Question

Sweatshirts at the school store cost $30. They currently sell about 4 shirts per month. They have decided to decrease the price of the shirts. They found the for each $1.50 decrease, they will sell 2 more shirts per month. What is the maximum income for the school store.A. $30B. $9C. $363D. $16.50

Answer 1

For this question the store earnings are:

`\begin{gathered} G(x)=(30-1.5\cdot x)\cdot(4+2x) \\ \text{Where x is the number of times that the price is lowered \$}1.5 \end{gathered}`

Now, we use the first and second derivative criteria to maximize G(x).

`\begin{gathered} G^{}(x)=-3x^2+54x+120 \\ G^(\prime)(x)=-6x+54 \\ G^{^(\prime)^(\prime)}(x)=-6 \end{gathered}`

Setting G'(x)=0 and solving for x we get:

`\begin{gathered} -6x+54=0 \\ 6x=54 \\ x=9 \end{gathered}`

Since G''(9)=-6<0, the function has a maximum when x=9. Finally, the maximum income for the school store is:

`G(9)=-3(9)^2+54(9)+120=363`