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PROBLEMI DI SCELTA in condizioni di certezza

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PROBLEMI DI SCELTA in condizioni di certezza

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scelta tra più alternative; scorte

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Look this video on supply, demand and equilibrium point

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Look this video on supply, demand and equilibrium point

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Price-Demand Equations and Equilibrium Points

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Supply and Demand Equations Introduction

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Uploaded by Alissa Carter on 2014-03-19.

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An introduction to the demand curve and factors that may cause a demand shift.

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Linear Demand and Supply Functions-model

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Linear Demand and Supply Functions-model

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An introduction to the supply curve and factors that may cause a shift in supply.

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An introduction to supply and demand, illustrated by the crossing of the supply curve and the demand curve.

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esercizi1-rappresentare funz. domanda e dominio economico

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esercizi1-rappresentare funz. domanda e dominio economico

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no es. 20-21-22-23-26-27

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esercizi2-trovare funz. offerta e prezzo equilibrio

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esercizi2-trovare funz. offerta e prezzo equilibrio

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no es. 42-43

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no es74

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no es 58-59-62-63

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Answers to these quiz questions are also available.

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Funzione costo, costo medio, costo marginale; funzione ricavo

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Funzione costo, costo medio, costo marginale; funzione ricavo

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solo teoria

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Problemi di scelta tra due o più alternative

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esercizi guidati di base

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Linear Programming 1: Maximization -Extreme/Corner Points

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Linear Programming 1: Maximization -Extreme/Corner Points (LP)

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Linear Programming 2: Graphical Solution - Minimization Problem

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Linear Programming 2: Graphical Solution - Minimization Problem

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exercises with solutions

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Linear Programming: Word Problem Examples

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Provides worked examples of linear programming word problems.

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exercises with solutions

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Linear programming word problems solved step by step, optimization, linear functions, objective function, constraints, feasible solution, optimal solution, graphs, activities, exercises and problems with solutions.

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exercises with solutions

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OR-Notes are a series of introductory notes on topics that fall under the broad heading of the field of operations research (OR). They were originally used by me in an introductory OR course I give at Imperial College. They are now available for use by any students and teachers interested in OR subject to the following conditions.

A full list of the topics available in OR-Notes can be found here.

Linear programming solution examples

Linear programming example 1997 UG exam

A company makes two products (X and Y) using two machines (A and B). Each unit of X that is produced requires 50 minutes processing time on machine A and 30 minutes processing time on machine B. Each unit of Y that is produced requires 24 minutes processing time on machine A and 33 minutes processing time on machine B.

At the start of the current week there are 30 units of X and 90 units of Y in stock. Available processing time on machine A is forecast to be 40 hours and on machine B is forecast to be 35 hours.

The demand for X in the current week is forecast to be 75 units and for Y is forecast to be 95 units. Company policy is to maximise the combined sum of the units of X and the units of Y in stock at the end of the week.

Formulate the problem of deciding how much of each product to make in the current week as a linear program.

Solve this linear program graphically.

Solution

Let

x be the number of units of X produced in the current week

y be the number of units of Y produced in the current week

then the constraints are:

50x + 24y <= 40(60) machine A time

30x + 33y <= 35(60) machine B time

x >= 75 - 30

i.e. x >= 45 so production of X >= demand (75) - initial stock (30), which ensures we meet demand

y >= 95 - 90

i.e. y >= 5 so production of Y >= demand (95) - initial stock (90), which ensures we meet demand

The objective is: maximise (x+30-75) + (y+90-95) = (x+y-50)
i.e. to maximise the number of units left in stock at the end of the week

It is plain from the diagram below that the maximum occurs at the intersection of x=45 and 50x + 24y = 2400

Solving simultaneously, rather than by reading values off the graph, we have that x=45 and y=6.25 with the value of the objective function being 1.25

Linear programming example 1995 UG exam

The demand for two products in each of the last four weeks is shown below.

                         Week
                      1  2  3  4
Demand - product 1    23 27 34 40
Demand - product 2    11 13 15 14

Apply exponential smoothing with a smoothing constant of 0.7 to generate a forecast for the demand for these products in week 5.

These products are produced using two machines, X and Y. Each unit of product 1 that is produced requires 15 minutes processing on machine X and 25 minutes processing on machine Y. Each unit of product 2 that is produced requires 7 minutes processing on machine X and 45 minutes processing on machine Y. The available time on machine X in week 5 is forecast to be 20 hours and on machine Y in week 5 is forecast to be 15 hours. Each unit of product 1 sold in week 5 gives a contribution to profit of £10 and each unit of product 2 sold in week 5 gives a contribution to profit of £4.

