This is what is called Linear Programming
Construct a 2by2 table with 2 rows labelled Model A or x, and Model B or y
2 columns labelled Line1 and Line2
in the first row enter 15/60 or 1/4 and 1/6 <----- converted to hours
in the 2nd row enter 1/6 and 1/5
equation 1: (1/4)x + 1/6y ≤ 25
times 12 ---> 3x + 2y ≤ 300
equation 2: (1/6)x + (1/5)y ≤ 22
times 30 ---> 5x + 6y ≤ 660
https://www.wolframalpha.com/input/?i=3x+%2B+2y+%E2%89%A4+300+%2C+5x+%2B+6y+%E2%89%A4+660
You have 3 critical points
- (100,0), the x-intercept of Line2
- (0,110), the y-intercept of Line1
- (60,60) , the intersection of the two lines
Profit = 14x + 11y
at (100,0), profit = 14(60) + 0 = 840
at (0,110), profit = 0 + 11(110) = 1210
at (60,60), profit = 14(60) + 11(60) = ....
So what do you think ?
Soundex produces x Model A radios and y Model B radios. Model A requires 15 min of work on Assembly Line I and 10 min of work on Assembly Line II. Model B requires 10 min of work on Assembly Line I and 12 min of work on Assembly Line II. At most, 25 labor-hours of assembly time on Line I and 22 labor-hours of assembly time on Line II are available each day. It is anticipated that Soundex will realize a profit of $11 on model A and $14 on model B. How many clock radios of each model should be produced each day in order to maximize Soundex's profit?
(x, y) =
What is the optimal profit?
1 answer