Russ P. answered • 10/25/14

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Michael,

Constraints usually apply to limitations on resources: labor, materials, production capacity, investment, etc. So let's make a table of the data in this problem to make things more obvious:

Flexo Velocity Constraints

$ 200 $ 200 None on the racquet selling prices

$ 60 $ 50 None on the profit contribution of each racquet type

f t Unknown #s of fiberglass Flexos (f) and titanium Velocitys (t) to produce

4 oz None Fiberglass material per Flexo, so constraint is f <= 5000/4 = 1250 Flexo units maximum possible

None 2 oz Titanium material per Velocity, so constraint is t <= 1000/2 = 500 Velocity units max possible

x 2x Unknown x, number of labor hours for a Flexo, so Labor constraint is x(f + 2t) <= 3500 hrs.

$ 20/hr $10/hr Direct labor cost per hour in making each racquet

? ? Material costs for the Fiberglass and Titanium per ounce - Not specified in problem

Total Profit P(f,t) from making and selling f Flexos and t Velocitys is just the sum of their units times profit contributions.

P(f,t) = 60f + 50t

So you want to maximize P(f,t) subject to following constraints:

From availability of materials : f <= 1250 Flexos; t <= 500 Velocitys

From labor hours on same production line: x(f + 2t) <= 3500 total hours

From the unknown x = # hours needed to make a Flexo. (x = 7 hrs, see below)

This last one comes from looking at the profit contribution, p, of each Flexo:

Flexo profit = selling price - labor cost - material cost

p = $60 = $200 - (20/hr)(x hrs.) - Unspecified (so take it as zero)

Therefore, x = 140/20 = 7 hrs to make a Flexo. Note: this overstates it since material cost had to be assumed zero.

The Velocity takes 2x or 14 hours to make.

The triangular solution space is all positive values of f & t subject to the materials and labor constraints:

0<=f<=1250, 0<=t<=500, and 0<= (f + 2t) <= 500. This last labor constraint is most limiting of all. Labor limits Flexos (when t=0) to 500 whereas the material limitation is at 1250. Similarly, Velocity units are limited by labor (when f=0) to 250 whereas the material limitation is 500.

Thus, the maximum total profit P(f,t) will occur somewhere on that (f + 2t) = 500 line

Hence, P(f,t) = 60f + 50t = 60(500 - 2t) + 50t = 30,000 - 120t + 50t = 30,000 -70t

For maximum, t = 0 units & f = 500 units.

Since, the profit contribution of each t is only $50 which is less than the Flexo's $60, you want to produce as many Flexos as possible subject to the constraints on materials and labor, and 500 units meets both those constraints.

It's pretty straightforward when you organize the data in table form.