Calculus 1 4.5 HW problems.

Question

1. (a) Ornithologists have determined that some species of birds tend to avoid flights over large bodies of water during daylight hours. It is believed that more energy is required to fly over water than land because air generally rises over land and falls over water during the day. A bird with these tendencies is released from an island that is 7 km from the nearest point B on the shoreline, flies to a point C on the shoreline, and then flies along the shoreline to its nesting area D. Assume that the bird instinctively chooses a path that will minimize its energy expenditure. Points B and D are 14 km apart. (Round your answers to two decimal places.)

(a) In general, if it takes 1.2 times as much energy to fly over water as land, to what point C should the bird fly in order to minimize the total energy expended in returning to its nesting area?

(b) Let W and L denote the energy (in joules) per kilometer flown over water and land, respectively. Assuming the bird's energy expenditure is minimized, determine a function for the ratio W/L in terms of x, the distance from B to C.

(c) What should the value of W/L be in order for the bird to fly directly to its nesting area D?

(d) If the ornithologists observe that birds of a certain species reach the shore at a point 5 km from B, how many times more energy does it take a bird to fly over water than land?

1. (b) If C(x) is the cost of producing x units of a commodity, then the average cost per unit is c(x) = C(x)/x. Consider the cost function C(x) given below. C(x)= 24,000+220x+6x^3/2

(a) Find the total cost at a production level of 1000 units. (Round your answer to the nearest cent.)

(b) Find the average cost at a production level of 1000 units. (Round your answer to the nearest cent.)

(c) Find the marginal cost at a production level of 1000 units. (Round your answer to the nearest cent.)

(d) Find the production level that will minimize the average cost. (Round your answer to the nearest whole number.)

(e) What is the minimum average cost? (Round your answer to the nearest dollar.)

Answer 1

Let's break down each part of the problem:

(a) In general, if it takes 1.2 times as much energy to fly over water as land, to what point C should the bird fly in order to minimize the total energy expended in returning to its nesting area?

Let x be the distance from point B to point C on the shoreline. The energy expended flying over land is L, and the energy expended flying over water is 1.2L. The total energy expended is given by:

E(x) = L * x + 1.2L * (14 - x)

To minimize E(x), we can take its derivative with respect to x and set it equal to zero:

E'(x) = 0

L - 1.2L = 0

0.2L = 0.2L = 0.2L = 0.2L

So, the bird should fly to point C that is 5.83 km from point B (rounded to two decimal places) to minimize the total energy expended.

(b) Let W and L denote the energy (in joules) per kilometer flown over water and land, respectively. Assuming the bird's energy expenditure is minimized, determine a function for the ratio W/L in terms of x, the distance from B to C.

The ratio W/L can be expressed as:

W/L = (1.2L) / L = 1.2

So, the ratio W/L is a constant 1.2 regardless of the distance x.

(c) What should the value of W/L be in order for the bird to fly directly to its nesting area D?

To fly directly to nesting area D, the bird should not fly over water, so W/L should be equal to 1. In other words, the energy cost over water should be the same as over land.

(d) If the ornithologists observe that birds of a certain species reach the shore at a point 5 km from B, how many times more energy does it take a bird to fly over water than land?

If the bird reaches the shore at a point 5 km from B, it means x = 5 km. The ratio W/L is constant at 1.2 as determined in part (b). Therefore, it takes 1.2 times more energy for the bird to fly over water than land.

Now, let's move on to the second part of the problem:

(a) Find the total cost at a production level of 1000 units.

Total cost (C(x)) is given by the formula:

C(x) = 24,000 + 220x + 6x^(3/2)

Plug in x = 1000:

C(1000) = 24,000 + 220(1000) + 6(1000)^(3/2)

C(1000) = 24,000 + 220,000 + 6(31.62) ≈ 24,000 + 220,000 + 189.72 ≈ 244,189.72

The total cost at a production level of 1000 units is approximately $244,189.72.

(b) Find the average cost at a production level of 1000 units.

Average cost (c(x)) is given by:

c(x) = C(x) / x

c(1000) = 244,189.72 / 1000 ≈ 244.19

The average cost at a production level of 1000 units is approximately $244.19 per unit.

(c) Find the marginal cost at a production level of 1000 units.

The marginal cost is the derivative of the total cost function:

C'(x) = 220 + 9x^(1/2)

Plug in x = 1000:

C'(1000) = 220 + 9(31.62) ≈ 220 + 284.58 ≈ 504.58

The marginal cost at a production level of 1000 units is approximately $504.58.

(d) Find the production level that will minimize the average cost.

To find the production level that minimizes the average cost, we need to find the value of x that minimizes c(x). Take the derivative of c(x) and set it equal to zero:

c'(x) = 0

220 + 9x^(1/2) = 0

9x^(1/2) = -220

x^(1/2) = -220/9

x = (-220/9)^2 ≈ 548.89

So, the production level that minimizes the average cost is approximately 549 units.

(e) What is the minimum average cost?

To find the minimum average cost, plug the value of x back into the average cost function:

c(549) = C(549) / 549 ≈ 244,189.72 / 549 ≈ $444.35

The minimum average cost is approximately $444.35 per unit.