Linear Programming Word Problem

Question

Mrs. Green wants to crochet beach hats and baby afghans for a church fundraising bazaar. She needs 7 hours to make a hat and 4 hours to make an afghan and she has 68 hours available. She wants to make no more than 14 items and no more than 11 afghans. The bazaar will sell the hats for $21 each and the afghans for $9 each. How many of each should she make to maximize the income for the bazaar?

Answer 1

X is the number of beach hats

Y is the number of afghans

 

7X + 4Y = 68 --->  Y = (68-7x)/4 = (-7/4)X + 17

X+Y <=14  -------> Y <= 14 - X = -x + 14

Y<=11

 

F(x,y) = $21X + $9Y is the revenue function

 

The region is bounded above by the line y=11.

The region is bounded to the right by the line y = -x +14

   Then by the linbe y = (-7/4)X + 17.

The region is bounded below by the x-axis and to the left

by the y-axis, since all quantities are positive and region

must be in the 1st quadrant.

 

(You can get a good picture of the critical region by

graphing these lines using a graphing calculator

or going to www.desmos.com and graphing them there)

 

The critical points are (0,11), (3,11), (4,10), abnd ( 68/7, 0)

 

Plugging these into the revenue function:

 

F(0,11) = 21*0 + 9*11 = 99

F(3,11) = 21*3 + 9*11 = 63+99 = 162

F(4,10) = 21*4 + 9*10 = 84 + 90 = 174

F(68/7,0) = 21*68/7 + 9*0 = 3*68 = 204

 

68/7 = 9 and 5/7 hats

 

She cannot make 10 hats,as that would exceed the 68 hour limit by 2 hours.

9 hats yields $189 in 63 hours.

With only 5 hours remaining, she can only make 1 afghan.

That would yield another $9, for a total of $198

 

9 hats and 1 afghan will yield $198 in 67 hours.