Question

Alex is a writer who writes poems and short stories. For an upcoming writer's workshop Alex wants to write some new works. He needs to determine how many poems and short stories he will have ready for the workshop. Alex figures that each poem will take him 30 hours and each short story will take him 70 hours. For the workshop Alex wants to have at least 4 poems and 3 short stories to display. If Alex has 840 hours of work time available, what solution represents the maximum number of works that he could write while staying in his time budget

Answer 1

Answer:

The maximum number of works that he can write while staying in his time budget is 24.

21 poems and 3 short stories

Step-by-step explanation:

In order to solve this problem we must first determine what our variables are. In this case it's the number of poems and short stories he can write.

p = # of poems

s = # of short stories

Next, we must build our objective function which will represent the total number of works he can write.

N=p+s

where N is the number of works.

Next, we must write the constrains based on the information provided by the problem.

The problem tells us that it takes him 30 hours to write a poem and 70 hours to write a short story and that he has 840 hours available to write them, so that constrain will be the following:

$30p+70s \leq 840$

it also tells us that he wants to write at least 4 poems and 3 short stories so there we have our other two constrains.

$p \geq 4$

$s \geq 3$

once we got our constrains we can go ahead and graph them to see how they will behave. (See attached picture)

In the graph p is the horizontal axis and s is the vertical axis.

On the graph we can see a polygon that is formed by the restriction. The vertices of the polygon will represent the optimal conditions for this linear programming problem. There are three optimal solutions there, so we need to test them to see which will return the greatest number of works he can write while keeping the given conditions.

Option 1:

4 poems and 3 short stories

N=4+3

N= 7 works

Option 2:

4  poems and 10 short stories

N=4+10

N=14 works

Option 3:

21 poems and 3 short stories

N=21+3

N=24 works

So the optimal solution will be given by option 3 with 21 poems and 3 short stories.