# A sports memorabilia store makes $6 profit on each football it sells and$5.50 profit on each baseball it sells. In a typical

A sports memorabilia store makes $6 profit on each football it sells and$5.50 profit on each baseball it sells. In a typical month, it sells between 35 and 45 footballs and between 40 and 55 baseballs. The store can stock no more than 80 balls total during a single month. What is the maximum profit the store can make from selling footballs and baseballs in a typical month?

## This Post Has 8 Comments

1. theojw says:

b

Step-by-step explanation:

460.00

2. sobell7725 says:

We are given with $6 profit per football sold and$5.50 profit per baseball sold. There 35-45 footballs and 40-55 baseballs sold but there are only 80 balls stock in the store. The maximum profit is incurred from selling more of football and less of baseball, not the vise versa. Equal number could also be possible. Having 40 footballs and 40 baseballs can be a possibility. However, having 39 footballs and 41 baseballs has smaller profit than the equal number. Likewise, 41 footballs leads to 39 baseballs which is not within the range. Hence answer is 40 footballs and 40 baseballs with a total profit of $460. 3. Bloom247 says: 460 Step-by-step explanation: 4. eweqwoewoji says: Since footballs yield more profit than baseballs, the store would aim to sell more footballs. It would like to sell the maximum number of 45 footballs. However, only a maximum of 80 footballs and baseballs can be kept in stock, and at least 40 of those are baseballs. This means that there are only 40 footballs at most, and the maximum profit for the store occurs if they sell 40 footballs and 40 baseballs. This gives a total profit of 40($6) + 40($5.5) =$460.

5. vicki1234 says:

6.00×45=270.00
5.50×55=302.50

so 270.00 +302.50 = 572.50

6. hannahkharel2 says:

The answer is option B : $460. Step-by-step explanation: A sports memorabilia store makes$6 profit on each football it sells and $5.50 profit on each baseball it sells. Let there be x footballs and y baseballs, then we have $p=6x+5.5y$ Now footballs sells between 35 and 45 so, range is : $35\leq x\leq 45$ And baseballs sells between 40 and 55 so, range is : $40\leq y\leq 55$ The store has no more than 80 footballs and baseballs in stock during the month, this gives , $x+y\leq 80$ Now out of 80, at least 40 are baseballs. So, there are only 40 footballs at most. The store gets maximum if they sell 40 footballs and 40 baseballs. This gives a total profit of $40(6)+40(5.5)=460$ dollars. Therefore, the answer is option B :$460.

7. brooke012002 says:

We are given with $6 profit per football sold and$5.50 profit per baseball sold. There are 35-45 footballs and 40-55 baseballs sold but there are only 80 balls stock in the store. The maximum profit is incurred from selling more of football and less of baseball. The equal number could also be possible. Having 40 footballs and 40 baseballs can be a possibility. However, having 39 footballs and 41 baseballs has smaller profit than the equal number. Likewise, 41 footballs lead to 39 baseballs which are not within the range. The answer is 40 footballs and 40 baseballs with a total profit of $460. 8. willveloz4 says: The answer to this problem is$460.00