Lesson Minimax linear problems to solve MENTALLY based on common sense
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<H2>Minimax linear problems to solve MENTALLY based on common sense</H2> In this lesson, I collected so called " false LP-problems ". From the first glance, they look like the problems to be solved by the Linear Programming method. But in fact, they are a class of entertainment problems that can be solved mentally by applying the common sense. The goal of this lesson is to teach you to distinct this difference between real LP-problems and joking and entertainment " false LP-problems". <H3>Problem 1</H3>A farmer has a field of 70 acres in which he plants potatoes and corn. The seed for potatoes costs $20 per acre, the seed for corn costs $60 per acre, and the farmer has set aside $3000 to spend on seed. The profit per acre of potatoes is $150 and the profit per acre of corn is $50. How many acres of each should the farmer plant? What is the maximum profit? <B>Solution</B> <pre> The seed of potatoes ($20/acre) is cheaper than the seed of corn ($60/acre). The profit per acre of potatoes ($150) is greater than the profit per acre of corn ($50). It makes it clear that the most aggressive strategy should be applied, which is to plant the potatoes as much as possible. There are enough money to buy the potatoes seed for 70 acres: 70*20 = 1400 dollars. So, the optimal strategy is to plant 70 acres of potatoes and do not plant the corn, at all. <U>ANSWER</U>. The farmer should plant 70 acres of potatoes and do not plant the corn, at all. The profit is 70*150 = 10500 dollars. </pre> <H3>Problem 2</H3>A farmer is going to divide her 60 acre farm between two crops. Seed for crop A costs $20 per acre. Seed for crop B costs $40 per acre. The farmer can spend at most $1400 on seed. If crop B brings in a profit of $270 per acre, and crop A brings in a profit of $120 per acre, how many acres of each crop should the farmer plant to maximize her profit? <B>Solution</B> <pre> Seed for crop B is twice as expensive as seed for crop A, but gives the profit which is more than twice as big as the profit of crop A (compare 270/120 = 2.25 with 40/20 = 2 !). +---------------------------------------------------------------+ | THEREFORE, it is clear, that the most aggressive strategy | | is to sow as much area with crop B as possible. | +---------------------------------------------------------------+ The possible area to sow with crop B, within the budget, is 1400/40 = 35 acres. So, the most profitable solution is to sow 35 acres with crop B and do not sow crop A, at all. The expected profit is then 35*270 = 9450 dollars, and it is maximal possible profit at given conditions. </pre> <H3>Problem 3</H3>A fertilizer producing company has two plants where the product is made. Plant A can make at most 30 tons per month and plant B can make at most 40 tons per month. The company wants to make at least 50 tons per month. The process produces a particulate matter in the atmosphere as observed over a nearby town. It is found that 20 pounds of the particulate matter for each ton made by plant A and 30 pounds for each ton of the product made at plant B. How many tons should be made each month from each plant to minimize the amount of particulate matter in the atmosphere. <B>Solution</B> This problem is for common sense and mental solution. <pre> From the condition, it is clear that as many as possible production should be made at the plant A, since it is more ecologically clean, and the rest can be (and should be) made at the plane B. <U>ANSWER</U>. 30 tons at plant A and the rest 20 tons at plant B. </pre> <H3>Problem 4</H3>A computer company has two type of laptop products, its flagship TURBO and the newer laptop called DELUXE. Every unit of Turbo has a profit of P1,000 while every unit of Deluxe has a profit of P6,000. The demand is limited to at most 200 Turbo and at most 300 of Deluxe. The current workforce can produce not more than 400 unit of both models of laptop. How much of each should it produce to maximize profits? <B>Solution</B> There are two ways to solve this problem. One way is to place the problem into the framework of usual/typical linear programming problem and solve it formally. Another approach is to apply common sense. I will present the solution based on common sense. <pre> The right strategy