2.An industrial recycling center uses two scrap aluminum metals, A and B, to produce a special alloyScrap A contains 6% aluminum, 3% silicon, and 4% carbonScrap B has 3% aluminum, 6% silicon and 3% carbonThe costs per ton for scraps A and B are $100 and $80, respectivelyThe specifications of the special alloy are as follows:1.The aluminum content must be at least 3% and at most 6%2.The silicon content must lie between 3% and 5%3.The carbon content must be between 3% and 7%Formulate a linear program that can be used to determine the amounts of scrap A and B that should be used to minimize the cost of creating 1000 tons of the special alloy.
What is the test? Because this is a tiny capacitor, perhaps 100 pF, and the amount of charge is also tiny, specially at the low 9 volts so it all depends on the testBut don't expect muchYou would need a special, very high impedance voltmeter to measure the voltageYou will not get a spark, the voltage is too lowAnd the charge is too low for any current to register on a ammeter, unless it is very sensitiveedit: the other problem is your handsIf you touch the terminals at any time, that will discharge the cap immediatelyOr if you touch the metal parts of the clip leads while connected to the cap.
because you touch yourself at night.
You have two inputs of metal, which have been helpfully named A and BThe decision variables A and B here will be measured in tonsIt's important to get the units of measure explicit at the beginning, because it helps make clearer the steps that happen next, rather than just say loosely that A and B are the variablesThe cost is 100A + 80B, which is the objective function to minimizeYou could minimize the cost to zero, except that the number of tons to produce, which can come only from these two inputs, must be 1000, thus your first constraint is: A + B 1000 Now, the quality constraintsIn the final product, the aluminum content must be at least 3%If you are making 1000 tons, then that means at least 30 tons of aluminum is containedWhere can it come from? Well, each ton of scrap A contains 6% aluminum, or 0.06 tonsLikewise for scrap B, each ton contains 0.03 tons of aluminumMultiplying by the numbers of tons of the scraps, you have: 0.06A + 0.03B 30 There are five more constraints that are similar to this, and I'll leave it to you to completeThe point here is to have a repeatable methodology for formulating these problemsYou should follow this general pattern (specify what your decision variables mean, construct your objective function, then look around for the reasons you can't push that objective function to extremes) each time.
