a sports memorabilia store makes $6 profit on each football it sells and $5.50 profit on each baseball it…

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? $457.50 $460.00 $462.50 $572.50

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? $457.50 $460.00 $462.50 $572.50

Answer

Explanation:

Step1: Define variables

Let $x$ be the number of footballs and $y$ be the number of baseballs.

Step2: Set up constraints

$35\leq x\leq45$, $40\leq y\leq55$, $x + y\leq80$.

Step3: Define profit function

The profit function $P=6x + 5.5y$.

Step4: Analyze the constraints for maximum profit

Since the profit per - football ($$6$) is higher than the profit per - baseball ($$5.5$), we want to sell as many footballs as the constraints allow. The maximum number of footballs we can sell while still satisfying the total - stock constraint is $45$ (because if $x = 45$, then $y$ can be at most $80 - 45=35$, but $y$ must be at least $40$. So we adjust $x$). Since $x + y\leq80$, to maximize $P = 6x+5.5y$, we note that the upper - bound of $x$ is limited by the total number of balls. If we take $x = 25$ (because $y$ has a lower bound of $40$ and $x + y\leq80$, so $x=80 - 40 = 40$ is also a candidate, but we calculate the profit for different valid combinations). When $x = 25$ and $y = 55$: $P=6\times25+5.5\times55$ $P = 150+302.5$ $P=452.5$ (not the maximum). When $x = 40$ and $y = 40$: $P=6\times40 + 5.5\times40$ $P=(6 + 5.5)\times40$ $P = 11.5\times40=460$ When $x = 45$ and $y=35$ (not valid as $y$ should be at least $40$).

Answer:

$$460.00$