1) a contractor builds boathouses in two basic models: theatlantic and the pacif

Question

1) a contractor builds boathouses in two basic models: theatlantic and the pacific. each atlantic model requires $1000 feetof framing lumber, $3000 cubic feet of concrete, and $2000 foradvertising. each pacific model requires $2000 feet of framinglumber, $3000 cubic feet of concrete, and $3000 foradvertising. contracts call for using at least $8000 feet offraming lumber, $18000 cubic feet of concrete and $15000 worthof advertising. if the total spent on each atlantic model is $3000and the total spent on pacific model is $4000. how many odeach model should be built to minimize costs?

Explanation / Answer

Not sure what you mean by "If the total spent on each AtlanticModel is $3,000 and the total spent on Pacific Model is $4,000, howmany of each model should be built to minimize costs" - you alreadyestablished that each Atlantic Model costs at least $6,000, andeach Pacific Model costs at least $8,000. That said, we have the following equations: AM = 1000L, 3000C, 2000A PM = 2000L, 3000C, 3000A If you have to spend at least 8000L, 18000C and 15000A, what thequestion is really asking is: xAM + yPM >= 8000L, 180000, 15000A So, you have the following equations [1.1] x(1000L) + y(2000L) = 8000L, or x + 2y = 8 [1.2] x(3000C) +y(3000C) = 18000C, or 3x + 3y = 18; x + y =6 [1.3] x(2000A) +y(3000A) = 15000A, or 2x + 3y = 15 You need to solve for the x and y among the three possibilities(using 1.1 and 1.2, using 1.2 and 1.3 and using 1.1 and 1.3) tofind the minimum x and y that totals at least 8000L, 18000C and15000A (and that is closest to $41000 total spent) From the equations [1.2] and [1.3], x + y = 6 2x + 3y = 15 So, x = 6 - y In equation [1.3], that means 2(6-y) + 3y = 15 12 -2y + 3y = 15 12 + y = 15 y = 3Therefore x = 3. By building 3 AM and 3PM you meet the minimum contract requirements(overspending an extra $1000 on lumber:) 3 AM = 3000L, 9000C, 6000A 3 PM = 6000L, 9000C, 9000A Total spent: 9,000L, 18,000C and 15, 000A, for a total of$42000 (Solving for x and y using equations [1.1] and [1.2], or [1.1] and[1.3] gives you different results and these are not the costminimizing proportions): From [1.1] and [1.2], you get: [1.1]    x + 2y = 8 and [1.2]    x + y = 6, therefore, y = 6 - x Inserting this into [1.1], you get: x + 2(6-x) = 8 x + 12 - 2x = 8 12 - x = 8 x = 4, so y = 2. The costs for 4AM and 2PM would be 4000L, 12000C, 8000A and 4000L,6000C and 6000A, respectively, or a total of 8000L, 18000C and14000A. This is 1000A less than what the contract calls for(15,000A). Therefore, you need to build at least one more AMor PM. If you build another PM, you could have built one lessAM, as we showed in the first solution set (i.e. 3AM and 3PM ischeaper than 4AM and 3PM). If you build another AM, you wouldbe spending an additional 6000 (1000L, 3000C and 2000A, when youonly needed 1000A total.   (i.e. you would have to spenda total of $46,000) Solving [1.1] and [1.3] [1.1]    x + 2y = 8, or x = 8 - 2y [1.3]    2x + 3y = 15 So, in [1.3], we get 2(8-2y) + 3y = 15 16 - 4y + 3y = 15 16 - y = 15 y = 1, so x = 6 That means you're making 6AM and 1PM, or 6000L, 18000C, 12000A and2000L, 3000C, 3000A, respectively, for a total of8000L, 21000C and 15000A. While this meets the contract,you've overspent by 3000C above what was necessary ($44,000). Therefore 3AM and 3PM is an aggregate of $2,000 cheaper than 6AMand 1PM.

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