Wednesday, August 17, 2016
Sunday, July 10, 2016
PROBABILITY QUESTIONS WITH SOLUTIONS
Question 1
A die is rolled, find the probability that an even number is obtained
- Let us first write the sample space S of the experiment
S = {1,2,3,4,5,6}
- We now use the formula of the classical probability
P(E) = n(E) / n(S) = 3 / 6 = 1 / 2
Question 2
Two coins are tossed, find the probability that two heads are obtained
Note: Each coin has two possible outcomes H (heads) and T (Tails)
- The sample space S is given by
S = {(H,T),(H,H),(T,H),(T,T)} - Let E be the event "two heads are obtained"
E = {(H,H)}
- We use the formula of the classical probability
P(E) = n(E) / n(S) = 1 / 4
Question 3
Which of these numbers cannot be a probability?
a) -0.00001
b) 0.5
c) 1.001
d) 0
e) 1
f) 20%
b) 0.5
c) 1.001
d) 0
e) 1
f) 20%
A probability is always greater than or equal to 0 and less than or equal to 1, hence only a) and c) above cannot represent probabilities: -0.00010 is less than 0 and 1.001 is greater than 1
Question 4
a) A die is rolled, find the probability that the number obtained is greater than 4.
b) Two coins are tossed, find the probability that one head only is obtained.
c) Two dice are rolled, find the probability that the sum is equal to 5.
d) A card is drawn at random from a deck of cards. Find the probability of getting the King of heart
b) Two coins are tossed, find the probability that one head only is obtained.
c) Two dice are rolled, find the probability that the sum is equal to 5.
d) A card is drawn at random from a deck of cards. Find the probability of getting the King of heart
Answers:
a) 2 / 6 = 1 / 3
b) 2 / 4 = 1 / 2
c) 4 / 36 = 1 / 9
d) 1 / 52
b) 2 / 4 = 1 / 2
c) 4 / 36 = 1 / 9
d) 1 / 52
Question 5
A card is drawn at random from a deck of cards. Find the probability of getting the 3 of diamond
P(E) = 1 / 52
Reference:
SETS QUESTIONS & ANSWERS
1. If U = {1, 3, 5, 7, 9, 11, 13}, then which of the following are subsets of U.
B = {2, 4}
A = {0}
C = {1, 9, 5, 13}
D = {5, 11, 1}
E = {13, 7, 9, 11, 5, 3, 1}
F = {2, 3, 4, 5}
2. Let A = {2, 3, 4, 5, 6, 7} B = {2, 4, 7, 8) C = {2, 4}. Fill in the blanks by ⊂ or ⊄ to make the resulting statements true.
(a) B __ A
(b) C __ A
(c) B __ C
(d) ∅ __ B
(e) C __ C
(f) C __ B
3. Which of the following sets is a universal set for the other four sets?
(a) The set of even natural numbers
(b) The set of odd natural numbers
(c) The set of natural numbers
(d) The set of negative numbers
(e) The set of integers
4. Write all the subsets for the following.
(a) {3}
(b) {6, 11}
(c) {2, 5, 9}
(d) {1, 2, 6, 7}
(e) {a, b, c}
(f) ∅
(g) {p, q, r, s}
5. Write down all the possible proper subsets for each of the following.
(a) {a, b, c, d}
(b) {1, 2, 3}
(c) {p, q, r}
(d) {5, 10}
(e) {x}
(f) ∅
6. Find the number of subsets for set
(a) containing 3 elements
(b) whose cardinal number is 5
7. Find the number of proper subsets of a set
(a) containing 6 elements
(a) containing 6 elements
(b) whose cardinal number is 4
8. Show with an example that if the number of elements in a set is ‘n’, then
(a) the number of subsets is 2n
(b) the number of proper subsets is 2n - 1.
9. Write the universal set for the following.
(a) P = {4, 6, 8} Q = {1, 3, 9} R = {0, 2, 5} S = {7}
(b) X = {a, b, c} Y = {c, b, f} Z = {e, g}
(c) Prime numbers less than 10, even numbers less than 10, multiples of 3 less than 10.
