Posted by **LIZ** on Friday, January 28, 2011 at 10:04pm.

The Megabuck Hospital Corp. is to build a state-subsidized nursing home catering to homeless patients as well as high-income patients. State regulations require that every subsidized nursing home must house a minimum of 770 homeless patients and no more than 900 high-income patients in order to qualify for state subsidies. The overall capacity of the hospital is to be 2,100 patients. The board of directors, under pressure from a neighborhood group, insists that the number of homeless patients should not exceed twice the number of high-income patients. Due to the state subsidy, the hospital will make an average profit of $9700 per month for every homeless patient it houses, whereas the profit per high-income patient is estimated at $7900 per month. How many of each type of patient should it house in order to maximize profit? HINT [See Example 3.] (If an answer does not exist, enter DNE.)

high-income patients 1Your answer is incorrect.

homeless patients 2Your answer is incorrect.

profit

- MATH -
**MathMate**, Friday, January 28, 2011 at 10:41pm
h=number of homeless patients

H=number of high-income patients

Constraints:

1. h≥770 (minimum)

2. H≤900 (maximum)

3. H≤2h (maximum ratio)

Capacity

4. H+h=2100

Z=profit = 9700h + 7900H

Obviously the objective function Z is maximized when h is at its maximum possible.

The response can be found by inspection, namely h=1400, H=700, which satisfies all 4 constraints.

If the answer is not obvious, try equal numbers (1050 each) and change for as many homeless as possible without violating any of the constraints.

- MATH -
**LIZ**, Friday, January 28, 2011 at 10:53pm
I had these as the equations but how do u set them up to find how many of each to maximise profit

- MATH -
**MathMate**, Friday, January 28, 2011 at 11:08pm
It depends on what your teacher expects you to know. Have you done simplex method, graphics, or trial and error?

You can plot the constraints on a graph as a line, and shade the non-feasible regions (usually on one side of the line). After all four constraints have been constructed and the appropriate sides shaded, the remaining unshaded region is the feasible region.

Evaluate the objective function at each of the corners of the feasible polygon, and find the maximum.

P.S.

I left out two other (trivial) constraints, namely h≥0 and H≥0.

- MATH -
**LIZ**, Friday, January 28, 2011 at 11:49pm
noo

- MATH -
**MathMate**, Saturday, January 29, 2011 at 9:12am
Sorry, there was a mistake in the formulation of the constraints.

The line H≤2h should have read h≤2H to reflect "the number of homeless patients should not exceed twice the number of high-income patients"

So a summary of the constraints now reads:

1. h≥770 (minimum)

2. H≥0 (minimum H)

3. H≤900 (maximum)

4. H≥h/2 (minimum ratio for H, maximum ratio for h)

5. H+h=2100 (capacity)

Z=profit = 9700h + 7900H

Z is to be evaluated at corners of the polygon.

Check:

(Broken Link Removed)

Legend:

red - max value of H

blue - capacity of 2100

green - min. ratio H/h

We see that the corners of the polygon of feasibility are at

(770,900), (1200,900), (1400,700), (770,385).

Evaluate the objective function Z at these points and make your choice.

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