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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?

Question

The school athletics department is raising money for new gym equipment. To raise 7,200byspringtraining,thedepartmentdecidestosellTshirtsduringthefallsemester.Basedonpreviousfundraisingdrives,theexpression10p+560canbeusedtogaugehowmanyshirtsthedepartmentwillselldependingonthepriceofashirt,p.Whatisthelowestshirtpricetheathleticsdepartmentcanusetoraiseexactly7,200 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?

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Solution

To solve this problem, we first need to understand that the revenue generated from selling the T-shirts is the product of the price per shirt (p) and the number of shirts sold. The number of shirts sold is given by the expression -10p + 560.

So, the revenue R can be expressed as R = p(-10p + 560).

We know that the athletics department wants to raise exactly $7200. So, we can set up the equation:

7200 = p(-10p + 560)

Solving this equation will give us the price per shirt that the athletics department needs to set to raise exactly $7200.

First, distribute p to the terms inside the parentheses:

7200 = -10p^2 + 560p

Rearrange the equation to set it equal to zero:

10p^2 - 560p + 7200 = 0

Divide the entire equation by 10 to simplify:

p^2 - 56p + 720 = 0

This is a quadratic equation in the form of ax^2 + bx + c = 0. We can solve it using the quadratic formula, which is x = [-b ± sqrt(b^2 - 4ac)] / (2a). In this case, a = 1, b = -56, and c = 720.

Substitute these values into the quadratic formula:

p = [56 ± sqrt((-56)^2 - 41720)] / (2*1) p = [56 ± sqrt(3136 - 2880)] / 2 p = [56 ± sqrt(256)] / 2 p = [56 ± 16] / 2

This gives us two possible solutions: p = 36 and p = 20.

However, the expression -10p + 560 for the number of shirts sold implies that as the price increases, the number of shirts sold decreases. Therefore, the lowest shirt price the athletics department can use to raise exactly 7200inrevenueis7200 in revenue is 20.

This problem has been solved

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