10. If ΞΎ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {2, 4, 6, 8}
B = {3, 5, 7}
C = {1, 5, 7, 8, 9}
Find (a) A’ (b) B’ (c) C’
11. State whether true or false.
(a) Quadrilateral ⊆ polygon
(b) {1} ↔ {0}
(c) Whole numbers ⊆ natural numbers
(d) {a} ∈ {d, e, f, a}
(e) Natural numbers ⊆ whole numbers
(f) Integers ⊆ natural numbers
(g) 0 ∈ ∅
(h) ∅ ∈ {1 , 2, 3 }
12. Let the set of integer be the universal set and let A = set of whole numbers, then what is A’?
13. Let A {x : x = n — 2, n < 5}. Find A when
(a) n = W, n ∈ W
(b) n = N, n ∈ N
(c) n ∈ I = I
14. If U = {2, 3, 4, 5, 6, 7, 8, 9} X = {3, 5, 7, 9} Y = {2, 4, 6, 8}
Show that X = Y’ and Y = X’
15. Let P = {3, 5, 7, 9, 11} Q = {9, 11, 13} R = {3, 5, 9} S = {13, 11}
Write Yes or No for the following.
(a) R ⊂ P
(b) Q ⊂ P
(c) R ⊂ S
(d) S ⊂ Q
(e) S ⊂ P
(f) P ⊄ Q
(g) Q ⊄ R
(h) S ⊄ Q
Answers:
1. C, D, E
2. (a) ⊄
2. (a) ⊄
(b) ⊂
(c) ⊄
(d) ⊂
(e) ⊂
(f) ⊂
3. (e)
4. (a) d, {3}
3. (e)
4. (a) d, {3}
(b) d, {6}, {11}, {6, 11}
(c) d, {2}, {5}, {9}, {2, 5}, {2, 9}, {5, 9}, {2, 5, 9}
(d) d, {1}, {2}, {6}, {7}, {1, 2}, {1, 6}, {1, 7}, {2, 6}, {2, 7}, {6, 7}, {1, 2, 6}, {1, 2, 7}, {1, 6, 7}, {2, 6, 7}, {1, 2, 6, 7}
(e) {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}, d
(f) d
(g) d, {p}, {q}, {r}, {s}, {p, q}, {p, r}, {p, s}, {q, r}, {q, s}, {r, s}, {p, q, r } {p, q, s}, {p, r, s }, {q, r, s}, {p, q, r, s}
5. (a) d, {a}, {b}, {c}, {d}, {a, b}, {a, c}, {a, d}, {b, c}, {b, d}, {c, d}, {a, b, c}, {a, b, d}, {a, c, d}, {b, c, d}
5. (a) d, {a}, {b}, {c}, {d}, {a, b}, {a, c}, {a, d}, {b, c}, {b, d}, {c, d}, {a, b, c}, {a, b, d}, {a, c, d}, {b, c, d}
(b) d, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}
(c) d {p}, {q}, {r}, {p, q}, {p, r}, {q, r}
(d) d, {5}, {10}
(e) d
(f) none
6. (a) 8
(b) 32
7. (a) 63
(b) 15
9. (a) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
(b) 15
9. (a) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
(b) {a, b, c, e, f, g}
(c) {2, 3, 4, 5, 6, 7, 8, 9, 10}
10. (a) {1, 3, 5, 7, 9, 10}
10. (a) {1, 3, 5, 7, 9, 10}
(b) {1, 2, 4, 6, 8, 9, 10}
(c) {2, 3, 4, 6, 10}
11. (a) True
11. (a) True
(b) True
(c) False
(d) False
(e) True
(f) False
(g) False
(h) False
12. set of negative integers
13. (a) {0, 1, 2}
12. set of negative integers
13. (a) {0, 1, 2}
(b) {1, 2}
(c) {... -3, -2, -1, 0, 1, 2}
15. (a) Yes
15. (a) Yes
(b) No
(c) No
(d) Yes
(e) No
(f) Yes
(g) Yes
(h) No
Reference:
Saturday, July 9, 2016
SOLVING INEQUALITIES
Solve the inequality z + 5 ≥ 3 and represent its solutions using a number line
Answer
z ≥ -2
Example 2
Solve the inequality 4(x – 1) < 3(x + 1) and represent its solutions using a number line
Answer
x < 7
Example 3
Solve the inequality 2x + 3 ≤ x and represent its solutions using a number line
Solve the inequality 2x + 3 ≤ x and represent its solutions using a number line
Answer
x ≤ -3Example 4
Solve the inequality 5(x + 2) – 3x ≤ 4 + 2x + 3(x – 1). Write the solution as an inequality, and show the solution on a number line
Answer
3 ≤ x Example 5
Solve the inequality -12x < 144. Write the solution as an inequality, and show the solution on a number line
Answer
x > -12Example 6
Solve the inequality 5[x + 3(1 – x) + 2] ≥ 7. Write the solution as an inequality, and show the solution on a number line
Answer
Solve the inequality 7y + 2 ≤ 65. Write the solution as an inequality, and show the solution on a number line
Answer
Solve the inequality z + z[3 – 2(1 + z)] < 5[1 + 2(z – 3)] – 2z2. Write the solution as an inequality, and show the solution on a number line
Answer
Reference:
LINEAR PROGRAMMING PROBLEMS & SOLUTIONS
Example 1
A farmer can plant up to 8 acres of land with wheat and barley. He can earn $5,000 for every acre he plants with wheat and $3,000 for every acre he plants with barley. His use of a necessary pesticide is limited by federal regulations to 10 gallons for his entire 8 acres. Wheat requires 2 gallons of pesticide for every acre planted and barley requires just 1 gallon per acre
What is the maximum profit he can make?
