BF has hired you as a business consultant to design an employment policy that would be consistent with its goal of providing the maximum possible service given its limited financial resources. You have determined that the service (Z) provided by BF is a function of its medical staff input (M) and social worker staff input (S) which is given by: Z = M + .5S + .5 MS - S2 BF’s staff budget for the coming year is $1,200,000. Annual employment costs are $30,000 for each social worker staff member (S) and $60,000 for each medical staff member (M).

(a) Using the Lagrangean multiplier approach calculate the optimal (i.e., service maximizing) combination of medical and social worker staff. Determine the optimal amount of service provided by BF.

Please explain step by step how to get medical and social worker staff. Thanks!

Answer 1

It is a maximizing problem. Sevice provided by BF depends upon number of medical staff (M) and number of social worker (S). Cost of employment of M is $60,000 and for S $30,000. Total money available to hire them is $1,200,000. BF will maximize Z subject to budget costraint.

Thus problem is-

\textup{Maximize}-

Z=M+0.5S+0.5MS-S^2

\textup{Subject to}-

60,000M+30,000S=1,200,000

Multiply lagrangian constant lambda with budget constraint and add it with objective function to develop understated Lagragian function-

\phi =M+0.5S+0.5MS-S^2+\lambda \left ( 60,000M+30,000M-1,200,000 \right )

Partially derivate phi with respect to M and S and take them as zero to get-

\frac{\partial \phi }{\partial M}\phi =\frac{\partial }{\partial M}\left[M+0.5S+0.5MS-S^2+\lambda \left ( 60,000M+30,000S-1,200,000 \right )\right]=0

=1+0.5S+60,000\lambda =0

\therefore \lambda =-\frac{1+0.5S}{60,000}.............................(1)

Similarly,

\frac{\partial \phi }{\partial S}\phi =\frac{\partial }{\partial S}\left[M+0.5S+0.5MS-S^2+\lambda \left ( 60,000M+30,000S-1,200,000 \right )\right]=0

0.5+0.5M-2S+30,000\lambda =0

\therefore 30,000\lambda =2S-0.5M-0.5

\therefore \lambda =\frac{2S-0.5M-0.5}{30,000}......................(2)

Substitute lamda value of (1) in (2)-

\frac{-1-0.5S}{60,000} =\frac{2S-0.5M-0.5}{30,000}......................(3)

\frac{-1-0.5S}{2} =2S-0.5M-0.5

-1-0.5S =4S-M-1

4.5S =M

\therefore S =\frac{M}{9}

Substitute it in budget constraint-

60,000M+30,000\times \frac{M}{9}=1,200,000

540,000M+30,000M=10,800,000

570,000M=10,800,000

\therefore M=18.95

\therefore S=\frac{M}{9}=\frac{18.95}{9}=2.11

Therefore optimal amount of service preovided is-

Z=M+0.55+0.5MS-S^2

Z=18.95+0.5\times 2.11+0.5\times 18.95\times 2.11-2.11\times 2.11

Z=18.95+1.55+19.99-4.45=36.04

Answer: oiptimal combination is 18.95 (19 rounded off) numbers of M and 2.11(2 rounded off) numbers of S to provide optimal benefit of 36.04 (36 rounded off).