Pauline Wong spends 4 hours selling a used car and 6 hours selling a new car. She works no more than 40 hours per week. In order to receive a bonus, she must sell at least one used car and four new cars each week. In that case, she recives a bonus of $180.00 for each used car and $290.00 for each new car. How many used and how many new cars should she try to sell to maximize her bonus?
Just from the data, you can see that she gets $180/4 or $45/hour worked for the used cars and $290/6 or $48.33/hour worked for the new cars. In this case, I would sell one used car and the rest new cars — this would mean one used car and 6 new cars.
4 + 6X = 40
I hope this helps. Thanks for asking.
To solve this problem, we can use a combination of algebra and logical reasoning. Let's break it down step by step:
1. We know that Pauline spends 4 hours selling a used car and 6 hours selling a new car. So, in order to maximize her bonus, she should work only 4 hours on a used car and the remaining time on selling new cars.
2. The total number of hours she can work in a week is limited to 40 hours.
3. Let's assume she sells x used cars and y new cars.
4. From step 1, we know that she spends 4 hours on each used car, so the total time spent on selling used cars will be 4x hours.
5. The time spent on selling new cars will be 6y hours.
6. From steps 4 and 5, we can set up the equation: 4x + 6y = 40.
7. We also know that in order to receive a bonus, she must sell at least one used car and four new cars each week.
8. So, we can set up another equation based on the requirement: x ≥ 1 and y ≥ 4.
9. Now, we need to determine the values of x and y that maximize her bonus.
10. In this case, the bonus for selling a used car is $180.00 and the bonus for selling a new car is $290.00.
11. To maximize her bonus, she should aim to maximize the total amount earned for selling new cars.
12. As per the given data, she earns $180.00 selling each used car and $290.00 selling each new car.
13. We can calculate the amount earned for selling used cars: 180x.
14. We can calculate the amount earned for selling new cars: 290y.
15. The total bonus earned can be represented as: Total Bonus = 180x + 290y.
16. To maximize the total bonus earned, we need to maximize the value of Total Bonus.
17. Therefore, our objective is to maximize the function f(x, y) = 180x + 290y.
18. Now, we have two equations:
- 4x + 6y = 40 (from step 6)
- x ≥ 1 and y ≥ 4 (from step 8)
19. To find the values of x and y that maximize the function f(x, y), we can solve the system of equations formed by steps 6 and 8 using either substitution or elimination method.
20. Once we find the values of x and y, we can calculate the corresponding bonus using the formulas mentioned in step 10.
In conclusion, to maximize her bonus, Pauline should aim to sell one used car and six new cars, which will result in a bonus of $180.00 for the used car and $290.00 for each new car.
To maximize Pauline Wong's bonus, let's solve the problem step-by-step:
Let's assume Pauline sells X used cars and Y new cars to maximize her bonus.
According to the given information:
- Selling one used car earns her a bonus of $180.
- Selling one new car earns her a bonus of $290.
Therefore, the total bonus earned by selling X used cars and Y new cars is:
Bonus = 180X + 290Y
Now, let's consider the time constraint:
- Pauline spends 4 hours selling a used car and 6 hours selling a new car.
- She works no more than 40 hours per week.
The time constraint equation can be written as:
4X + 6Y ≤ 40
Next, we need to analyze the sales requirement for the bonus:
- She must sell at least one used car and four new cars each week to receive a bonus.
Therefore, the sales requirement can be written as:
X ≥ 1 (At least 1 used car)
Y ≥ 4 (At least 4 new cars)
To maximize the bonus, we need to find the maximum value of the bonus function within the given constraints.
Now, we can solve the problem using linear programming techniques or substitution.
Let's substitute the value of X in the time constraint equation:
4(1) + 6Y ≤ 40
4 + 6Y ≤ 40
6Y ≤ 40 - 4
6Y ≤ 36
Y ≤ 36/6
Y ≤ 6
Therefore, the maximum value of Y is 6.
Now, let's substitute the value of Y in the sales requirement equation for X:
X ≥ 1
Since X must exceed 1 and maximize the bonus, the maximum value of X will be 1.
Therefore, the maximum values of X and Y are:
X = 1 used car
Y = 6 new cars
By selling 1 used car and 6 new cars, Pauline will maximize her bonus.
I hope this helps! Let me know if you have any further questions.