Can anyone help me with these two optimization problems?
A boat leaves a dock at 9:00 P.M. and travels due south at a speed of 20 km/h. Another boat has been heading due east at 15 km/h and reaches the same dock at 10:00 P.M. How many minutes after 9:00 P.M. were the two boats closest together? Round your answer to the nearest minute.
An oil refinery is located on the north bank of a straight river that is 2 km wide. A pipeline is to be constructed from the refinery to storage tanks located on the south bank of the river 3 km east of the refinery. The cost of laying pipe is $300,000 per km over land to a point P on the north bank and $600,000 per km under the river to the tanks. To minimize the cost of the pipeline, how far from the refinery should P be located? (Give your answer correct to two decimal places.)
write the position of each boat as a function of time. Then, write the distance equation between those positions. Then, minimize.
On the second, make a diagram,going up the river L km, then across the river some distance. WRite the cost function as a function of L.
Sure, I can help you with these two optimization problems step-by-step.
Problem 1: Boat distance optimization
To find the minimum distance between the two boats, we can start by determining their positions at any given time after 9:00 PM. Let's say we are considering the time t hours after 9:00 PM.
The southbound boat will have traveled a distance of 20t km from the dock, while the eastbound boat will have traveled a distance of 15(t - 1) km from the dock, as it leaves at 10:00 PM.
To find the distance between the two boats, we can use the Pythagorean theorem, which states that in a right-angled triangle, the sum of the squares of the two shorter sides is equal to the square of the longest side (the hypotenuse).
Let's define the distance between the boats as D(t). Using the Pythagorean theorem, we can write:
D(t)^2 = (20t)^2 + (15(t - 1))^2
D(t)^2 = 400t^2 + 225(t^2 - 2t + 1)
D(t)^2 = 400t^2 + 225t^2 - 450t + 225
Simplifying this equation, we get:
D(t)^2 = 625t^2 - 450t + 225
To find the minimum distance, we need to find the value of t that minimizes D(t). We can achieve this by finding the derivative of D(t) with respect to t and equating it to zero.
Differentiating D(t) with respect to t, we get:
dD(t)/dt = 1250t - 450
Setting dD(t)/dt = 0 and solving for t, we get:
1250t - 450 = 0
1250t = 450
t = 450/1250
t = 0.36 hours
To convert this into minutes, we multiply by 60:
0.36 hours * 60 minutes/hour ≈ 22 minutes
Therefore, the two boats are closest together approximately 22 minutes after 9:00 PM.
Problem 2: Pipeline cost optimization
To minimize the cost of the pipeline, we need to determine the location P on the north bank where the pipeline should be directed.
Let's assume P is located x km east of the refinery on the north bank. The pipeline will run straight from the refinery to P and then drop vertically under the river to the tanks.
The total cost of the pipeline C(x) can be calculated by adding the cost of laying the pipe over land and the cost of laying the pipe under the river.
Cost over land = 300,000 * x (since the cost is $300,000 per km)
Cost under river = 600,000 * sqrt(x^2 + 2^2) (since the cost is $600,000 per km and the distance under the river is the hypotenuse of a right triangle with sides x and 2)
Thus, the total cost is given by:
C(x) = Cost over land + Cost under river
= 300,000 * x + 600,000 * sqrt(x^2 + 4)
To minimize the cost, we differentiate C(x) with respect to x and equate it to zero.
Differentiating C(x) with respect to x, we get:
dC(x)/dx = 300,000 + 600,000(1/2)(2x)(x^2 + 4)^(-1/2)
dC(x)/dx = 300,000 + 600,000x/(sqrt(x^2 + 4))
Setting dC(x)/dx = 0 and solving for x, we get:
300,000 + 600,000x/(sqrt(x^2 + 4)) = 0
600,000x/(sqrt(x^2 + 4)) = -300,000
600,000x = -300,000(sqrt(x^2 + 4))
Square both sides:
(600,000x)^2 = (-300,000)^2(x^2 + 4)
360,000,000x^2 = 90,000,000x^2 + 1,200,000,000
270,000,000x^2 = 1,200,000,000
x^2 = 1,200,000,000 / 270,000,000
x^2 = 4.44
x = √4.44
x ≈ 2.11 km
Therefore, to minimize the cost, point P should be located approximately 2.11 km east of the refinery on the north bank.
Sure! I'd be happy to help you with these optimization problems.
For the first problem, we want to find the time at which the two boats are closest together. Let's start by setting up a coordinate system, with the starting point of the first boat as the origin (0,0). Since the first boat is traveling due south at a speed of 20 km/h, its position at any time t can be represented by the point (0, -20t), where t is the time in hours since 9:00 P.M.
The second boat is heading due east at a speed of 15 km/h and reaches the dock at 10:00 P.M. This implies that the position of the second boat at any time t can be represented by the point (15(t-1), 0).
To find the distance between the two boats at any time t, we can use the distance formula:
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
In this case, x1 = 0, y1 = -20t, x2 = 15(t-1), and y2 = 0. Plugging these values into the distance formula, we get:
Distance = sqrt((15(t-1))^2 + (-20t)^2)
= sqrt(225(t-1)^2 + 400t^2)
= sqrt(225t^2 - 450t + 225 + 400t^2)
= sqrt(625t^2 - 450t + 225)
To find the minimum distance between the two boats, we need to find the value of t that minimizes the distance function. We can do this by taking the derivative of the distance function with respect to t, setting it equal to zero, and solving for t:
d(distance)/dt = 0
625t - 450 = 0
t = 450/625 = 0.72 hours
To convert this to minutes, we multiply by 60 (since there are 60 minutes in an hour):
t = 0.72 * 60 = 43.2 minutes
Rounding to the nearest minute, the two boats are closest together approximately 43 minutes after 9:00 P.M.
Now, let's move on to the second problem.
To minimize the cost of the pipeline, we want to find the location P on the north bank that minimizes the overall cost. Let's set up a coordinate system, with the refinery located at the origin (0,0). The storage tanks are located 3 km east of the refinery, so their position is (3,0).
We need to find the x-coordinate of the point P on the north bank. Let's call this x. The cost of laying the pipeline over land is $300,000 per km, so the cost of laying the pipeline from the refinery to point P is 300,000x. The cost of laying the pipeline under the river is $600,000 per km, so the cost of laying the pipeline from point P to the tanks is 600,000(3-x). Hence, the total cost C(x) of the pipeline is given by:
C(x) = 300,000x + 600,000(3-x)
To minimize this cost, we need to find the value of x that minimizes the function C(x). We can do this by taking the derivative of C(x) with respect to x, setting it equal to zero, and solving for x:
d(C)/dx = 0
300,000 - 600,000 = 0
-300,000 = 300,000x - 600,000x
300,000 = 300,000x
x = 1
Therefore, the optimal position P on the north bank should be located at x = 1 km from the refinery.
I hope this helps! Let me know if you have any further questions.