is the most aggressive strategy to maximize profit: It is to produce and cell as many as possible product units, what bring the maximum profit, and then add to it as many as possible less profitable units, to be and to remain in the frame of given restrictions. Following this strategy, the company should produce 300 units of DELUXE and 400-300 = 100 units of TURBO. It will have then the maximum profit of 300 * P6000 + 100 * P1000 = P 1,900,000. <U>ANSWER</U> </pre> <H3>Problem 5</H3>Tala is a semi-professional photographer in her spare time. As a wedding photographer she can earn 10 Jordan Dinars (JOD) per hour. As a studio photographer she can earn 8 JOD per hour. She wants to spend at least 6 hours as a studio photographer. She can only work for a maximum of 20 hours a week. Find her maximum possible weekly earnings. <B>Solution</B> <pre> Tala wants to earn maximum money. For it, she should work at the greater hour salary as long as possible: as long as the restrictions do allow. The restrictions allow her to work 20-6 = 14 hours at the rate of 10 JOD per hour. The remaining 6 hours she wants to work at the rate of 8 JOD per hour. It is the solution and the <U>ANSWER</U> to the problem: At given conditions, the maximum earning is 14*10 + 6*8 = 188 JOD. It is achieved at 14 hours working at 10 JOD per hour and 6 hours working at 8 JOD per hour. </pre> <H3>Problem 6</H3>A company makes two types of biscuits: Jumbo and Regular. The oven can cook at most 400 biscuits per day. Each Jumbo biscuit requires 2 oz of flour, each Regular biscuit requires 1 oz of flour, and there is 600 oz of flour available. The profit from each Jumbo biscuit is $0.07 and from each Regular biscuit is $0.12. How many of each size biscuit should be made to maximize profit ? What is the maximum profit ? How many Jumbo and how many Regular biscuits the company should make ? <B>Solution</B> <pre> As you read the problem, you see that each Jumbo biscuit requires more flour than each Regular biscuit, but provides lesser profit than Regular biscuit. It means that the most aggressive strategy should work, when the company produces Regular biscuits ONLY and does not produce Jumbo biscuits, at all. It is so OBVIOUS, that I will not spend my and your time for explanations. And since we just chose the strategy, we simply need to divide 600 oz by 1 to get the number of possible Regular biscuits: it is 600/1 = 600. But at this point, other restriction turns on: the number of biscuits can not be greater than 400. +---------------------------------------------------------------+ | Thus the optimal solution/answer is as follows: | | | | 400 Regular biscuits should be produced; no Jumbo biscuits; | | optimal (maximum) profit is 400*0.12 = 48 dollars. | +---------------------------------------------------------------+ </pre> <H3>Problem 7</H3>A factory makes two types of beds, type A and type B. Each month, a number of type A and a number of type B are produced. The following constraints control monthly production: No more than 50 beds of Type A and no more than 40 beds of type B can be made. At least 60 beds in all must be made. The maximum number of beds that can be produced is 80. The profit on type A is Php300 and on type B is Php150. How many beds on both types must be produced to maximize the profit? What is the maximum profit? ~~~~~~~~~~~~~~~~~~~~ <B>Solution</B> In this problem, both the solution and the answer can be obtained {{{highlight(highlight(MENTALLY))}}}, since they both are {{{highlight(highlight(OBVIOUS))}}}, from the common sense point of view. <pre> The solution for maximum profit is OBVIOUS: apply the most aggressive strategy and make as many A-beds as possible under given restrictions, i.e. 50 A-beds (since they create the maximum profit) and complement it by 30 B-beds, making them as many as possible, still remaining under the given restrictions. So, the answer in this case is (50 A and 30 B). </pre> <H3>Problem 8</H3>Sophia is organizing a movie night and wants to provide popcorn and soda to her guests. A bag of popcorn costs her $2 to buy, and a can of soda costs her $1. She only has room to store 24 cans of soda and 50 bags of popcorn at her house. She needs to have at least 60 items total to satisfy her guests. What is the least amount of money she can spend on popcorn and soda? <B>Solution</B> The problem can