At noon, ship A is 150 km west of ship B. Ship A is sailing east at 35 km/h and ship B is sailing north at 25 km/h. How fast is the distance between the ships changing at 4:00 pm.
Find the intervals of increase or decrease
find the local maxiumum and minimum values
find the intervals of concavity and the inflection points.
b(x) = 3x^2/3 - x
a rectangular storage container with an open top is to have a volume of 10 m^3. the length of its base is twice the width. Material for the base costs $10 per square meter. Material for the sides cost $6 per square meter. Find the cost of the materials for the cheapest such container.
There are three totally different questions posted here. I will be happy to respond to any of them in which work is shown by you.
As drwls indicated, we are not here to merely do the work for you.
You have posted 3 routine calculus questions.
I will get you going on the first one which deals with "rates of change"
After t hours, distance traveled by ship A is 35t km and that of ship B is 25t km
Did you make a diagram?
If you let the distance between them be y km, then
y^2 = (150-35t)^2 + (25t)^2
find dy/dt and sub in t=4 (4:00 pm)
Let me know what answer you got.
(2) Intervals of increase are where db/dx > 0
Intervals of decrease are where db/dx < 0
db/dx = 2x - 1
d^2b/dx^2 = 2
Inflection points are where d^b/dx^2 = 0. There appear to be no such points.
Intervals of concavity (upward) occurwhere d^b/dx^2 > 0
Maximum and minimum values occur where db/dx = 0. Use the second derivative test to determime if it is a maximum or a minimum.
To find the rate at which the distance between the ships is changing at 4:00 pm, we can use the concept of related rates.
Let's consider the position of ship A at time t. Ship A is initially 150 km west of ship B, so we can represent its position as (150 + 35t, 0), where t is the time in hours.
The position of ship B at time t is (0, 25t), as it is initially at the origin and is sailing north at a constant rate of 25 km/h.
The distance between the ships can be found using the distance formula: √[(x2 - x1)^2 + (y2 - y1)^2].
So, the distance d(t) between the two ships at time t is given by:
d(t) = √[(150 + 35t - 0)^2 + (0 - 25t)^2]
= √[(150 + 35t)^2 + (25t)^2]
= √[22500 + 10500t + 1225t^2 + 625t^2]
= √(1700t^2 + 10500t + 22500)
To find the rate at which the distance is changing with respect to time, we need to find d'(t). We differentiate d(t) with respect to t, using the chain rule:
d'(t) = (1/2) * (1700t^2 + 10500t + 22500)^(-1/2) * (3400t + 10500)
Now, to find the rate at 4:00 pm, we substitute t = 4 into the equation:
d'(4) = (1/2) * (1700(4)^2 + 10500(4) + 22500)^(-1/2) * (3400(4) + 10500)
= (1/2) * (27200 + 42000 + 22500)^(-1/2) * (13600 + 10500)
= (1/2) * (91700)^(-1/2) * 24100
≈ 0.024 km/h
Therefore, the distance between the ships is changing at a rate of approximately 0.024 km/h at 4:00 pm.
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To find the intervals of increase or decrease for the function b(x) = (3x^(2/3)) - x, we need to find the derivative of b(x).
Taking the derivative, we get:
b'(x) = (2/3)(3x^(2/3 - 1)) - 1
= 2x^(-1/3) - 1
To determine the intervals of increase or decrease, we need to find where b'(x) is positive or negative.
Setting b'(x) greater than zero, we solve for x:
2x^(-1/3) - 1 > 0
2x^(-1/3) > 1
x^(-1/3) > 1/2
(1/x)^(1/3) > 1/2
1/x > (1/2)^3
1/x > 1/8
8 > x
So, b'(x) is positive for x < 8.
Setting b'(x) less than zero, we solve for x:
2x^(-1/3) - 1 < 0
2x^(-1/3) < 1
x^(-1/3) < 1/2
(1/x)^(1/3) < 1/2
1/x < (1/2)^3
1/x < 1/8
8 < x
Hence, b'(x) is negative for x > 8.
Therefore, the interval of increase for b(x) is (negative infinity, 8), and the interval of decrease is (8, positive infinity).
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To find the local maximum and minimum values of b(x) = (3x^(2/3)) - x, we need to find the critical points.
The critical points occur when b'(x) = 0 or is undefined.
For b'(x) = 2x^(-1/3) - 1 = 0, we solve for x:
2x^(-1/3) = 1
x^(-1/3) = 1/2
(1/x)^(1/3) = 1/2
1/x = (1/2)^3
1/x = 1/8
8 = x
So, x = 8 is a critical point.
To find if there are any other points where b'(x) is undefined, we check for discontinuities. However, since b'(x) is defined for all real numbers except x = 0 (as the function contains x^(-1/3)), there are no other critical points.
To find the corresponding y-values for the critical point(s), we substitute x = 8 into the original function:
b(8) = (3(8)^(2/3)) - 8
= (3(2)) - 8
= 6 - 8
= -2
Therefore, the local maximum value is -2.
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To find the intervals of concavity and the inflection point(s) of b(x) = (3x^(2/3)) - x, we need to find the second derivative of b(x).
Taking the second derivative, we get:
b''(x) = d/dx(2x^(-1/3) - 1)
= -2/3x^(-4/3)
To determine the concavity of b(x), we need to find where b''(x) is positive or negative.
Setting b''(x) greater than zero, we solve for x:
-2/3x^(-4/3) > 0
x^(-4/3) < 0
However, x^(-4/3) cannot be negative, so there are no intervals of concavity for b''(x).
To find the potential inflection point(s), we set b''(x) = 0:
-2/3x^(-4/3) = 0
Since b''(x) is undefined for x = 0, we can consider x = 0 as a potential inflection point. However, since the function b(x) = (3x^(2/3)) - x is not defined for negative values of x, there are no valid inflection points.
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To find the cost of the materials for the cheapest rectangular storage container, we need to determine the dimensions that minimize the cost.
Let's assume the width of the base of the container is w meters. Therefore, the length of the base will be 2w meters, according to the given conditions.
The height of the container can be determined by dividing the volume of the container by the area of the base:
Height = Volume / Area of base
Height = 10 / (w * 2w)
Height = 10 / (2w^2)
Height = 5 / w^2
Now, let's determine the cost of the materials in terms of the width, w:
Cost = Cost of base + Cost of sides
Cost = (Area of base * Cost per square meter) + (Perimeter of base * Height * Cost per square meter)
Cost = [(2w * w) * 10] + [(2w + 2(2w)) * (5 / w^2) * 6]
Cost = 20w^2 + (14w / w^2)
Cost = 20w^2 + 14 / w
To minimize the cost, we can differentiate the cost function with respect to w and set it equal to zero:
dCost/dw = 40w - 14/w^2 = 0
40w = 14/w^2
40w^3 = 14
w^3 = 14/40
w^3 = 7/20
w = (7/20)^(1/3)
w ≈ 0.858 meters
Substituting this value of w back into the cost function, we can find the minimum cost:
Cost ≈ 20(0.858)^2 + 14 / (0.858)
Cost ≈ 12.685 + 16.318
Cost ≈ 29.003
Therefore, the cost of materials for the cheapest rectangular storage container is approximately $29.003.