The local salon offers floral hairbands for adults and for children. The salon manager can make a total of 50 hairbands each day. The adult hairbands use 5 flowers and the children hairbands only use 3 flowers. There are 75 flowers available. The adult hairbands give an income of $5.00 and the children hairbands give an income of $3.00. How many of each of the adult hairbands and children hairbands should be made to maximize income. Solution needs to include the following: Objective Function Constraints Graph Vertices
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Plot two lines on the same graph: A=50-C (constraint on numbers of hair bands) and A=(75-3C)/5 (from 5A+3C=75, the flower constraint), where A=number of adult bands and C=the number of child bands. The area under the first graph (a straight line joining the A intercept 50 and C intercept 50) includes the area under the second graph (A intercept 15, C intercept 25), so the more severe constraint is the second graph. Its area is enclosed by the axes and the line. The income equation R (revenue)=5A+3C matches the flower constraint, as it happens.

No matter how many bands of either type are made the income will always be $75 if all the available flowers are used. So 15 adult bands will fetch 15*5=$75 as will 25 child bands=25*3=$75. If we take 9 adult bands, for example, we need 45 flowers, leaving 30 flowers from which 10 child bands can be made. The income is $45+$30=$75. The total number of bands is 19, well below the capacity of 50. The number of child bands will always be a multiple of 5 and the number of adult bands a multiple of 3 to maximise income at $75 (C=0, 5, 10, 15, 20, 25 with A=15, 12, 9, 6, 3, 0, so A+C=15, 17, 19, 21, 23, 25).

The graph vertices are defined by the line A=(75-3C)/5 as (C,A)=(0,15), (0,0) and (25,0). These vertices enclose an area which is enclosed by the vertices (0,50), (0,0) and (50,0). These cover the constraints outlined in the question.

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