the school athletics department is raising money for new gym equipment. to raise $7,200 by spring training…

the school athletics department is raising money for new gym equipment. to raise $7,200 by spring training, the department decides to sell t - shirts during the fall semester. based on previous fundraising drives, the expression - 10p + 560 can be used to gauge how many shirts the department will sell depending on the price of a shirt, p. what is the lowest shirt price the athletics department can use to raise exactly $7,200 in revenue?

the school athletics department is raising money for new gym equipment. to raise $7,200 by spring training, the department decides to sell t - shirts during the fall semester. based on previous fundraising drives, the expression - 10p + 560 can be used to gauge how many shirts the department will sell depending on the price of a shirt, p. what is the lowest shirt price the athletics department can use to raise exactly $7,200 in revenue?

Answer

Explanation:

Step1: Set up revenue - equation

Revenue = Price×Quantity. Quantity of shirts is $- 10p + 560$ and price is $p$, and revenue is $7200$. So, $p(-10p + 560)=7200$.

Step2: Expand the equation

Expand $p(-10p + 560)=7200$ to get $-10p^{2}+560p - 7200 = 0$. Divide through by $-10$ to simplify: $p^{2}-56p + 720=0$.

Step3: Use the quadratic formula

For a quadratic equation $ax^{2}+bx + c = 0$ ($a = 1$, $b=-56$, $c = 720$), the quadratic formula is $p=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-56)^{2}-4\times1\times720=3136 - 2880 = 256$.

Step4: Find the values of $p$

$p=\frac{56\pm\sqrt{256}}{2}=\frac{56\pm16}{2}$. We have two solutions: $p_1=\frac{56 + 16}{2}=\frac{72}{2}=36$ and $p_2=\frac{56-16}{2}=\frac{40}{2}=20$.

Answer:

$20$