It may not be possible to produce enough to meet your forecast demand for these products in week 5 and each unit of unsatisfied demand for product 1 costs £3, each unit of unsatisfied demand for product 2 costs £1.

Formulate the problem of deciding how much of each product to make in week 5 as a linear program.

Solve this linear program graphically.

Solution

Note that the first part of the question is a forecasting question so it is solved below.

For product 1 applying exponential smoothing with a smoothing constant of 0.7 we get:

M ₁ = Y ₁ = 23
M ₂ = 0.7Y ₂ + 0.3M ₁ = 0.7(27) + 0.3(23) = 25.80
M ₃ = 0.7Y ₃ + 0.3M ₂ = 0.7(34) + 0.3(25.80) = 31.54
M ₄ = 0.7Y ₄ + 0.3M ₃ = 0.7(40) + 0.3(31.54) = 37.46

The forecast for week five is just the average for week 4 = M ₄ = 37.46 = 31 (as we cannot have fractional demand).

For product 2 applying exponential smoothing with a smoothing constant of 0.7 we get:

M ₁ = Y ₁ = 11
M ₂ = 0.7Y ₂ + 0.3M ₁ = 0.7(13) + 0.3(11) = 12.40
M ₃ = 0.7Y ₃ + 0.3M ₂ = 0.7(15) + 0.3(12.40) = 14.22
M ₄ = 0.7Y ₄ + 0.3M ₃ = 0.7(14) + 0.3(14.22) = 14.07

The forecast for week five is just the average for week 4 = M ₄ = 14.07 = 14 (as we cannot have fractional demand).

We can now formulate the LP for week 5 using the two demand figures (37 for product 1 and 14 for product 2) derived above.

Let

x ₁ be the number of units of product 1 produced

x ₂ be the number of units of product 2 produced

where x ₁, x ₂>=0

The constraints are:

15x ₁ + 7x ₂ <= 20(60) machine X

25x ₁ + 45x ₂ <= 15(60) machine Y

x ₁ <= 37 demand for product 1

x ₂ <= 14 demand for product 2

The objective is to maximise profit, i.e.

maximise 10x ₁ + 4x ₂ - 3(37- x ₁) - 1(14-x ₂)

i.e. maximise 13x ₁ + 5x ₂ - 125

The graph is shown below, from the graph we have that the solution occurs on the horizontal axis (x ₂=0) at x ₁=36 at which point the maximum profit is 13(36) + 5(0) - 125 = £343

Linear programming example 1994 UG exam

A company is involved in the production of two items (X and Y). The resources need to produce X and Y are twofold, namely machine time for automatic processing and craftsman time for hand finishing. The table below gives the number of minutes required for each item:

         Machine time Craftsman time
Item X   13           20
     Y   19           29

The company has 40 hours of machine time available in the next working week but only 35 hours of craftsman time. Machine time is costed at £10 per hour worked and craftsman time is costed at £2 per hour worked. Both machine and craftsman idle times incur no costs. The revenue received for each item produced (all production is sold) is £20 for X and £30 for Y. The company has a specific contract to produce 10 items of X per week for a particular customer.

Formulate the problem of deciding how much to produce per week as a linear program.

Solve this linear program graphically.

Solution

Let

x be the number of items of X

y be the number of items of Y

then the LP is:

maximise

20x + 30y - 10(machine time worked) - 2(craftsman time worked)

subject to:

13x + 19y <= 40(60) machine time

20x + 29y <= 35(60) craftsman time

x >= 10 contract

x,y >= 0

so that the objective function becomes

maximise

20x + 30y - 10(13x + 19y)/60 - 2(20x + 29y)/60

i.e. maximise

subject to:

13x + 19y <= 2400

20x + 29y <= 2100

x >= 10

x,y >= 0

It is plain from the diagram below that the maximum occurs at the intersection of x=10 and 20x + 29y <= 2100

Solving simultaneously, rather than by reading values off the graph, we have that x=10 and y=65.52 with the value of the objective function being £1866.5

Linear programming example 1992 UG exam

A company manufactures two products (A and B) and the profit per unit sold is £3 and £5 respectively. Each product has to be assembled on a particular machine, each unit of product A taking 12 minutes of assembly time and each unit of product B 25 minutes of assembly time. The company estimates that the machine used for assembly has an effective working week of only 30 hours (due to maintenance/breakdown).

