Math application Help Plz? :)?

Question

Hi, I'm having trouble with my calculus... If someone can help me with either one of these problems, it would be very helpful.

a.) The cost of operating a bus between Moose Jaw and Saskatoon is 100+5x, where x is the number of passengers. If a ticket costs $20 there will be 10 passengers, while if a tickets costs $15, there will be 20 passengers. Assuming the demand function is linear, what is the Maximum profit?

b.) A lakefront runs east-west. A man in a rowboat is 5 miles due north of point A on the shor. He wishes to get to B, 5 miles due east of A, in the least time. He is able to row 3 miles per hour and walk 5 hour. What is the minimum time for the trip in minutes?

Answer 1

Only one problem per question please.

a.) The cost of operating a bus between Moose Jaw and Saskatoon is 100 + 5x, where x is the number of passengers. If a ticket costs $20 there will be 10 passengers, while if a tickets costs $15, there will be 20 passengers. Assuming the demand function is linear, what is the Maximum profit?
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Let
t = ticket price
x = number passengers
c = cost
p= profit

Find x(t).

Slope = ∆x/∆t = (10 - 20)/(20 - 15) = -10/5 = -2

x - 20 = -2(t - 15) = -2t + 30
x = -2t + 50

Cost.

c = 100 + 5x
c = 100 + 5(-2t + 50) = 100 - 10t + 250 = 350 - 10t

Profit.

p = xt - c = (-2t + 50)t - (350 - 10t)
p = -2t² + 50t - 350 + 10t
p = -2t² + 60t - 350

Take the derivative and set it equal to zero to find the critical points.

dp/dt = -4t + 60 = 0
4t = 60
t = 15

Take the second derivative to find the nature of the critical point.

d²p/dt² = -4 < 0
This implies a relative maximum which is what we want.

Maximum profit occurs when the ticked price is $15.

p = -2t² + 60t - 350 = -2(15²) + 60*15 - 350
p = -450 + 900 - 350 = 100

Maximum profit is $100.
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Answer 2

Why not do your own homework, son?