a clothing store is holding a buy one, get one half off sale for tshirts, where the second tshirt purchased is half the original price.
a; let b be the cost to the store to buy the shirts, and let s be the original selling price. write an inequality relating the two variables such that the store profits more from a customer purchasing two T-shirts then from a customer purchasing one.(hint: the expression s-b represents the profit when one shirt is purchased. What is an expression for the profit when two shirts are purchased?)
b; isolate b in the inequality. explain what the inequality says about the buying and selling prices
c; if the store buys tshirts for 12$, what selling price range will make selling two tshirts more profitable that selling one shirt
5 years ago
Answered By Leonardo F
The profit is the selling price for the t shirt minus the buying price. The profit made for the first t shirt will be:
s-b
The profit made for the second t shirt, since it's 50% off, will be:
0.5s-b
Hence, the profit made for selling two t shirts is:
(s-b)+(0.5s-b)
a) The inequality will be:
(s-b)+(0.5s-b)>(s-b)
b) If we isolate b in the inequality above:
We can cancel (s-b) on both sides. This leaves us with:
(0.5s-b)>0
b<0.5s
This means that the store must buy the t shirts at a cost that is less than half the selling price of the first t shirt for selling two t shirts to be more profitable than just selling one. Otherwise, selling one will be more advantageous.
c) b=12
Hence:
0.5s>12
s>12/0.5
s>24
This means that the selling price must be greater than $24 to make selling two t shirts to be more profitable than just selling one.
5 years ago
Answered By Emily H
A) Since the profit per t-shirt at the standard price is $p_1=\left(s-b\right)$p1=(s−b) , we can then represent profit of the second, half-off shirt as $p_2=\left(\frac{1}{2}s-b\right)$p2=(12 s−b) and the profit from a sale can be written as $P=\frac{3}{2}s-2b$P=32 s−2b .
Thus, the inequality we need can be written as
$\left(s-b\right)+\left(\frac{1}{2}s-b\right)>\left(s-b\right)$(s−b)+(12 s−b)>(s−b) or $\frac{3}{2}s-2b>s-b$32 s−2b>s−b
B) Then, if we simplify the above equation, we get
$\frac{1}{2}s-b>0$12 s−b>0
And isolating for b:
$\frac{1}{2}s>b$12 s>b
Which indicates that either the buying price must be less than half the selling price in order for the store to not take a loss on the second shirt. However, Its easier to see a more complete picture is to look at the profit equation above, so
$P=\frac{3}{2}s-2b$P=32 s−2b
As we want to have a profit, we can rewrite this as
$P>0$P>0
$\frac{3}{2}s-2b>0$32 s−2b>0
$\frac{3}{2}s>2b$32 s>2b
$\frac{3}{4}s>b$34 s>b
So we can see that the buying price must be less than three quarters of the selling price in order to make a profit in the sale.
C) Using the equations we identified above, we can use $p_2>0$p2>0 to solve this question.
$b=12$b=12
$\frac{1}{2}s>12$12 s>12
$s>24$s>24
So if each shirt is at least $24, then the sale of two shirts will make a greater profit than that of a single shirt.