Technological constraints mean that for every five units of product A produced at least two units of product B must be produced.

Formulate the problem of how much of each product to produce as a linear program.

Solve this linear program graphically.

The company has been offered the chance to hire an extra machine, thereby doubling the effective assembly time available. What is the maximum amount you would be prepared to pay (per week) for the hire of this machine and why?

Solution

Let

x A = number of units of A produced

x B = number of units of B produced

then the constraints are:

12x A + 25x B <= 30(60) (assembly time)

x B >= 2(x A/5)

i.e. x B - 0.4x A >= 0

i.e. 5x B >= 2x A (technological)

where x A, x B >= 0

and the objective is

maximise 3x A + 5x B

It is plain from the diagram below that the maximum occurs at the intersection of 12x A + 25x B = 1800 and x B - 0.4x A = 0

Solving simultaneously, rather than by reading values off the graph, we have that:

x A= (1800/22) = 81.8

x B= 0.4x A = 32.7

with the value of the objective function being £408.9

Doubling the assembly time available means that the assembly time constraint (currently 12x A + 25x B <= 1800) becomes 12x A + 25x B <= 2(1800) This new constraint will be parallel to the existing assembly time constraint so that the new optimal solution will lie at the intersection of 12x A + 25x B = 3600 and x B - 0.4x A = 0

i.e. at x A = (3600/22) = 163.6

x B= 0.4x A = 65.4

with the value of the objective function being £817.8

Hence we have made an additional profit of £(817.8-408.9) = £408.9 and this is the maximum amount we would be prepared to pay for the hire of the machine for doubling the assembly time.

This is because if we pay more than this amount then we will reduce our maximum profit below the £408.9 we would have made without the new machine.

Linear programming example 1988 UG exam

Solve

minimise

subject to

a + b >= 11

a - b <= 5

c - a - b = 0

7a >= 35 - 12b

a >= 0 b >= 0 c >= 0

Solution

To solve this LP we use the equation c-a-b=0 to put c=a+b (>= 0 as a >= 0 and b >= 0) and so the LP is reduced to

minimise

subject to

a + b >= 11

a - b <= 5

7a + 12b >= 35

a >= 0 b >= 0

From the diagram below the minimum occurs at the intersection of a - b = 5 and a + b = 11

i.e. a = 8 and b = 3 with c (= a + b) = 11 and the value of the objective function 10a + 11b = 80 + 33 = 113.

Linear programming example 1987 UG exam

Solve the following linear program:

maximise 5x ₁ + 6x ₂

subject to

x ₁ + x ₂ <= 10

x ₁ - x ₂ >= 3

5x ₁ + 4x ₂ <= 35

x ₁ >= 0

x ₂ >= 0

Solution

It is plain from the diagram below that the maximum occurs at the intersection of

5x ₁ + 4x ₂ = 35 and

x ₁ - x ₂ = 3

Solving simultaneously, rather than by reading values off the graph, we have that

5(3 + x ₂) + 4x ₂ = 35

i.e. 15 + 9x ₂ = 35

i.e. x ₂ = (20/9) = 2.222 and

x ₁ = 3 + x ₂ = (47/9) = 5.222

The maximum value is 5(47/9) + 6(20/9) = (355/9) = 39.444

Linear programming example 1986 UG exam

A carpenter makes tables and chairs. Each table can be sold for a profit of £30 and each chair for a profit of £10. The carpenter can afford to spend up to 40 hours per week working and takes six hours to make a table and three hours to make a chair. Customer demand requires that he makes at least three times as many chairs as tables. Tables take up four times as much storage space as chairs and there is room for at most four tables each week.

Formulate this problem as a linear programming problem and solve it graphically.

Solution

Variables

Let

x T = number of tables made per week

x C = number of chairs made per week

Constraints

6x T + 3x C <= 40

x C >= 3x T

(x C/4) + x T <= 4

Objective

maximise 30x T + 10x C

The graphical representation of the problem is given below and from that we have that the solution lies at the intersection of

(x C/4) + x T = 4 and 6x T + 3x C = 40

Solving these two equations simultaneously we get x C = 10.667, x T = 1.333 and the corresponding profit = £146.667

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OR-Notes are a series of introductory notes on topics that fall under the broad heading of the field of operations research (OR). They were originally used by me in an introductory OR course I give at Imperial College. They are now available for use by any students and teachers interested in OR subject to the following conditions.