be solved using logical reasoning and common sense. <pre> Sophia wants to minimize her spending, still satisfying all constraints. Let's apply the most economic strategy. Sophia should take as much cheep items (sodas) as possible (24 cans), and then complement it with an appropriate number of more expensive items (popcorn bags). So, she needs 60-24 = 36 popcorn bags. Her spending will be $2*36 + $1*24 = $96, and, obviously, it is the minimum spending. Notice that this solution satisfies all the constraints. </pre> <H3>Problem 9</H3>A picnic basket can hold a maximum of 30 apples and 20 sandwiches. Each apple takes up 1 unit of space, and each sandwich takes up 2 units of space. The picnic basket has a total capacity of 60 units of space. What is the maximum number of apples and sandwiches that can be packed in the basket? <B>Solution</B> This problem can be solved using logical reasoning and common sense. <pre> We want maximize the number of apples and sandwiches (more precisely, the sum of them), satisfying the other given constraints. Let's apply the most aggressive strategy. We will take as much apples as possible (30 apples), since they take only 1 unit of space, each. Then we complement it with an appropriate number of sandwiches, which is {{{(60-30)/2}}} = 15 (the remaining space divided by 2). So, the solution is 30 apples and 15 sandwiches, which gives the total of 30 + 15 = 45 items. Notice that this solution satisfies all the constraints. </pre> <H3>Problem 10</H3>Sarah wants to plant a combination of tulips and roses in her garden. Each tulip requires 2 square feet of space, and each rose requires 3 square feet of space. The garden has a total area of 30 square feet. Sarah does not want more tulips than roses in her garden. What is the maximum number of tulips and roses? <B>Solution</B> I will solve it using logical reasoning plus common sense. <pre> We want maximize the number of flowers in the garden (= maximize the sum of tulips and roses), satisfying the imposed constraints. Let's apply the most aggressive strategy. Since the number of roses should be greater than or equal to the number of tulips, we will take as much pairs (rose, tulip) as possible. Let the number of pairs be p >= 0. Each such pair requires 2+3 = 5 square feet of the garden area. Then we will complement the pairs with separate roses. Let "r" be the numbers of separate roses. Now the restriction regarding the area is 5p + 3r <= 30 square feet. The pairs will provide that the number of roses is not less than the number of tulips; and each pair provides 2 flowers; so we are interested to maximize p and minimize r >= 0. Thus, in this inequality, we want to have integer r as small as possible, and due to this, to have p as great as possible. Take r = 0. Then p = 30/5 = 6. So, the optimal solution under given restriction is to plant 6 roses and 6 tulips. Thus, the solution is 6 roses and 6 tulips, which gives the total of 6 + 6 = 12 flowers. Notice that this solution satisfies all the constraints. </pre> <H3>Problem 11</H3>Blue strand wire rope is the top seller of the company. TSD has to make a production schedule. To create product a blue strand wire rope, 3 machines are need. For wire rope #1, it needs 19 coils of wires and 3 hours of production, while for wire rope #2, 14 coils of wires and 4 hours of production. Wire rope #1 is sold for PhP 64,500 while wire rope #2 is sold for PhP 10,500. How many rolls of wire rope #1 and #2 does TSD has to schedule in the production to maximize the monthly sales while being limited to 6 machines, 2,006 coils of wires and 315 hours production? <B>Solution</B> From the first glance, this problem is the kind of the Linear Programming optimization. But in reality, it is for totally different method of solution: it is to apply the COMMON SENSE. <pre> First, let's organize the given info into this TABLE. m-hours stands for machine-hours. one product one product of wire roop #1 of wire roop #2 3 machine 3 machine 19 coils of wires 14 coils of wires 3 m-hours production 4 m-hours production sold for PhP 64500 each sold for PhP 10500 each Resources: 2006 coils of wires and 315 machine-hours. Let six machines be A, B, C, D, E and F. So, we have two triples of machines ABC and DEF. The rate of making money is three machines making product #1: {{{64500/(19*3)}}} = 1131.579 of PhP per coil