Let x = the number of acres of wheat
Let y = the number of acres of barley
Since the farmer earns $5,000 for each acre of wheat and $3,000 for each acre of barley, then the total profit the farmer can earn is 5000*x + 3000*y
Let p = total profit that can be earned. your equation for profit becomes:
p = 5000x + 3000y
That's your objective function. it's what you want to maximize
The constraints are:
Number of acres has to be greater than or equal to 0
Number of acres has to be less than or equal to 8
Amount of pesticide has to be less than or equal to 10
Your constraint equations are:
x >= 0
y >= 0
x + y <= 8
2x + y <= 10
To graph these equations, solve for y in those equations that have y in them and then graph the equality portion of those equations
x >= 0
y >= 0
y <= 8-x
y <= 10 - 2x
x = 0 is a vertical line that is the same line as the y-axis
y = 0 is a horizontal line that is the same line as the x-axis
The area of the graph that satisfies all the constraints is the region of feasibility
The maximum or minimum solutions to the problem will be at the intersection points of the lines that bound the region of feasibility
The graph of your equations looks like this:
The region of feasibility is the shaded area of the graph
You can see from this graph that the region of feasibility is bounded by the following (x,y) coordinate points:
(0,0)
(0,8)
(2,6)
(5,0)
The point (0,0) is the intersection of the line x-axis with the y-axis.
The point (0,8) is the intersection of the line y = 8 - x with the y-axis.
The point (5,0) is the intersection of the line y = 10 - 2x with the x-axis.
The point (2,6) is the intersection of the line y = 8 - x with the line y = 10 - 2x
The point (2,6) was solved for in the following manner:
Equations of the intersecting lines are:
y = 8 - x
y = 10 - 2x
Subtract the first equation from the second equation and you get:
0 = 2 - x
Add x to both sides of this equation and you get:
x = 2
substitute 2 for x in either equation to get y = 6
that makes the intersection point (x,y) = (2,6)
The objective equation is:
p = 5000x + 3000y
Profit will be maximum at the intersection points of the region of feasibility on the graph
The profit equation is evaluated at each of these points as shown in the following table
The maximum profit occurs when the farmer plants 2 acres of wheat and 6 acres of barley
Number of acres of wheat is 2 and number of acres of barley is 6 for a total of 8 acres which is the maximum number of acres available for planting
Number of gallons of pesticide used for wheat is 4 and number of gallons of pesticide used for barley is 6 for a total of 10 gallons of pesticide which is the maximum amount of pesticide that can be used
Example 2
A painter has exactly 32 units of yellow dye and 54 units of green dye
He plans to mix as many gallons as possible of color A and color B
Each gallon of color A requires 4 units of yellow dye and 1 unit of green dye
Each gallon of color B requires 1 unit of yellow dye and 6 units of green dye
Find the maximum number of gallons he can mix
The objective function is to determine the maximum number of gallons he can mix
The colors involved are color A and color B
Let x = the number of gallons of color A
Let y = the number of gallons of color B
If we let g = the maximum gallons the painter can make, then the objective function becomes:
g = x + y
Make a table for color A and color B to determine the amount of each dye required
your table will look like this:
Each gallon of color A or B will require:
Total units of yellow dye available are 32
Total units of green dye available are 54
Your constraint equations are:
x >= 0
y >= 0
4x + y <= 32
x + 6y <= 54
x and y have to each be greater than or equal to 0 because the number of gallons can't be negative
In order to graph these equations, you solve for y in those equations that have y in them
The equations for graphing are:
x >= 0
y >= 0
y <= 32 - 4x
y <= (54 - x)/6
x = 0 is a vertical line that is the same line as the y-axis.