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exercises with solutions

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Lesson LINEAR PROGRAMMING PROBLEMS AND SOLUTIONS 1

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This Lesson (LINEAR PROGRAMMING PROBLEMS AND SOLUTIONS 1) was created by by

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Padlet-costi e ricavi

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A demand function is defined by \(p = f(x)\text{,}\) where \(p\) measures the unit price and \(x\) measures the number of units of the commodity in question, and is generally characterized as a decreasing function of x; that is, \(p = f(x)\) decreases as \(x\) increases. Since both \(x\) and \(p\) assume only nonnegative values, the demand curve is that part of the graph of \(f(x)\) that lies in the first quadrant ( Figure 2.7).

A supply function defined by \(p = f(x)\) with \(p\) and \(x\) as before is generally characterized as an increasing function of \(x\text{;}\) that is, \(p = f(x)\) increases as \(x\) increases. Since both \(x\) and \(p\) assume only nonnegative values, the supply curve is that part of the graph of \(f(x)\) that lies in the first quadrant ( Figure 2.7)

The demand function for a certain commercial product is given by

\begin{equation*} p = d(x) = -0.01x^{2}-0.2x+8 \end{equation*}

and the corresponding supply function is given by

\begin{equation*} p = s(x) = 0.01x^{2} + 0.1x + 3 \end{equation*}

where \(p\) is expressed in dollars and \(x\) is measured in units of a hundred. Find the market equilibrium.

Solution

which is equivalent to

The linear cost function \(C\) is given by

\begin{equation*} C(x) = mx + b \end{equation*}

where \(m\) is the variable cost per unit, \(b\) is the overall fixed cost, and \(x\) is the number of units produced.

Suppose \(C(x)\) is the total cost function. Then the average cost function, denoted by \(\overline{C}(x)\) - read "\(C\) bar of \(x\)" - is given by

\begin{equation*} \overline{C}(x) = \dfrac{C(x)}{x}\text{.} \end{equation*}

Suppose a certain company has a monthly fixed cost of $\(10,000\) and a variable cost of

\begin{equation*} -0.0001x^{2} + 12x \ \ \ 0 \leq x \leq 40,000 \end{equation*}

dollars, where \(x\) denotes the number of units manufactured per month. Find a function \(C\) that gives the total cost incurred by the company in the manufacture of \(x\) units.

Suppose the total revenue realized by the same company from Example 2.48 from the sale of \(x\) units is given by the total revenue function

\begin{equation*} R(x) = -0.0005x^{2} + 22x \ \ \ 0 \leq x \leq 40,000 \end{equation*}

1. Find the total profit function - that is, the function that descrbies the total profit the company realizes in manufacturing and selling \(x\) units per month.

2. Find the profit when the level of production is \(10,000\) units.

If exactly \(1000\) people sign up for a charter flight, a certain travel agency charges \(600\) per person. However, if more than \(1000\) people sign up for the flight (assume this is the case), then each fare is reduced by $\(5\) for each additional person. Letting \(x\) denote the number of passengers above \(1000\text{,}\) find a function giving the revenue realized by the company.

A certain firm determines that the total cost \(C(x)\) in dollars of producing and selling \(x\) units is given by

\begin{equation*} C(x) = 20x + 100\text{.} \end{equation*}

Management plans to charge $\(24\) per unit.

1. How many units must be sold for the firm to break even?

2. What is the profit if \(100\) units of feed are sold?

3. How many units must be sold to produce a profit of $\(900\text{?}\)

Solution

Exercises for Section 2.7.

\(0 = -2x^{2}-p+80\) and \(0 = 15x-p+30\)
Answer Solution

To find the equilbrium quantity, we can equate the supply and demand equations.

\begin{equation*} \begin{split} -2x^{2} + 80 \amp = 15x + 30 \\ -2x^{2}-15x+50 \amp = 0 \\ -(x+10)(2x-5) \amp = 0 \end{split} \end{equation*}

This gives one nonnegative solution, \(x = 2.5\text{.}\) The corresponding price is

\begin{equation*} p = 15(2.5)+30 = 67.5\text{.} \end{equation*}

This tells us that the equilibrium quantity is \(2500\text{,}\) and the equilibrium price is $\(67.50\text{.}\)

1. Sketch the graph of \(p(t)\text{.}\)
Answer

2. What was the price of ivory at the beginning of \(1970\text{?}\)
Answer Solution

The beginning of \(1970\) corresponds to \(t = 0\text{.}\) Therefore, the price of ivory was \(p(0) =\) $ 7.44 per kg.