per hour. three machines making product #2: {{{10500/(14*4)}}} = 187.5 of PhP per coil per hour. It makes it clear that the most aggressive strategy is THIS +-------------------------------------------------------------------------------------------+ | 3 machine (A,B,C) work as long as possible and produce as many wire roop #1 as possible. | | (To produce one product #1, they consume 19 coils and 3 m-hours). | | | | 3 machine (D,E,F) work as long as possible and produce as many wire roop #1 as possible. | | (To produce one product #1, they consume 19 coils and 3 m-hours). | +-------------------------------------------------------------------------------------------+ Thus, doing this way, 6 machines produce 2 products #1 and consume 19+19 = 38 coils and 3+3 = 6 m-hours. Which resource will run out first - machine-hours or coils ? coils: 2006/38 = 52 + 30/38 ( <<<---=== division with the remainder ) m-hours: 315/6 = 52 1/2 ( <<<---=== division with the remainder ) Thus, coils of wires will run out first. +--------------------------------------------------------------------------------+ | Let's call this cycle "consuming 3+3 = 6 m-hours, 6 machines produce | | 2 products #1 and consume 19+19 = 38 coils of wires" | | as "one basic cycle". | +--------------------------------------------------------------------------------+ Thus we found out that the "basic cycle" can be repeated (cycled) 52 times. After that, the remaining resources are 30 coils of wires and 315 - 6*52 = 315 - 312 = 3 m-hours. With these remaining resources, we can (and we will) run only 3 machines ABC to produce one #1 product. They will spend 19 coils of wire and 3 m-hours. The remaining resource is 30-19 = 11 coils, and the time is just used ion full. The final conclusion and the <U>ANSWER</U> is that we run two triples of the machines only to produce 2*52 + 1 = 105 of products #1; we do not make product #2, at all. The earning of this strategy is 105*64500 = 6772500 dollars. The remaining unused resource is 2006 - 19*105 = 11 coils of wires; the time is used in full. </pre> <H3>Problem 12</H3>A western shop wishes to purchase 470 felt and 370 straw cowboy hats. Bids have been received from three wholesalers. Texas hatters has agreed to supply not more than 370, Lone Star Hatters not more than 463, and Lariat Ranch Wear not more than 185. The owner of the shop has estimated that his profit per hat sold from Texas Hatters would be $3/felt and $4/straw, from Lone Star Hatters would be $3,80/felt and $3,50/straw and from Lariat Ranch Wear $4/felt and $3,60/straw. Solve this to maximize the owner's profit, and calculate that profit. <B>Solution</B> <pre> To present information in compact form, I prepared this table T A B L E Texas Lone Lariat To purchase felt 3 3.80 4 470 cowboy 4 3.50 3.60 370 maximum 370 463 185 Now imagine that you are aggressive buyer/seller, who wants to get maximum profit. You look at the table, and you see that among felt hats, you will get maximum profit of $4/hat from Lariat. So you take as many felt hats as possible from Lariat: 185. After that, you still need 470-185 = 285 felt hats. You look at the Table again, and you see that next the most profitable felt hat is from Lone at $3.80 - so you buy these 285 felt hats from Lone: the Lone's restriction of 463 hats allows you to do it. At this point, you completed with felt hats and provided a profit of 185*4 + 285*3.80 = 1823 dollars. Next, we will work/analyze/buy cowboy hats. You look at the table again, and you see that among cowboy hats, you will get maximum profit of $4/hat from Texas. So you take as many cowboy hats as possible from Texas: 370. After that, you just completed with the cowboy hats and provided your profit of 370*4 = 1480 dollars. RESUME: you buy 185 felt hats from Lariat and 285 felt hats from Lone; you buy 370 cowboy hats from Texas. Doing this way, you provide total profit of 1823 + 1480 = 3303 dollars, and it is maximum possible (OPTIMUM) profit under imposed conditions. </pre> <H3>Problem 13</H3>Maximize P = 3x-y subject to 4x+y <= 16, x+2y <= 9, x, y ≥ 0 using the simplex method. <B>Solution</B> This problem is a kind of joke problems on Simplex Method and can be solved MENTALLY, as I will show it to you in this post. Read it to the end. <pre> The feasible domain for this problem is a quadrilateral in QI under straight line 4x + y = 16 (1) and under straight line x + 2y = 9. (2) It has vertices A = (0,0), B = (0,4.5) ( y-intercept of line (2) ) C = ({{{23/7}}},{{{20/7}}}) ( intersection of lines (1) and (2) ) D = (4,0) ( x-intercept of line (1) ) Next, the objective function increases as x increases and decreases as y increases. +------------------------------------------------------+ | THEREFORE, the objective function has the maximum | | in the feasibility quadrilateral | | when x is the maximum and y is the minimum. | +------------------------------------------------------+ This point is, OBVIOUSLY, point D, where (x,y) = (4,0) and P(x,y) = 3x-y = 3*12 - 0 = 12. <U>ANSWER</U>. The maximum point is (4,0) with the objective function value of 12. </pre> <H3>Problem 14</H3>TMA manufactures 37-in. high-definition LCD televisions in two separate locations: Location I and Location II. The output at Location I is at most 5,800 televisions/month, whereas the output at Location II is at most 5,100 televisions/month. TMA is the main supplier of televisions to Pulsar Corporation, its holding company, which has priority in having all its requirements met. In a certain month, Pulsar placed orders for 2,900 and 4,000 televisions to be shipped to two of its factories located in City A and City B, respectively. The shipping costs (in dollars) per television from the two TMA plants to the two Pulsar factories are as follows: <pre> To Pulsar Factories City A City B From TMA Location I $5 $5 From TMA Location II $8 $8 </pre>TMA will ship x televisions from Location I to City A and y televisions from Location I to City B. Find a shipping schedule that meets the requirements of both companies while keeping costs, C (in dollars), to a minimum. <B>Solution</B> <pre> Since the shipping cost per unit from Location I to A and to B is the same $5, and since the shipping cost per unit from Location II to A and to B is the same $8, it MEANS that the most reasonable strategy is (1) to ship as many TVs as possible from Location I (5800) for cheaper shipping price to cities A and B, (2) and then to ship the rest of TVs from Location II (2900 + 4000 - 5800 = 6900 - 5800 = 1100) for more expensive shipping price to cities A and B. How these quantities 5800 from Location I and 1100 from Location II will be distributed between the cities A and B, does not matter for the total shipping cost. Such strategy provides the minimum total shipping cost. </pre> <H3>Problem 15</H3>The dean of Faculty of Management and Information Technology must plan the faculty’s course offerings for the first semester, 2020/2021. Student demands make it necessary to offer at least 30 undergraduate and 20 graduate courses in the term. Faculty contracts also dictate that at least 60 courses be offered in total. Each undergraduate course taught course the faculty an average of RM2,500 in faculty wages, and each graduate course costs RM3,000. Find the number of undergraduate and graduate courses should be taught in the first semester so that the total faculty salaries are kept to a minimum. <B>Solution</B> Surely, this problem can be solved using Linear Programming method. But it also can be solved, using logical reasoning and common sense, practically mentally, in your mind. <pre> Let U be the number of undergraduate courses, and G be the number of graduate courses. First, notice that if the conditions "at least 30 U and at least 20 G" are satisfied and we have MORE than 60 courses, than the total number of courses can be diminished by 1 such way that we still will be in the feasible domain. Indeed, if U >= 30 and G >= 20, and U + G > 60, it means that at least one number U or G is greater than their corresponding lower boundaries, so this course can be eliminated. Repeating this reasoning as many times as required, we will get the condition U >= 30, G >= 20, U + G = 60 so the total number of courses is precisely 60. Eliminating excessive courses, we diminish the total cost, so it is good, from the problem's point of view. Now, if the number of undergraduate courses is U, then the number of undergraduate courses is U = 60-G. The total cost is then 2500*U + 3000*G = 2500*(60 - G) + 3000*G = 150000 + 500*G, and we want to make it as small as it is possible under restrictions. From the last formula, it is clear that the total cost is minimum when the number of graduate courses G is at