y = 0 is a horizontal line that is the same line as the x-axis
The graph will look like this:
The region of feasibility is the shaded area of the graph. All points within the feasibility region meet the constraint of the problem
The intersection points of the region of feasibility are:
(0,0)
(0,9)
(6,8)
(8,0)
The maximum or minimum value of the objective function will be at these points of intersection
Solve the objective function at each of these intersection points to determine which point contains the maximum number of gallons
The objective function is:
g = x + y
The table with the value of g at each of these intersection points is shown below:
The maximum gallons of paint for color A and B, given the constraints, is equal to 14
This is comprised of 6 gallons of color A and 8 gallons of color B
6 gallons of color A uses 24 gallons of yellow dye and 8 gallons of color B uses 8 gallons of yellow dye for a total of 32 gallons of yellow dye which is the maximum amount of yellow dye that can be used
6 gallons of color A user 6 gallons of green dye and 8 gallons of color B uses 48 gallons of green dye for a total of 54 gallons of green dye which is the maximum amount of green dye that can be used
Example 3
The Bead Store sells material for customers to make their own jewelry. Customer can select beads from various bins. Grace wants to design her own Halloween necklace from orange and black beads. She wants to make a necklace that is at least 12 inches long, but no more than 24 inches long. Grace also wants her necklace to contain black beads that are at least twice the length of orange beads. Finally, she wants her necklace to have at least 5 inches of black beads
Find the constraints, sketch the problem and find the vertices (intersection points)
Let x = the number of inches of black beads
Let y = the number of inches of orange beads
The objective function is the length of the necklace
There is a maximum length and a minimum length
If you let n equal the length of the necklace, then the objective function becomes:
n = x + y
Since the problem is looking for the number of inches of black beads and the number of inches of orange beads, we will let:
The constraint equations for this problem are:
x >= 0
y >= 0
x + y >= 12
x + y <= 24
x >= 2y
x >= 5
x >= 0 is there because the number of inches of black beads can't be negative
y >= 0 is there because the number of inches of orange beads can't be negative
x + y >= 12 is there because the total length of the necklace has to be greater than or equal to 12 inches
x + y <= 24 is there because the total length of the necklace has to be less than or equal to 24 inches
x >= 2y is there because the length of the black beads has to be greater than or equal to twice the length of the orange beads
x >= 5 is there because the number of inches of black beads has to be greater than or equal to 5
To graph these equations, we have to solve for y in each equation that has y in it and then graph the equality portion of each of them
Your equations for graphing are:
x >= 0
y >= 0
y >= 12 - x
y <= 24 - x
y <= x/2
x >= 5
x = 0 is a vertical line that is the same line as the y-axis
y = 0 is a horizontal line that is the same line as the x-axis
x = 5 is a vertical line at x = 5
A graph of you equations is shown below:
The region of feasibility is the shaded area of the graph. All points within the region of feasibility meet the constraint requirements of the problem
The intersection points bounding the region of feasibility are:
(8,4)
(12,0)
(16,8)
(24,0)
(8,4) is the intersection of the lines y = x/2 and y = 12 - x
To find the point of intersection, set x/2 and 12-x equal to each other and solve for xx
you get:
x/2 = 12-x
Multiply both sides of the equation by 2 to get:
x = 24-2x
Add 2x to both sides of the equation to get:
3x = 24
Divide both sides of the equation by 3 to get:
x = 8
Substitute 8 for x in either equation to get y = 4
(12,0) is the intersection of the line y = 12 - x with the x-axis
(24,0) is the intersection of the line y = 24 - x with the x-axis
To find the point of intersection, set y equal to 0 in each equation and solve for x
(16,8) is the intersection of the lines y = x/2 and y = 24 - x
To find the intersection point, set x/2 equal to 24-x and solve for x
you get:
x/2 = 24-x
Multiply both sides of this equation by 2 to get:
x = 48 - 2x
Add 2x to both sides of this equation to get:
3x = 48
Divide both sides of this equation by 3 to get:
x = 16
Substitute 16 for x in either equation to get:
y = 8
The maximum / minimum necklace length will be at the intersection points of the boundaries of the region of feasibility
Evaluation of the objective function at these intersections yields the following:
objective function is:
x + y = n where n is the length of the necklace in inches
The number of inches of black beads is at least twice the number of inches of orange beads
The number of inches of black beads is at least 5
The total length of the necklace is greater than or equal to 12 inches or less than or equal to 24 inches
All the constraints have been met
The maximum length the necklace can be and still meet the constraints is 24 inches
The minimum length the necklace can be and still meet the constraints is 12 inches
Example 4
A garden shop wishes to prepare a supply of special fertilizer at a minimal cost by mixing two fertilizers, A and B
The mixture is to contain:
At least 45 units of phosphate
At least 36 units of nitrate
At least 40 units of ammonium
Fertilizer A costs the shop $.97 per pound
Fertilizer B costs the shop $1.89 per pound
Fertilizer A contains 5 units of phosphate and 2 units of nitrate and 2 units of ammonium
Fertilizer B contains 3 units of phosphate and 3 units of nitrate and 5 units of ammonium
How many pounds of each fertilizer should the shop use in order to minimize their cost?
