Linear Programming Practice - Algebra 2
Authored by Amy Granier
Mathematics
10th Grade
Used 7+ times
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13 questions
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1.
MULTIPLE CHOICE QUESTION
2 mins • 1 pt
A factory makes purses and shoes. For each purses, you make a $100 profit. For each pair of shoes, you make $50 profit. A purse takes 3 machine hours and 5 man hours. A pair of shoes requires 4 machine hours and 2 man hours. The machine can operate for up to 50 hours. The total number of man hours is 60. How many purses and pairs of shoes should the factory make in order to maximize profits?
What are the variables for this situation
X = profit
Y = cost
X = machine hours
Y = man hours
X = purses
Y = pairs of shoes
X = man hours
Y = Profits
Answer explanation
The variables represent the quantities produced: X = purses and Y = pairs of shoes. This choice aligns with the goal of maximizing profits based on the production constraints given in the problem.
2.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
A factory makes purses and shoes. For each purses, you make a $100 profit. For each pair of shoes, you make $50 profit. A purse takes 3 machine hours and 5 man hours. A pair of shoes requires 4 machine hours and 2 man hours. The machine can operate for up to 50 hours. The total number of man hours is 60. How many purses and pairs of shoes should the factory make in order to maximize profits?
Given that X = # of Purses and that Y = # of pairs of shoes
What is the objective function?
P = 4x + 3y
P = 100X + 50y
P = 3x + 5y
P = 4x + 2y
Answer explanation
The objective function represents total profit. For purses, profit is $100 per unit (X), and for shoes, it's $50 per unit (Y). Thus, the correct objective function is P = 100X + 50Y.
3.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
A factory makes purses and shoes. For each purses, you make a $100 profit. For each pair of shoes, you make $50 profit. A purse takes 3 machine hours and 5 man hours. A pair of shoes requires 4 machine hours and 2 man hours. The machine can operate for up to 50 hours. The total number of man hours is 60. How many purses and pairs of shoes should the factory make in order to maximize profits?
Given that x = # of Purses and y = # of pairs of shoes
What inequality represents the constraint of man hours?
4x + 2y < 50
5x + 2y < 60
3x + 5y < 50
5x + 2y < 50
Answer explanation
The constraint for man hours is represented by the total man hours used for purses and shoes. Each purse requires 5 man hours and each pair of shoes requires 2 man hours, leading to the inequality 5x + 2y < 60.
4.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
A factory makes purses and shoes. For each purses, you make a $100 profit. For each pair of shoes, you make $50 profit. A purse takes 3 machine hours and 5 man hours. A pair of shoes requires 4 machine hours and 2 man hours. The machine can operate for up to 50 hours. The total number of man hours is 60. How many purses and pairs of shoes should the factory make in order to maximize profits?
Given that x = # of Purses and y = # of pairs of shoes
What inequality represents the constraint of machine hours?
4x + 3y < 50
3x + 4y < 60
3x + 4y < 50
5x + 2y < 60
Answer explanation
The constraint for machine hours is based on the hours required for purses and shoes. Each purse requires 3 machine hours and each pair of shoes requires 4 machine hours. Thus, the correct inequality is 3x + 4y < 50.
5.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
The Wily Monka Chocolate Factory makes 2 kinds of candies: Everlasting Goobstoopers and Monka Bars. The profit for a case of Everlasting Goob Stoopers is $20. The profit for a case of Monka Bars is $35. The goobstoopers have to have 4 machine hours and 5 man hours. The Monka bar has to have 5 machine hours and 10 man hours. There are only 410 machine hours and 700 Man hours available. How many cases of each kind of candy should the factory make to maximize profit?
How would you define your varaibles for this problem?
x = Profit
y = cst
x = cases of Everlasting Goobstoopers
Y = cases of Monka Bars
x = candies
y = hours
x = man hours
y = machine hours
Answer explanation
The correct choice defines x as the cases of Everlasting Goobstoopers and y as the cases of Monka Bars. This aligns with the goal of maximizing profit based on the number of cases produced.
6.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
The Wily Monka Chocolate Factory makes 2 kinds of candies: Everlasting Goobstoopers and Monka Bars. The profit for a case of Everlasting Goob Stoopers is $20. The profit for a case of Monka Bars is $35. The goobstoopers have to have 4 machine hours and 5 man hours. The Monka bar has to have 5 machine hours and 10 man hours. There are only 410 machine hours and 700 Man hours available. How many cases of each kind of candy should the factory make to maximize profit?
Given that x = cases od Everlastign Goobstoopers and y = cases of Monka Bars
What is the objective function?
P = 20x + 35y
P = 4x + 5y
P = 35x + 20y
P = 5x + 10y
Answer explanation
The objective function represents the total profit. For Everlasting Goobstoopers, the profit is $20 per case (20x), and for Monka Bars, it's $35 per case (35y). Thus, the correct objective function is P = 20x + 35y.
7.
MULTIPLE SELECT QUESTION
45 sec • 1 pt
The Wily Monka Chocolate Factory makes 2 kinds of candies: Everlasting Goobstoopers and Monka Bars. The profit for a case of Everlasting Goob Stoopers is $20. The profit for a case of Monka Bars is $35. The goobstoopers have to have 4 machine hours and 5 man hours. The Monka bar has to have 5 machine hours and 10 man hours. There are only 410 machine hours and 700 Man hours available. How many cases of each kind of candy should the factory make to maximize profit?
Given that x = cases od Everlastign Goobstoopers and y = cases of Monka Bars
What are the TWO constraint inequalities?
4x + 5y < 410
4x + 5y < 700
5x + 10y < 410
5x + 10y < 700
Answer explanation
The first constraint, 4x + 5y < 410, represents machine hours, while the second constraint, 5x + 10y < 700, represents man hours. These inequalities ensure the production does not exceed available resources.
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