The operating cost, C, in dollars per hour, for an airplane cruising at a height of h metres and an air speed of 200km/h is given by:

C = 4000 + h/15 + 15000000/h

for the domain 1000 ≤ h ≤ 20000.

Determine the height at which the operating cost is at a minimum, and find the operating cost per hour at this height.

So far this is what I've tried:

C' = 4000 + 1/15 -15000000/h^2

I have no clue what to do next lol plz help

1. find derivative with respect to h:

C'(h) = 1/15-15000000/h²
2. equate C'(h) to zero and solve for h.
2. 1/15-15000000/h² = 0
h²=225000000
h=15000
3. verify that it is a minimum (not maximum) by taking second derivative, and see that C"(15000)>0.
(if C"(15000)<0, then it is a maximum).
C"(h)=30000000/h³
C"(15000)=1/112500 > 0
so h=15000 is a minimum.

Well, it looks like you're on the right track with finding the derivative of the operating cost function. To find the minimum, we can set the derivative equal to zero:

C' = 4000 + 1/15 - 15000000/h^2 = 0

Now, let's solve this equation for h:

4000 + 1/15 - 15000000/h^2 = 0
15000000/h^2 = -4000 - 1/15
15000000/h^2 = -60001/15

To simplify the equation, we can cross-multiply:

h^2 = (15 * 15000000) / (60001)
h^2 = 3750
h = √3750

Since h should be positive, we take the positive square root:

h ≈ 61.24

So, the height at which the operating cost is at a minimum is approximately 61.24 meters.

To find the operating cost per hour at this height, we substitute this value of h back into the original equation:

C = 4000 + h/15 + 15000000/h
C ≈ 4000 + 61.24/15 + 15000000/61.24
C ≈ 4000 + 4.08 + 244666.25
C ≈ 248670.33

Therefore, the operating cost per hour at the height of approximately 61.24 meters is approximately $248,670.33.

I hope this helps! Let me know if you have any other questions.

To find the height at which the operating cost is at a minimum, we need to find the critical points of the function C(h).

To do that, we set the derivative C'(h) equal to zero and solve for h:

C'(h) = 4000 + 1/15 - 15000000/h^2 = 0

To simplify this equation, we can multiply through by 15h^2 to get rid of the denominators:

15h^2(C'(h)) = (4000h^2 + h - 15000000) = 0

Now, we can solve this quadratic equation for h. We have:

4000h^2 + h - 15000000 = 0

We can solve this using the quadratic formula. The quadratic formula states:

For an equation of the form ax^2 + bx + c = 0, the solutions are given by:

x = (-b ± √(b^2 - 4ac)) / (2a)

Using this formula, we can calculate the values of h.

a = 4000, b = 1, c = -15000000

h = (-1 ± √(1^2 - 4 * 4000 * (-15000000))) / (2 * 4000)

h = (-1 ± √(1 + 2400000000)) / 8000

h = (-1 ± √(2400000001)) / 8000

Now, we can calculate the values for h. We have two possible solutions:

1. h = (-1 + √(2400000001)) / 8000
2. h = (-1 - √(2400000001)) / 8000

Since the domain is given as 1000 ≤ h ≤ 20000, we can eliminate the second solution because it falls outside the given range.

So, the height at which the operating cost is at a minimum is approximately:

h = (-1 + √(2400000001)) / 8000

To find the operating cost per hour at this height, substitute this value of h into the equation C(h):

C = 4000 + h/15 + 15000000/h

C = 4000 + ((-1 + √(2400000001)) / 8000) / 15 + 15000000 / ((-1 + √(2400000001)) / 8000)

Now, simplify this expression to find the operating cost per hour at this height.

To find the height at which the operating cost is at a minimum, you need to find the critical points of the function C with respect to h, and determine whether these points correspond to a minimum or maximum.

To find the critical points, we need to find where the derivative C' equals 0 or does not exist.

You correctly found the derivative of the function C as:

C' = 4000 + 1/15 - 15000000/h^2

Now, set C' equal to 0 and solve for h:

4000 + 1/15 - 15000000/h^2 = 0

To simplify this equation, we can find a common denominator:

(60000h^2 + h - 15000000)/15h^2 = 0

Next, multiply both sides of the equation by 15h^2 to eliminate the denominator:

60000h^2 + h - 15000000 = 0

This equation is a quadratic equation in terms of h. We can solve this equation using the quadratic formula:

h = (-b ± √(b^2 - 4ac)) / (2a)

In this case, the quadratic equation is in the form of ah^2 + bh + c = 0, where:

a = 60000
b = 1
c = -15000000

Plugging these values into the quadratic formula, we get:

h = (-(1) ± √((1)^2 - 4(60000)(-15000000))) / (2(60000))

Simplifying further, we have:

h = (-1 ± √(1 + 3600000000)) / 120000

Now, we can evaluate the two possible values of h:

1. h = (-1 + √(3600000001)) / 120000
2. h = (-1 - √(3600000001)) / 120000

Since the domain of h is given as 1000 ≤ h ≤ 20000, we need to verify whether these values fall within this range. Only the first value satisfies the domain requirement:

h ≈ 1000.361

Now that we have determined the value of h at which the operating cost is at a minimum, we can substitute this value back into the original cost function C to find the operating cost per hour at this height.

Plugging h ≈ 1000.361 into the cost function:

C ≈ 4000 + (1000.361)/15 + 15000000/(1000.361)

Evaluating this expression, we can find the operating cost per hour at the determined height.