Let x = the number of pounds of fertilizer A
Let y = the number of pounds of fertilizer B
The objective function is to minimize the cost
The objective function becomes:
c = .97x + 1.89y
The constraint equations are:
Since the number of pounds of each fertilizer can't be negative, 2 of the constraint equations become:
x >= 0
y >= 0
Since the number of units of phosphate has to be at least 45, the constraint equation for phosphate becomes:
5x + 3y >= 45
Since the number of units of nitrate must be at least 36, the constraint equation for nitrates becomes:
2x + 3y >= 36
Since the number of units of ammonium must be at least 40, the constraint equation for ammonium becomes:
2x + 5y >= 40
The constraint equations for this problem become:
x >= 0
y >= 0
5x + 3y >= 45
2x + 3y >= 36
2x + 5y >= 40
In order to graph these equations, you have to solve for y in each equation that has y in it and then graph the equality portion of those equations
The equations to be graphed are:
x >= 0
y >= 0
y >= (45-5x)/3
y >= (36 - 2x)/3
y >= (40-2x)/5
x = 0 is a vertical line that is the same line as the y-axis
y = 0 is a horizontal line that i the same line as the x-axis
A graph of your equation is shown below:
The feasibility region is the area of the graph that is shaded. All points within the region of feasibility meet the constraint requirements of the problem
The intersection points at the boundaries of the feasibility region are:
The intersection points of the boundaries of the region of feasibility contain the minimum cost solution for the objective function in this problem
Now that you have the intersection points, you can solve for the minimum cost equation which is the objective function of:
c = .97x + 1.89y
The following table shows the value of the cost equation at each of the intersection points
The table suggests that we have a minimum cost solution when the value of x is equal to 15 and the value of y is equal 2
When x = 15 and y = 2, the number of pounds of potassium, nitrates, and ammonium are:
phosphate = 5x + 3y = 5*15 + 3*2 = 75 + 6 = 81
nitrate = 2x + 3y = 2*15 + 3*2 = 30 + 6 = 36
ammonium = 2x + 5y = 2*15 + 5*2 = 30 + 10 = 40
All the constraints associated with the minimum cost objective have been met
Reference:
https://www.algebra.com/algebra/homework/coordinate/word/THEO-2011-08-28-03.lesson
Example 1
A farmer can plant up to 8 acres of land with wheat and barley. He can earn $5,000 for every acre he plants with wheat and $3,000 for every acre he plants with barley. His use of a necessary pesticide is limited by federal regulations to 10 gallons for his entire 8 acres. Wheat requires 2 gallons of pesticide for every acre planted and barley requires just 1 gallon per acre
What is the maximum profit he can make?
Let x = the number of acres of wheat
Let y = the number of acres of barley
Since the farmer earns $5,000 for each acre of wheat and $3,000 for each acre of barley, then the total profit the farmer can earn is 5000*x + 3000*y
Let p = total profit that can be earned. your equation for profit becomes:
p = 5000x + 3000y
That's your objective function. it's what you want to maximize
The constraints are:
Number of acres has to be greater than or equal to 0
Number of acres has to be less than or equal to 8
Amount of pesticide has to be less than or equal to 10
Your constraint equations are:
x >= 0
y >= 0
x + y <= 8
2x + y <= 10
To graph these equations, solve for y in those equations that have y in them and then graph the equality portion of those equations
x >= 0
y >= 0
y <= 8-x
y <= 10 - 2x
x = 0 is a vertical line that is the same line as the y-axis
y = 0 is a horizontal line that is the same line as the x-axis
The area of the graph that satisfies all the constraints is the region of feasibility
The maximum or minimum solutions to the problem will be at the intersection points of the lines that bound the region of feasibility
The graph of your equations looks like this:
You can see from this graph that the region of feasibility is bounded by the following (x,y) coordinate points:
(0,0)
(0,8)
(2,6)
(5,0)
The point (0,0) is the intersection of the line x-axis with the y-axis.
