Suppose a local vendor charges $2 per hot dog and that the number of hot dogs sold per hour x is given by x(t)=−4t^2+20t+92, where t is the number of hours since 10 AM, 0≤t≤4.

Question

d. If the price were raised to $3 per hot dog with no change in the x(t) equation, which hour would produce the most revenue? Why?

e. If the price were dropped to $1 per hot dog, but that price drop caused the number of sales to increase according to the function x(t)=−9t2+22t+138, would the vendor make more money at the original $2 price, or at the $1 price?

(when answering pls label the variables (1,2,3.. or a,b,c...) when writing the answers, THANK YOU)

Answer 1

d)

Since raising the price to $3 and x(t) is unchanged, we can still use x(t) to determine the time.

This is an inverted parabola, as in at²+bt + c = 0. The vertex of a parabola is at t = -b/2a, or

t = - 20/-8 = 20/8 = 5/2 = 2.5

So the hour will be 10:00 + 2:30 = 12:30 PM

e)

Old number = x(2.5) = -4(2.5)^2 + 4(2.5) + 92 = 77

Old Revenue = 2(77) = $154

The max t for the new formula

t = -b/2a = -22/(-18) = 22/18 = 11/9

x(11/9) = -9(11/9)² +22(11/9) +138 = -121/9 + 242/9 +138 = 121/9 + 1242/9 = 151.44

New revenue = $1 X 151.44 = $151.44

Selling at $2 is better.