its lower bound G = 20. Thus the optimal solution is G = 20 graduate courses and U = 60-20 = 40 undergraduate courses, giving the minimum possible cost 40*2500 + 20*3000 = 100000 + 60000 = RM160,000. </pre> The idea is to assign as few of expensive courses as possible under restrictions, and then to add as many of cheaper courses to get other restrictions. <H3>Problem 16</H3>An oil refinery refines types A and B of crude oil and can refine as much as 4000 barrels each week. Type A crude has 2 kg of impurities per barrel, type B has 3 kg of impurities per barrel, and the refinery can handle no more than 9000 kg of these impurities each week. How much of each type should be refined in order to maximize profits, if the profit is R25/barrel for type A and R30/barrel for type B? <B>Solution</B> <pre> Let x be the amount of the oil type A to refine (in barrels). Then the amount of oil type B to refine is 4000-x barrels. We want to maximize profit P(x) = 25x + 30*(4000-x) = 25x + 120000 - 35x = 120000 - 5x. (1) under this restriction 2x + 3*(4000-x) =<= 9000, or 2x + 12000 - 3x <= 9000, or x >= 3000. (2) +------------------------------------------------------+ | According to (1), we want to subtract from 120000 | | the minimal possible value of 5x. | | | | According to (2), x must be at least 3000. | +------------------------------------------------------+ So, it is logical to take for x the minimal possible value, which is 3000. So, the <U>ANSWER</U> to the problem is this: +-----------------------------------------------------------------------+ | Under given restrictions, the optimal solution is | | | | x = 3000 barrels for oil type A | | and 4000-x = 4000-3000 = 1000 barrels of oil type B. | | | | The profit is then P = 25*3000 + 30*1000 = 75000 + 30000 = 105000. | +-----------------------------------------------------------------------+ </pre> <H3>Problem 17</H3>A company has $14,830 available per month for advertising. Newspaper ads cost $190 each and can't run more than 24 times per month. Radio ads cost $590 each and can't run more than 32 times per month at this price. Each newspaper ad reaches 5700 potential customers, and each radio ad reaches 6700 potential customers. The company wants to maximize the number of ad exposures to potential customers. Determine the most profitable / (effective) way to do it. <B>Solution</B> <pre> To find the maximum number of ad exposures, let's formulate the problem in terms of objective function and constraints. Let x be the number of newspaper ads and y be the number of radio ads. The objective function is P = 5700*x + 6700*y. (1) It is the number of possible expositions, and we want to maximize it. The constraints are: 190*x + 590*y ≤ 14830 (the budget), (2) x ≤ 24 (newspaper ad limit), y ≤ 32 (radio ad limit), (3) x ≥ 0, y ≥ 0 (non-negativity). (4) Now it is presented as a typical Linear Programming problem. But it can be easily solved MENTALLY using "the most aggressive" logical strategy/methodology. From expression (1) for the objective function, we see that the contribution of each single newspaper ad (in terms of the number of potential expositions, 5700) is comparable with (or distinct insignificantly from) the contribution of each single radio ad (6700). But each newspaper ad is much cheaper ($190) than each radio ad ($590). So, it is clear that the most profitable strategy is to make as many newspaper ads as possible (x=24), and then to spend the rest of the budget for the radio ads. Thus the most effective solution is to make 24 newspaper ads, spending 24*190 = 4560 dollars for it. The rest of the budget is then $14830 - $4560 = #10270. This amount can be / (should be) spent for radio ads. It provides the number of radio ads y = {{{10270/590}}} = 17.40678, and we should round this decimal number to the closest lesser integer number, which is y = 17. So, the answer to the problem's question is THIS: 24 newspaper ads and 17 radio ads provide the greatest possible number of expositions (~ potential customers), which is then 5700*24 + 6700*17 = 250700. </pre> <H3>Post-solution note</H3> In this concrete problem, the presented method has one important advantage comparing with the traditional form geometric solution of Linear Programming problems. Working in the frame of traditional Linear Programming geometric method, you will get the solution with