The point (0,8) is the intersection of the line y = 8 - x with the y-axis.
The point (5,0) is the intersection of the line y = 10 - 2x with the x-axis.
The point (2,6) is the intersection of the line y = 8 - x with the line y = 10 - 2x
The point (2,6) was solved for in the following manner:
Equations of the intersecting lines are:
y = 8 - x
y = 10 - 2x
Subtract the first equation from the second equation and you get:
0 = 2 - x
Add x to both sides of this equation and you get:
x = 2
substitute 2 for x in either equation to get y = 6
that makes the intersection point (x,y) = (2,6)
The objective equation is:
p = 5000x + 3000y
Profit will be maximum at the intersection points of the region of feasibility on the graph
The profit equation is evaluated at each of these points as shown in the following table
intersection point of (x,y) p
(0,0) $0
(0,8) $24,000
(2,6) $28,000 *****
(5,0) $25,000
The maximum profit occurs when the farmer plants 2 acres of wheat and 6 acres of barley
Number of acres of wheat is 2 and number of acres of barley is 6 for a total of 8 acres which is the maximum number of acres available for planting
Number of gallons of pesticide used for wheat is 4 and number of gallons of pesticide used for barley is 6 for a total of 10 gallons of pesticide which is the maximum amount of pesticide that can be used
Example 2
A painter has exactly 32 units of yellow dye and 54 units of green dye
He plans to mix as many gallons as possible of color A and color B
Each gallon of color A requires 4 units of yellow dye and 1 unit of green dye
Each gallon of color B requires 1 unit of yellow dye and 6 units of green dye
Find the maximum number of gallons he can mix
The objective function is to determine the maximum number of gallons he can mix
The colors involved are color A and color B
Let x = the number of gallons of color A
Let y = the number of gallons of color B
If we let g = the maximum gallons the painter can make, then the objective function becomes:
g = x + y
Make a table for color A and color B to determine the amount of each dye required
your table will look like this:
Each gallon of color A or B will require:
units of yellow dye units of green dye
color A 4 1
color B 1 6
Total units of yellow dye available are 32
Total units of green dye available are 54
Your constraint equations are:
x >= 0
y >= 0
4x + y <= 32
x + 6y <= 54
x and y have to each be greater than or equal to 0 because the number of gallons can't be negative
In order to graph these equations, you solve for y in those equations that have y in them
The equations for graphing are:
x >= 0
y >= 0
y <= 32 - 4x
y <= (54 - x)/6
x = 0 is a vertical line that is the same line as the y-axis.
y = 0 is a horizontal line that is the same line as the x-axis
The graph will look like this:
The intersection points of the region of feasibility are:
(0,0)
(0,9)
(6,8)
(8,0)
The maximum or minimum value of the objective function will be at these points of intersection
Solve the objective function at each of these intersection points to determine which point contains the maximum number of gallons
The objective function is:
g = x + y
The table with the value of g at each of these intersection points is shown below:
intersection point (x,y) gallons of paint (0,0) 0 (0,9) 9 (6,8) 14 ***** (8,0) 8
The maximum gallons of paint for color A and B, given the constraints, is equal to 14
This is comprised of 6 gallons of color A and 8 gallons of color B
6 gallons of color A uses 24 gallons of yellow dye and 8 gallons of color B uses 8 gallons of yellow dye for a total of 32 gallons of yellow dye which is the maximum amount of yellow dye that can be used
6 gallons of color A user 6 gallons of green dye and 8 gallons of color B uses 48 gallons of green dye for a total of 54 gallons of green dye which is the maximum amount of green dye that can be used
Example 3
The Bead Store sells material for customers to make their own jewelry. Customer can select beads from various bins. Grace wants to design her own Halloween necklace from orange and black beads. She wants to make a necklace that is at least 12 inches long, but no more than 24 inches long. Grace also wants her necklace to contain black beads that are at least twice the length of orange beads. Finally, she wants her necklace to have at least 5 inches of black beads
