## anonymous one year ago ) A hotel finds that its revenue is given by R = 8000 + 760x - 30x2 when it charges 80 + 10x dollars for a room. To the nearest dollar, what is the maximum revenue it can earn?

1. Jack1

$$\large R = 8000 + 760x - 30x^2$$ maximum occurs at the turning point of this line can you work out the turning point of a parabola?

2. anonymous

i tried using the -b/2(a) and then plugging that result for x but thats where i get lost

3. Jack1

have you worked with derivatives yet?

4. anonymous

no.

5. Jack1

damn... k hang on

6. Jack1

For $$\Large y = ax^2 + bx + c$$ $$\Large (c - \frac{b^2}{4a}) = ymax~~ or~~ ymin$$ value at the max point

7. Jack1

so ur equation is $$\large R = 8000 + 760x - 30x^2$$ $$\large y = ax^2 + bx + c$$ so a = -30 b = 760 c = 8000 no can you solve for ymax?

8. anonymous

this was the answer in the help video but i think it's wrong. R=(80+10x)(100-3x) 8000-24x+1000x-30x^2 R=-30x^2+976+8000 X= -b/2a= -976/2(-30) and with all of that i was suppposed to end up with $12813 for the maximum 9. anonymous the answer i got was 11400 10. anonymous these were the options A)$9,920 B) $13,530 C)$12,813 D) no maximum

11. Jack1

hmm... 12813 is correct using ymax = c - (b^2)/4a ymax = 8000 - 760^2/4(-30) ymax = 8000 - 57760/-120 ymax = 8000 + 4813.33 = \$12813 ... does this kinda make ssense tho?

12. anonymous

yes, i have just never used that formula before that's why. thank you :)

13. Jack1

welcomes ;)