3. What was the price of ivory at the beginning of \(1990\text{?}\)
Answer Solution

The beginning of \(1990\) corresponds to \(t = 20\text{.}\) Looking at the definition of \(p\text{,}\) we see that \(p(20) = 2.84(20)+51.68 =\) $108.48 per kg.

Solution

Answer

\(p = 10\)

Answer

1. Sketch \(D(q)\)
Answer

2. Sketch \(S(q)\) on the same plot as for part (g)
Answer

1. \(C(x) = 105x + 6000\text{;}\) \(R(x) = 250x\text{;}\) no more than \(400\) units can be sold.
Answer

Break-even quantity is about \(41\) units; produce. \(P(x) = 145x - 6000\text{.}\)

Solution

We find the break-even point by setting \(C(x) = R(x)\text{:}\)

\begin{equation*} \begin{split} 105x+6000 \amp = 250x \\ 145x \amp = 6000\\ x \amp \approx 41 \end{split} \end{equation*}

The corresponding profit function is

\begin{equation*} P(x) = R(x) - C(x) = 145x - 6000 \ \ \ 0 \leq x \leq 400\text{.} \end{equation*}

We would therefore decide to produce this product, since there is opportunity to profit.

2. \(C(x) = 1000x + 5000\text{;}\) \(R(x) = 900x\text{.}\)
Answer

Break-even quantity is \(-50\) units; don't produce. Impossible to make a profit when \(C(x) > R(x)\) for all \(x > 0\text{.}\) \(P(x) = -100x - 5000\) (always a loss).

Solution

Let's first find the profit function.

\begin{equation*} \begin{split} P(x) \amp = 900x - (1000x + 5000) \\ \amp = -100x - 5000 \\ \amp = -(100x+5000) \end{split} \end{equation*}

The break even point corresponds to \(P(x) = 0 \implies x = -50\text{.}\) We can see that, for all \(x \geq 0\text{,}\) \(P(x) \leq 0\text{.}\) Therefore, we would decide to not produce.

For the following demand equations, where \(x\) is the quantity demanded in units of a thousand and \(p\) is the unit price in dollars, sketch the demand curve and determine the quantity demanded at the given unit price.

For the following supply equations, where \(x\) is the quantity supplied in units of a thousand and \(p\) is the unit price in dollars, sketch the supply curve and determine the price at the given number of units.

For the following pairs of supply and demand equations, where \(x\) represents the quantity demanded in units of a thousand and \(p\) is the unit price in dollars, find the equilbrium point.

\(0 = -2x^{2}-p+80\) and \(0 = 15x-p+30\)

To find the equilbrium quantity, we can equate the supply and demand equations.

\begin{equation*} \begin{split} -2x^{2} + 80 \amp = 15x + 30 \\ -2x^{2}-15x+50 \amp = 0 \\ -(x+10)(2x-5) \amp = 0 \end{split} \end{equation*}

This gives one nonnegative solution, \(x = 2.5\text{.}\) The corresponding price is

\begin{equation*} p = 15(2.5)+30 = 67.5\text{.} \end{equation*}

This tells us that the equilibrium quantity is \(2500\text{,}\) and the equilibrium price is $\(67.50\text{.}\)

A production company has a monthly fixed cost of $\(100,000\) and a variable cost of $\(14\) for each unit produced. The commodity sells for $\(20\) per unit.

The price of ivory is determined by a variety of legal market sources and illegal ivory trades. The World Wildlife Fund, an international non-governmental organization trying to preserve wilderness, determined that the ivory price can be approximated by the piecewise-defined function

\begin{equation*} p(t) = \begin{cases} 8.37t + 7.44 \amp \text{ if } 0 \leq t \leq 8 \\ 2.84t + 51.68 \amp \text{ if } 8 \lt t \leq 30 \end{cases} \end{equation*}

where \(t\) is measured in years, with \(t = 0\) corresponding to the beginning of the year \(1970\text{,}\) and \(p(t)\) is measured in dollars per kilogram.

1. Sketch the graph of \(p(t)\text{.}\)
Answer

2. What was the price of ivory at the beginning of \(1970\text{?}\)
Answer Solution

The beginning of \(1970\) corresponds to \(t = 0\text{.}\) Therefore, the price of ivory was \(p(0) =\) $ 7.44 per kg.

