Help with some quick math 11 problems

[webnewland]

Hi:
Could you guys clarify 3 math 11 problems for me? Tutoring someone today and sadly seem to have forgotten how to do these questions quickly. Please show some brief steps.

1. X^2 - kx + (k+8)

For what values of K does the equation produce 2 different real roots?


2. y= -(0.003x)^3 + 6x^2 - 62

When graph it, the two zeros near (0,0) can be seen easily, what can we conclude about the the other zero. Is it non-existent, negative, or positive?


3. A charter bus company want to make the most money, for 30 or less passengers they charge $80 per ticket, for each person over 30, the fare decrease by $2 per person. How many passengers should they have to make maximum revenue?

Thanks

 

[Legendary]

Originally posted by: webnewland
Hi:
Could you guys clarify 3 math 11 problems for me? Tutoring someone today and sadly seem to have forgotten how to do these questions quickly. Please show some brief steps.

1. X^2 - kx + (k+8)

For what values of K does the equation produce 2 different real roots?


2. y= -(0.003x)^3 + 6x^2 - 62

When graph it, the two zeros near (0,0) can be seen easily, what can we conclude about the the other zero. Is it non-existent, negative, or positive?


3. A charter bus company want to make the most money, for 30 or less passengers they charge $80 per ticket, for each person over 30, the fare decrease by $2 per person. How many passengers should they have to make maximum revenue?

Thanks

(a) the discriminant (b^2 - 4ac) must be greater than zero (zero is one real root)
a = 1
b = -k
c = (k+8)
k^2 - 4k + 32 > 0
(k-8)(k+4) > 0
k > 8 or k < -4 (anything in between will make one binomial positive and the other negative)

 

[Special K]

Here is some help for the third one:

Let r(x) be the revenue. In this case, r(x) would be:

r(x) = 80x for 0<=x<=30

and

r(x) = (80 - 2(x - 30))x for 31<=x<=inf.

80 is the base price for a ticket, x is the number of passengers. So the price of a ticket is 80 - 2 times (x - 30), or 2 times the number of passengers over 30. All of this is multiplied by x, the number of passengers on the bus. This simplifies to:

r(x) = -2x^2 + 140x

To find the number of passengers that would generate the most revenue, find the maximum value of r(x) on the 2 intervals. For the first function, the max value is r(30) = 2400 (since that price is only valid for up to 30 passengers). To find the maximum of the second function, take the derivative and set = to 0:

r'(x) = -4x + 140, -4x + 140 = 0, x = 35

r(35) = 2450

So they should have 35 passengers to generate max. revenue. I'm pretty sure that's correct.