Find the constraints, sketch the problem and find the vertices (intersection points)
Let x = the number of inches of black beads
Let y = the number of inches of orange beads
The objective function is the length of the necklace
There is a maximum length and a minimum length
If you let n equal the length of the necklace, then the objective function becomes:
n = x + y
Since the problem is looking for the number of inches of black beads and the number of inches of orange beads, we will let:
The constraint equations for this problem are:
x >= 0
y >= 0
x + y >= 12
x + y <= 24
x >= 2y
x >= 5
x >= 0 is there because the number of inches of black beads can't be negative
y >= 0 is there because the number of inches of orange beads can't be negative
x + y >= 12 is there because the total length of the necklace has to be greater than or equal to 12 inches
x + y <= 24 is there because the total length of the necklace has to be less than or equal to 24 inches
x >= 2y is there because the length of the black beads has to be greater than or equal to twice the length of the orange beads
x >= 5 is there because the number of inches of black beads has to be greater than or equal to 5
To graph these equations, we have to solve for y in each equation that has y in it and then graph the equality portion of each of them
Your equations for graphing are:
x >= 0
y >= 0
y >= 12 - x
y <= 24 - x
y <= x/2
x >= 5
x = 0 is a vertical line that is the same line as the y-axis
y = 0 is a horizontal line that is the same line as the x-axis
x = 5 is a vertical line at x = 5
A graph of you equations is shown below:
The region of feasibility is the shaded area of the graph. All points within the region of feasibility meet the constraint requirements of the problem
The intersection points bounding the region of feasibility are:
(8,4)
(12,0)
(16,8)
(24,0)
(8,4) is the intersection of the lines y = x/2 and y = 12 - x
To find the point of intersection, set x/2 and 12-x equal to each other and solve for xx
you get:
x/2 = 12-x
Multiply both sides of the equation by 2 to get:
x = 24-2x
Add 2x to both sides of the equation to get:
3x = 24
Divide both sides of the equation by 3 to get:
x = 8
Substitute 8 for x in either equation to get y = 4
(12,0) is the intersection of the line y = 12 - x with the x-axis
(24,0) is the intersection of the line y = 24 - x with the x-axis
To find the point of intersection, set y equal to 0 in each equation and solve for x
(16,8) is the intersection of the lines y = x/2 and y = 24 - x
To find the intersection point, set x/2 equal to 24-x and solve for x
you get:
x/2 = 24-x
Multiply both sides of this equation by 2 to get:
x = 48 - 2x
Add 2x to both sides of this equation to get:
3x = 48
Divide both sides of this equation by 3 to get:
x = 16
Substitute 16 for x in either equation to get:
y = 8
The maximum / minimum necklace length will be at the intersection points of the boundaries of the region of feasibility
Evaluation of the objective function at these intersections yields the following:
objective function is:
x + y = n where n is the length of the necklace in inches
intersection points number of inches
(8,4) 12
(12,0) 12
(16,8) 24
(24,0) 24
The number of inches of black beads is at least twice the number of inches of orange beads
The number of inches of black beads is at least 5
The total length of the necklace is greater than or equal to 12 inches or less than or equal to 24 inches
All the constraints have been met
The maximum length the necklace can be and still meet the constraints is 24 inches
The minimum length the necklace can be and still meet the constraints is 12 inches
Example 4
A garden shop wishes to prepare a supply of special fertilizer at a minimal cost by mixing two fertilizers, A and B
The mixture is to contain:
At least 45 units of phosphate
At least 36 units of nitrate
At least 40 units of ammonium
Fertilizer A costs the shop $.97 per pound
Fertilizer B costs the shop $1.89 per pound
Fertilizer A contains 5 units of phosphate and 2 units of nitrate and 2 units of ammonium
Fertilizer B contains 3 units of phosphate and 3 units of nitrate and 5 units of ammonium
How many pounds of each fertilizer should the shop use in order to minimize their cost?