non-integer decimal for 'y', so, you will be forced to complete your traditional solution with other arguments. Working in the way, presented here in the solution above, you will get the answer in integer numbers in logical form without any complications. <H3>Problem 18</H3>A box company makes small and large wooden boxes. Small boxes require 0.8 square meters of wood, while large ones require 1.4 square meters. All boxes require 0.5 hours of labor, regardless of size. Wood is limited to 42 square meters, and only 24 hours of labor are available. Due to warehouse space limitations, no more than 20 large boxes can be made each day. Also, demand by customers for small boxes is limited to a maximum of 30 boxes. Each small box yields a profit of R42.00 and each large box earns only R14.00. Find the strategy to maximize the profit. <B>Solution</B> This minimax problem is very special. It is special, because it can be easily solved MENTALLY, based on common sense ONLY, without using heavy artillery technique of linear programming. <pre> Indeed, after reading the post, it should be clear that the winning strategy is the most aggressive strategy making SO MANY small boxes as possible, until the restrictions allow do it; and when the restriction on small boxes is reached, then to switch making large boxes, until the restrictions on large boxes is reached. It is so, because the profit of each small box is greater than the profit of each large box (R42 against R14), while each small box requires less amount of material than each large box (0.8 sq. m against 1.4 sq. m). So, it is OBVIOUS that to produce small boxes is more profitable. The other restrictions (total wood and total working time) do not give a preference to any sort of production, and the limitation on the numbers of small boxes and large boxes (like the warehouse space limitation and the demand) are in favor of small boxes. So, following to this idea, as many of small boxes should be produced, making 30 small boxes. (limited by the demand). It will require 0.8*30 = 24 sq.m of wood, leaving 42-24 = 18 sq.m of wood for large boxes and will require 0.5*30 = 15 hours of labor, leaving 24-15 = 9 hours for large boxes. Now we determine the number of large boxes. The limitations are 18/1.4 = 12.86 for wood; 9/0.5 = 19 from labor hours; 20 from the warehouse space limitations. So, for large boxes the number is 12. The maximum profit is 30*42 + 12*14 = 1428 monetary units. <U>ANSWER</U>. The optimal strategy is to make 30 small boxes and 12 large boxes, making the maximum possible profit of R1428. </pre> My other additional lessons on Miscellaneous word problems (section 3) in this site are - <A HREF=https://www.algebra.com/algebra/homework/word/misc/More-complicated-problems-on-finding-number-of-elements-in-finite-subsets.lesson>More complicated problems on finding number of elements in finite subsets</A> - <A HREF=https://www.algebra.com/algebra/homework/word/misc/Solving-problems-by-the-Backward-method.lesson>Solving problems by the Backward method</A> - <A HREF=https://www.algebra.com/algebra/homework/word/misc/Solving-linear-optimization-problems-by-reduction-to-linear-function.lesson>Solving linear optimization problems without LP-method by reduction to linear function</A> - <A HREF=https://www.algebra.com/algebra/homework/word/misc/Solving-one-special-linear-minimax-problem-in-100-D-space.lesson>Solving one special linear minimax problem in 100-D space by the Linear Programming method</A> - <A HREF=https://www.algebra.com/algebra/homework/word/misc/Miscellaneous-logical-problems.lesson>Miscellaneous logical problems</A> - <A HREF=https://www.algebra.com/algebra/homework/word/misc/Upper-class-entertainment-Math-problems-for-all-ages.lesson>Upper class entertainment Math problems for all ages</A> - <A HREF=https://www.algebra.com/algebra/homework/word/misc/OVERVIEW-of-my-additional-lessons-on-Miscellaneous-word-problems-section-3.lesson>OVERVIEW of my additional lessons on Miscellaneous word problems, section 3</A> Use this file/link <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A> to navigate over all topics and lessons of the online textbook ALGEBRA-I. Use this file/link <A HREF=https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-II - YOUR ONLINE TEXTBOOK</A> to navigate over all topics and lessons of the online textbook ALGEBRA-II.
