SOLUTION: Jim is a fisherman. He varies the depth at which he fishes according to the following function: D(t)= -t^2+10t where t is measured in hours. Estimate the time when he fishes at the

Algebra ->  Systems-of-equations -> SOLUTION: Jim is a fisherman. He varies the depth at which he fishes according to the following function: D(t)= -t^2+10t where t is measured in hours. Estimate the time when he fishes at the      Log On


   

Question 65295: Jim is a fisherman. He varies the depth at which he fishes according to the following function: D(t)= -t^2+10t where t is measured in hours. Estimate the time when he fishes at the greatest depth and tell me that depth.
Found 2 solutions by josmiceli, Earlsdon:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
D(t)= -t^2+10t
If this is put in the form
D%28t%29+=+at%5E2+%2B+bt+%2B+c, the the maximum is at (-b%2F%282a%29,D)
b = +10
a = -1
-b/(2a) = -10/-2
-10/-2 = 5
D(5) = -(5)^2 + 10*5
D(5) = -25 + 50
D(5) = 25
so, the maximum is at (5,25)
He fishes at a depth of 25 in 5 hours. Is D in units of feet?
Problem didn't say.

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Since the equation for the depth as a function of time is a parabola that opens downward (-t%5E2), the vertex of the parabola occurs at the maximum value of the independent variable, t.
The t-coordinate of the vertex can be found by: t+=+%28-b%29%2F2a where: b = 10 and a = -1.
t+=+%28-10%29%2F2%28-1%29
t+=+5
The maximum depth, D, occurs at t = 5 or 5 hours.
To find the value of D at this time, substitute t=5 into the original equation and solve for D.
D+=+-%285%29%5E2+%2B+10%285%29
D+=+-25%2B50
D+=+25
The maximum depth is 25 feet(?).