Let x = the number of pounds of fertilizer A
Let y = the number of pounds of fertilizer B
The objective function is to minimize the cost
The objective function becomes:
c = .97x + 1.89y
The constraint equations are:
Since the number of pounds of each fertilizer can't be negative, 2 of the constraint equations become:
x >= 0
y >= 0
Since the number of units of phosphate has to be at least 45, the constraint equation for phosphate becomes:
5x + 3y >= 45
Since the number of units of nitrate must be at least 36, the constraint equation for nitrates becomes:
2x + 3y >= 36
Since the number of units of ammonium must be at least 40, the constraint equation for ammonium becomes:
2x + 5y >= 40
The constraint equations for this problem become:
x >= 0
y >= 0
5x + 3y >= 45
2x + 3y >= 36
2x + 5y >= 40
In order to graph these equations, you have to solve for y in each equation that has y in it and then graph the equality portion of those equations
The equations to be graphed are:
x >= 0
y >= 0
y >= (45-5x)/3
y >= (36 - 2x)/3
y >= (40-2x)/5
x = 0 is a vertical line that is the same line as the y-axis
y = 0 is a horizontal line that i the same line as the x-axis
A graph of your equation is shown below:
The feasibility region is the area of the graph that is shaded. All points within the region of feasibility meet the constraint requirements of the problem
The intersection points at the boundaries of the feasibility region are:
(0,15)
(3,10)
(15,2)
(20,0)
The intersection points of the boundaries of the region of feasibility contain the minimum cost solution for the objective function in this problem
Now that you have the intersection points, you can solve for the minimum cost equation which is the objective function of:
c = .97x + 1.89y
The following table shows the value of the cost equation at each of the intersection points
intersection points (x,y) c = .97x + 1.89y minimum solution
(0,15) 28.35
(3,10) 21.81
(15,2) 18.33 *****
(20,0) 19.40
The table suggests that we have a minimum cost solution when the value of x is equal to 15 and the value of y is equal 2
When x = 15 and y = 2, the number of pounds of potassium, nitrates, and ammonium are:
phosphate = 5x + 3y = 5*15 + 3*2 = 75 + 6 = 81
nitrate = 2x + 3y = 2*15 + 3*2 = 30 + 6 = 36
ammonium = 2x + 5y = 2*15 + 5*2 = 30 + 10 = 40
All the constraints associated with the minimum cost objective have been met
Reference:
https://www.algebra.com/algebra/homework/coordinate/word/THEO-2011-08-28-03.lesson
SOLVING LOGARITHM
Example 1
Solve the equation (1/2)2x + 1 = 1
Rewrite equation as (1/2)2x + 1 = (1/2)0
Leads to 2x + 1 = 0
Solve for x: x = -1/2
Solve for x: x = -1/2
Example 2
Solve x ym = y x3 for m
Solve x ym = y x3 for m
Divide all terms by x y and rewrite equation as: ym - 1 = x2
Take ln of both sides (m - 1) ln y = 2 ln x
Solve for m: m = 1 + 2 ln(x) / ln(y)
Solve for m: m = 1 + 2 ln(x) / ln(y)
Example 3
Given: log8(5) = b. Express log4(10) in terms of b
Use log rule of product: log4(10) = log4(2) + log4(5)
log4(2) = log4(41/2) = 1/2
Use change of base formula to write: log4(5) = log8(5) / log8(4) = b / (2/3) , since log8(4) = 2/3
log4(10) = log4(2) + log4(5) = (1 + 3b) / 2
log4(2) = log4(41/2) = 1/2
Use change of base formula to write: log4(5) = log8(5) / log8(4) = b / (2/3) , since log8(4) = 2/3
log4(10) = log4(2) + log4(5) = (1 + 3b) / 2
Example 4
Simplify without calculator: log6(216) + [ log(42) - log(6) ] / log(49)
log6(216) + [ log(42) - log(6) ] / log(49)
= log6(63) + log(42/6) / log(72)
= 3 + log(7) /2 log(7) = 3 + 1/2 = 7/2
= log6(63) + log(42/6) / log(72)
= 3 + log(7) /2 log(7) = 3 + 1/2 = 7/2
Example 5
Simplify without calculator: ((3-1 - 9-1) / 6)1/3
((3-1 - 9-1) / 6)1/3
= ((1/3 - 1/9) / 6)1/3
= ((6 / 27) / 6)1/3 = 1/3
= ((1/3 - 1/9) / 6)1/3
= ((6 / 27) / 6)1/3 = 1/3
Example 6
Express (logxa)(logab) as a single logarithm
Use change of base formula: (logxa)(logab)
= logxa (logxb / logxa) = logxb
= logxa (logxb / logxa) = logxb
Example 7
Find a so that the graph of y = logax passes through the point (e , 2)
2 = logae
a2 = e
ln(a2) = ln e
2 ln a = 1
a = e1/2
a2 = e
ln(a2) = ln e
2 ln a = 1
a = e1/2
Reference:
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