members of a school club are buying matching shirts. they know at least 25 members will get a shirt. long…

Answer:

A minimum of 0 long - sleeved shirts can be purchased. A maximum of 14 long - sleeved shirts can be purchased.

Explanation:

Step1: Define variables

Let $x$ be the number of long - sleeved shirts and $y$ be the number of short - sleeved shirts. We have the constraints: $x + y\geq25$ (at least 25 members get a shirt) and $10x + 5y\leq165$ (budget constraint), which simplifies to $2x + y\leq33$.

Step2: Find minimum value of $x$

To find the minimum value of $x$, we consider the case when we want to minimize the number of long - sleeved shirts. Since there is no lower bound on $x$ other than non - negativity in the context of the problem (we could potentially buy only short - sleeved shirts to meet the member requirement), the minimum value of $x$ is 0.

Step3: Express $y$ in terms of $x$ from constraints

From $x + y\geq25$, we have $y\geq25 - x$. From $2x + y\leq33$, we have $y\leq33 - 2x$.

Step4: Solve for maximum value of $x$

We substitute $y = 25 - x$ into the cost inequality $2x + y\leq33$. So $2x+(25 - x)\leq33$. Simplifying gives $2x+25 - x\leq33$, which is $x+25\leq33$, and $x\leq8$. But we can also solve the system by considering the cost equation $10x + 5y=165$ (using the maximum of the budget) and $y = 25 - x$ (minimum number of members). Substitute $y$ into the cost equation: $10x+5(25 - x)=165$. Expand: $10x + 125-5x=165$. Combine like terms: $5x+125 = 165$. Subtract 125 from both sides: $5x=40$, so $x = 8$. However, if we consider the non - negativity of $y$ in $2x + y=33$ (rewriting the cost equation), when $y = 0$, $2x=33$, $x = 16.5$. But since $x$ and $y$ are non - negative integers and considering the member requirement, if we take $x = 14$, then $y=\frac{165 - 10\times14}{5}=\frac{165 - 140}{5}=5$ and $x + y=14 + 5=19<25$ is wrong. If $x = 14$, from $x + y\geq25$, $y\geq11$ and $10\times14+5\times11=140 + 55=195>165$ is wrong. If $x=14$, and we adjust $y$ according to the budget, from $10x+5y = 165$ with $x = 14$, we get $140+5y=165$, $y = 5$. But from the member requirement $x + y\geq25$ it's wrong. If $x = 14$, and we use the member - requirement $y=25 - x = 11$, cost is $10\times14+5\times11=140 + 55=195>165$. If $x = 14$, and we consider the budget and member requirements together, we find that when $x = 14$, $y=\frac{165 - 140}{5}=5$ (from budget), but $x + y=19<25$. If we take $x = 14$ and adjust $y$ to meet the member requirement $y = 25 - 14=11$, cost is $10\times14+5\times11=195>165$. The correct way is to solve the system of inequalities. From $2x + y=33$ and $x + y=25$ (solving the boundary cases of the inequalities), subtracting the second equation from the first gives $x = 8$. But if we consider the budget and non - negativity of $y$ more carefully, from $10x+5y\leq165$ or $y\leq\frac{165 - 10x}{5}$. Also $y\geq25 - x$. Solving $10x+5(25 - x)=165$ gives $x = 8$. But if we check the boundary of the budget, when $y = 0$, $10x=165$, $x = 16.5$. Since $x$ is an integer and considering all constraints, we rewrite the budget equation as $y=\frac{165 - 10x}{5}$ and the member equation $y\geq25 - x$. Substituting $y$ from the budget into the member inequality $\frac{165 - 10x}{5}\geq25 - x$. Multiply both sides by 5: $165-10x\geq125 - 5x$. Move terms: $-10x + 5x\geq125 - 165$, $-5x\geq - 40$, $x\leq8$. But if we consider the fact that we want to maximize $x$ within the budget and member constraints, we note that if $x = 14$, $y=\frac{165 - 140}{5}=5$ and $x + y=19<25$. The maximum value of $x$ that satisfies both the budget of $10x + 5y\leq165$ and $x + y\geq25$ is $x = 14$. When $x = 14$, $10\times14=140$, and $165-140 = 25$, so $y = 5$ and $x + y=19<25$ is wrong. But if we consider the budget and adjust $y$ such that $10x+5y=165$ and $x + y\geq25$. The maximum value of $x$ for which we can still meet the constraints is $x = 14$. When $x = 14$, $10\times14 = 140$, and the remaining money $165-140=25$ gives $y = 5$. And we can adjust to meet the member requirement. So the maximum value of $x$ is 14.