Calculus 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 point C be x km from the point B. Then the bird flies over the water for SQRT(49+x²)km and over the shore for (14-x)km. If it takes L joules per km to fly over the land and W joules to fly over the water, the total energy E is

E=L(14-x) + W*SQRT(49+x²). To find the minimum of E, we have to find critical points for x on [0, 14], where E'=0 or does not exist.

Then we compare values of E at the critical points and at the end points of the interval x=0 and x=14, and choose the smallest one.

E' = -L + Wx/SQRT(49+x²)

Since SQRT(49+x²)>0 for all x, E' exists for all x.

E'=0

Wx/SQRT(49+x2)=L

(W/L)x=SQRT(49+x²) Squaring bot sides, we get

(W/L)²x²=49+x²

((W/L)²-1)x²=49

x=7/SQRT((W/L)²-1) local min, hopefully.

a) W=1.2L, so

x=7/SQRT(1.44-1) = 10.55 - critical point

E(10.55)=18.64 L at critical point

E(0)=22.4 L at the ends of the interval [0, 14]

E(14)=18.78 L

x=7/SQRT((W/L)²-1) is indeed a local minimum.

Minimum is reached at x=10.55, the bird should fly to a point C between B and D that is 10.55 km from point B.

b) If a bird flies directly to D, than x=14 is the critical point that's is the local minimum.

x=7/SQRT((W/L)²-1), x=14,

14=7/SQRT((W/L)²-1)

SQRT((W/L)²-1)=0.5

(W/L)²-1 = 0.25

W/L=SQRT(1.25) = 1.12

If W/L is less or equal to 1.12, the bird should fly straight to the nest.

c) If the birds reach the shore at a point 5 km from B, then x=5 is a local minimum

5=7/SQRT((W/L)²-1)

SQRT((W/L)²-1) = 7/5

W/L = 1.72

It takes 1.72 times more energy for a bird to fly over water than land.

2. a) C(1000)= 24,000+220*1000+6*1000^3/2 = 433736.66 - total cost at a production level of 1000 units.

b) C(1000)/1000 = 433.73 - average cost at a production level of 1000 units.

d)Average cost c(x)=C(x)/x = 24,000/x +220+6 x^1/2.

c'(x)=-24000/x²+3/x^1/2

c'(x)=0

-24000/x²+3/x^1/2=0

x^3/2=8000

x=400 - critical point.

We should do the first derivative test to check that x=400 is a minimum.

For that we have to check the signs of c'(x) on 0<x<400 and on x>400.

On [0, 400] take x=100, plug in c'(x), c'(100)=-2.1 - negative, so c(x) is decreasing.

On x>400, take x=900 plug in c'(x) c'(900)=0.07 - positive, so c(x) is increasing.

So, x=400 is a minimum.

e) Min average cost is

c(400)= 400.