Question

A company's monthly profit, P, from a product is given by P=−x2+105x−1050, where x is the price of the product in dollars. What is the lowest price of the product, in dollars, that gives a monthly profit of $1550?

Answer 1

just another quadratic formula
\[ 1550=−x^2+105x−1050 \]
can you work that out?

Answer 2

how do you work out 2,600=-x squared +105x

Answer 3

how about:
\[ 0=x^2-105x+2600 \]

Answer 4

I added 1050 to both sides
and got 2,600=-x squared +105x

Answer 5

yeah nah you want to equalize it by 0, so that you can then solve it

ok so we can see that \( a=1 ~~~ ; ~~~ b=-105 ~~~ ;~~~ c=2600 \)

plug that information into \( \frac{-b\pm\sqrt{b^2-4ac}}{2a} \)

Answer 6

ok

Answer 7

Thank you

Answer 8

I got 65 and 40

Answer 9

I got it right

Answer 10

youre right -- i will just do the latex cuz it took me a while and i dont want to delete it
\[ x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
\[ x= \frac{-(-105)\pm\sqrt{(-105)^2-4\cdot 1 \cdot 2600}}{2\cdot 1 } \]
\[x= \frac{-(-105)\pm\sqrt{(-105)^2-10400}}{2\cdot 1 } \]
\[ x= \frac{-(-105)\pm\sqrt{11025-10400}}{2\cdot 1 } \]
\[x= \frac{-(-105)\pm\sqrt{625}}{2\cdot 1 } \]
\[ x= \frac{-(-105)\pm\sqrt{625}}{2\cdot 1 } \]
\[ x=\frac{-(-105)\pm\sqrt{25^2}}{2\cdot 1 } \]
\[ x= \frac{105\pm25}{2\cdot 1 } \]
\[ x_1=65 ~~~~~~~~~~ x_2=40 \]
now back to the question "What is the `lowest price` of the product, in dollars, that gives a monthly profit of $1550? "
so the answer would be what?

Answer 11

40$

Answer 12

there you go!