the profit, P(x), of a video company, in thousands of dollars, is given by P(x)=-5x^2 + 550x - 5000. where x is the amount spent on advertising, in thousands.

a. determine the maximum profit that the company can make. i got $16,125.00 b. determine the amount spent on advertising that will result in a profit. i keep gettin a negative number which i know cannot b rite.bt i am using the quadratic formula. c. determine the amounts spent on advertising that will result in a profit of at least$4,000,000.00

2. Originally Posted by justine_0612
the profit, P(x), of a video company, in thousands of dollars, is given by P(x)=-5x^2 + 550x - 5000. where x is the amount spent on advertising, in thousands.

a. determine the maximum profit that the company can make. i got $16,125.00 this is incorrect. find the x-coordinate for the vertex and plug it into the formula. the result for P(x) is the desired profit. recall: For a quadratic$\displaystyle ax^2 + bx + c = 0$, the x-coordinate of the vertex is given by:$\displaystyle x = \frac {-b}{2a}$b. determine the amount spent on advertising that will result in a profit. i keep gettin a negative number which i know cannot b rite.bt i am using the quadratic formula. again, you are way off here. BOTH roots are positive, so you did something wrong. recall: for a quadratic$\displaystyle ax^2 + bx + c = 0$, the roots of the quadratic are given by:$\displaystyle x = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}$try again c. determine the amounts spent on advertising that will result in a profit of at least$4,000,000.00
are you sure you didn't mean to write 4000? because that is impossible

3. nope its 4,000,000.00
so the sheet seys...
bt im not quite sure wat u were talkin abt b4. can u elaborate or explain thouroughly. i'm still tryin to learn this.

4. I think there's a lot of confusion here over the 'in thousands' comment.
This is how I interpret it:

$\displaystyle -5x^2+550x-5000$

Where x represents money spent on advertising 'in thousands'.

Therefore, any output will be 'in thousands'.

So if you spend $50,000 on advertising, then x = 50$\displaystyle -5(50)^2+550(50)-5000=10000$So you gain$10,000,000 profit.

But remember this is just my interpretation.

For Question A complete the square on $\displaystyle P(x)=-5x^2+550x-500$. This will give you the turning point. And from its graph you will find the turning point is the place of highest profit. Remember that it is 'in thousands'.

For Question B use the quadratic formula as Jhevon indicated. From the shape of the graph you can see that the 'region of profit' will be in between the two roots. It is very similar to my solution to Question C, only this time set $\displaystyle -5x^2+500x-5000 \geq 0$

For Question C, set

$\displaystyle -5x^2 + 550x - 5000 \geq 4000$

$\displaystyle -5x^2+550x-9000 \geq 0$

$\displaystyle x^2-110+1800 \geq 0$

I would just cut straight to the quadratic formula here

$\displaystyle \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$\displaystyle x=\frac{-(-110) \pm \sqrt{(-110)^2 - 4 \times 1800\times 1}}{2 \times 1}$

$\displaystyle x=\frac{110\pm 70}{2}$

$\displaystyle x=55\pm 35$

$\displaystyle x=20$ or $\displaystyle x=90$ "thousands" on advertising.

So the places when x gives 4,000,000 profit is at x = 20 and x = 90. If you look at the shape of the graph of $\displaystyle x^2-110+1800$, you will see that it is an inverted parabola. This can be deduced from the fact that the $\displaystyle x^2$ coefficient is negative. Then it is greater than zero between x = 20 and x = 90.

Therefore, if you want a profit of at least 4,000,000

Then $\displaystyle 20 \geq x \geq 90$

5. i understand everything u have sed there...bt how did u interpret x=50?? i neva had a value of 50,000

6. Originally Posted by justine_0612
i understand everything u have sed there...bt how did u interpret x=50?? i neva had a value of 50,000
the units for the formula is thousands. so x = 50 means 50,000 was spent on advetising. now the last question makes more sense. the 4000000 should be 4000 for the "scale" we are using.

the formula i gave you for the vertex calculates the turning point a lot quicker and easier than completing the square does. but do it whichever way you understand