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It's me. I got an answer for the following question but it just looks wrong lol. It's a decimal.

In a factory, machine 1 produces 8-inch (in.) pliers at the rate of 60 units per hour (hr) and 6-in. pliers at the rate of 70 units/hr. Machine 2 produces 8-in. pliers at the rate of 40 units/hr and 6-in. pliers at the rate of 20 units/hr. It costs 50 dollars per hr to operate machine 1, and machine 2 costs 30 dollars per hr to operate. The production schedule requires that at least 240 units of 8-in. pliers and at least 140 units of 6-in. pliers be produced during each 10-hr day. Which combination of machines will cost the least money to operate?

This is what I got:

Let x be no. of units to produce in machine 1 and

Let y be no. of units to produce in machine 2

Objective Function: C=50x+30y

Constraints:

x ≥ 0

y ≥ 0

60x+40y ≥ 240

70x+20y ≥ 140

Then I found the intercepts, plotted the graph, and ended up with .5 and 5.15 as answers...that doesn't seem right. ._.

Any help is appreciated as always.

Regards,

Jason

icewolf461
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2 Answers2

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Double check your algebra. I got that the two production constraints intersect at $\left(\frac{1}{2},\frac{21}{4}\right)$. Non-integer answers can be interpreted as rates or frequencies. For instance, running half a machine doesn't make sense per se, but running one machine half the day would accomplish the same thing.

You should also double check that this corner is in fact the minimum of $C$. The other corners are $(4,0)$ and $(0,7)$.

Matthew Leingang
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Defining the variables is a really good idea. Unfortunately your definition is not right. Let´s have a look at your first constraint:

$60 \left[\frac{\text{units}}{\text{hour}}\right] \cdot x+40 \left[\frac{\text{units}}{\text{hour}}\right] \cdot y\geq 240 \ [\text{units}]$

The name of the unit has to be equal on both sides of the inequality. In this case the name of the unit on the RHS is just "units". Therefore $x,y$ have to have the unit "hour".

The optimal solution is $(x^*,y^*)=(0.5,5.25)$

Thus machine $1$ has to run $0.5$ hours and machine $2$ has to run $5.25$ hours. We can convert the decimal places of the run time of machine $2$ into minutes. $0.25$ hours are equal to $15$ minutes. It seems that you have mixed that up.

callculus42
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