3. What was the price of ivory at the beginning of \(1990\text{?}\)
Answer Solution

The beginning of \(1990\) corresponds to \(t = 20\text{.}\) Looking at the definition of \(p\text{,}\) we see that \(p(20) = 2.84(20)+51.68 =\) $108.48 per kg.

In the accompanying figure, \(L_{1}\) is the demand curve for the version \(X\) of cell phones, and \(L_{2}\) is the demand curve for the version \(Y\) of cell phones. Which line has greater slope? Interpret your results.

Solution

The slope of \(L_{2}\) is greater than the slope of \(L_{1}\text{.}\) For each drop of a dollar in price, the quantity demanded for version \(X\) is greater than for version \(Y\text{.}\)

The demand function for a certain product is given by

\begin{equation*} p = \dfrac{3000}{2x^{2} + 100} \ \ \ (0 \leq x \leq 10) \end{equation*}

where \(x\) (measured in units of a thousand) is the quantity demanded per week and \(p\) is the unit price in dollars. Sketch the graph of the demand function. What is the price that corresponds to a quantity demanded of \(10,000\) units?
Answer

\(p = 10\)

A certain manufacturer has determined that the weekly demand and supply functions for their product is given by

\begin{equation*} p(q) = 144-q^{2} \end{equation*}

\begin{equation*} p(q) = 48 + \frac{1}{2}q^{2} \end{equation*}

respectively, where \(p\) is measured in dollars and \(q\) is measured in units of a thousand. Find the equilibrium point.

Suppose that the demand and price for a certain model of wristwatch are related by

\begin{equation*} p = D(q) = 16-1.25q \end{equation*}

where \(p\) is the price in dollars and \(q\) is the demand in units of hundreds. The price and supply of the watch are related by

\begin{equation*} p = S(q) = 0.75q \end{equation*}

where \(p\) is the price in dollars and \(q\) is the supply of watches in units of hundreds.

1. Sketch \(D(q)\)
Answer

2. Sketch \(S(q)\) on the same plot as for part (g)
Answer

Joanne sells T-shirts at community festivals and creaft fairs. Her marginal cost to produce one T-shirt is $\(3.50\text{.}\) Her total cost to produce \(60\) T-shirts is $\(300\text{,}\) and she sells them for $\(9\) each.

You are the manager of a firm. You are considering the manufacture of a new product, so you ask the accounting department for cost estimates and the sales deparment for sales estimates. After you recieve the data, you must decide whether to go ahead with production of the new product. Analyze the given data (find a break-even quantity) and then decide what you would do in each case. Include the profit function.

1. \(C(x) = 105x + 6000\text{;}\) \(R(x) = 250x\text{;}\) no more than \(400\) units can be sold.

Break-even quantity is about \(41\) units; produce. \(P(x) = 145x - 6000\text{.}\)

We find the break-even point by setting \(C(x) = R(x)\text{:}\)

\begin{equation*} \begin{split} 105x+6000 \amp = 250x \\ 145x \amp = 6000\\ x \amp \approx 41 \end{split} \end{equation*}

The corresponding profit function is

\begin{equation*} P(x) = R(x) - C(x) = 145x - 6000 \ \ \ 0 \leq x \leq 400\text{.} \end{equation*}

We would therefore decide to produce this product, since there is opportunity to profit.

2. \(C(x) = 1000x + 5000\text{;}\) \(R(x) = 900x\text{.}\)

Break-even quantity is \(-50\) units; don't produce. Impossible to make a profit when \(C(x) > R(x)\) for all \(x > 0\text{.}\) \(P(x) = -100x - 5000\) (always a loss).

Let's first find the profit function.

\begin{equation*} \begin{split} P(x) \amp = 900x - (1000x + 5000) \\ \amp = -100x - 5000 \\ \amp = -(100x+5000) \end{split} \end{equation*}

The break even point corresponds to \(P(x) = 0 \implies x = -50\text{.}\) We can see that, for all \(x \geq 0\text{,}\) \(P(x) \leq 0\text{.}\) Therefore, we would decide to not produce.

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A demand function is defined by \(p = f(x)\text{,}\) where \(p\) measures the unit price and \(x\) measures the number of units of the commodity in question, and is generally characterized as a decreasing function of x; that is, \(p = f(x)\) decreases as \(x\) increases.

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solo esercizi 13-14-15