Ask question

Ask question

All categories

Oksi-84 [34.3K]

4 years ago

15

Suppose a firm is producing 2,475 units of output by hiring 50 workers (W = $20 per hour) and 25 units of capital (R = $10 per h

our). The marginal product of labor and marginal product of capital are 40 and 25, respectively. Is the firm minimizing the cost of producing 2,475 units of output?

1 answer:

Neko [114]4 years ago

3 0

Answer

given,

firm is producing = 2,475 units

output by hiring 50 workers W = $20 per hour

25 units of capital R = $10 per hour

marginal product of labor = 40

marginal product of capital = 25

$\dfrac{MP_l}{MP_c}=\dfrac{40}{25}$

$\dfrac{MP_l}{MP_c}=\dfrac{8}{5}$

$\dfrac{W}{R}=\dfrac{20}{10}$

$\dfrac{W}{R}=2$

$\dfrac{MP_l}{MP_c} < \dfrac{W}{R}$

Firm is not minimizing the cost because the firm use more capital and less labor.

You might be interested in

Hooke's law describes an ideal spring. Many real springs are better described by the restoring force (F Sp ) s =−kΔs−q(Δs) 3 (p)

Montano1993 [528]

Answer with Explanation:

We are given that

Restoring force, $(FS_p)s=-k\Delta s-q(\Delta s)^3$

$k=350N/m$

$q=750 N/m^3$

We have to find the work must you do to compress this spring 15 cm.

$\Delta s=15 cm=0.15 m$

Using 1 m=100 cm

Work done= $\int_{0}^{0.15}-Fd(\Delta s)$

W= $-\int_{0}^{0.15}(-k\Delta s-q(\Delta s)^3))d(\Delta s)$

$W=k[\frac{(\Delta s)^2}{2}]^{0.15}_{0}+q[\frac{(\Delta s)^4}{4}]^{0.15}_{0}$

$W=0.01125k+0.000127q=0.01125\times 350+0.000127\times 750$

$W=4.033 J$

Ideal spring work= $0.5k(\Delta s)^2=0.5\times 350\times (0.15)^2=3.938 J$

Percentage increase in work= $\frac{4.033-3.938}{3.928}\times 100=2.4$ %

6 0

4 years ago

. Orbit of a satellite It is advantageous to place communications satellites in a circular orbit so that their position is fixed

Alexxx [7]

Answer:

a) r = 4.22 10⁷ m, b) v = 3.07 10³ m / s and c) a = 0.224 m / s²

Explanation:

a) For this exercise we will use Newton's second law where acceleration is centripetal and force is gravitational force

F = m a

a = v² / r

F = G m M / r²

G m M / r² = m v² / r

G M / r = v²

The squared velocity is a scalar and this value is constant, so let's use the uniform motion relationships

v = d / t

As the orbit is circular the distance is the length of the circle in 24 h time

d = 2π r

t = 24 h (3600 s / 1 h) = 86400 s

Let's replace

G M / r = (2π r / t)²

G M = 4 π² r³ / t²

r = ∛(G M t² / (4π²)

r = ∛( 6.67 10⁻¹¹ 5.98 10²⁴ 86400² / (4π²)) = ∛( 75.4 10²¹)

r = 4.22 10⁷ m

b) the speed module is

v = √G M / r

v = √(6.67 10⁻¹¹ 5.98 10²⁴/ 4.22 10⁷

v = 3.07 10³ m / s

c) the acceleration is

a = G M / r²

a = 6.67 10⁻¹¹ 5.98 10²⁴ / (4.22 10⁷)²

a = 0.224 m / s²

5 0

3 years ago

In Greek philosophy, the gods were there to be helpful to humans.<br> A. True<br> B. False

ollegr [7]

Answer:

True, people would actually pray to Poseidon for safe sea travel

3 0

3 years ago

A motorboat accelerates uniformly from a velocity of 6.5 m/s west to a velocity of 1.5 m/s west. If its acceleration was 2.7 m/s

expeople1 [14]

Answer:

<em>The motorboat ends up 7.41 meters to the west of the initial position </em>

Explanation:

<u>Accelerated Motion </u>

The accelerated motion describes a situation where an object changes its velocity over time. If the acceleration is constant, then these formulas apply:

$\vec v_f=\vec v_o+\vec a.t$

$\displaystyle \vec r=\vec v_o.t+\frac{\vec a.t^2}{2}$

The problem provides the conditions of the motorboat's motion. The initial velocity is 6.5 m/s west. The final velocity is 1.5 m/s west, and the acceleration is $2.7 m/s^2$ to the east. Since all the movement takes place in one dimension, we can ignore the vectorial notation and work with the signs of the variables, according to a defined positive direction. We'll follow the rule that all the directional magnitudes are positive to the east and negative to the west. Rewriting the formulas:

$v_f=v_o+a.t$

$\displaystyle x=v_o.t+\frac{a.t^2}{2}$

Solving the first one for t

$\displaystyle t=\frac{v_f-v_o}{a}$

We have

$v_o=-6.5,\ v_f=-1.5,\ a=2.7$

Using these values

$\displaystyle t=\frac{-1.5+6.5}{2.7}=1.852\ s$

We now compute x

$\displaystyle x=(-6.5)(1.852)+\frac{(2.7)(1.852)^2}{2}$

$x=-7,41\ m$

The motorboat ends up 7.41 meters to the west of the initial position

5 0

3 years ago

12. Suppose a person uses a mechanical iack to lift half the weight of a car with a

Likurg_2 [28]

a) The force that must be applied is 73.5 N

b) The actual efficiency is 82 %

Explanation:

a)

Since the jack is 100% efficient, all the input work is converted into output work. So we can write:

$W_{in} = W_{out}\\F_{in} d_{in} = F_{out} d_{out}$

where:

$F_{in}$ is the input force applied on the jack

$d_{in}$ is the arm of the input force

$F_{out}$ is the output force applied on the jack

$d_{out}$ is the arm of the output force

Here we have:

$F_{out}=\frac{mg}{2}= \frac{(1200 kg)(9.8 m/s^2)}{2}=5880 N$ is the output force (half the weight of the car)

$d_{in} = 40.0 cm = 0.40 m$ is the arm of the input force

$d_{out} = 5.0 mm = 0.005 m$ is the arm of the output force

Solving for $F_{in}$ , we find the force that must be applied in input to lift the car:

$F_{in} = \frac{F_{out}d_{out}}{d_{in}}=\frac{(5880)(0.005)}{0.40}=73.5 N$

b)

The efficiency of the jack is given by the ratio between the output work and the input work:

$\eta = \frac{W_{out}}{W_{in}}=\frac{F_{out}d_{out}}{F_{in}d_{in}}$

where we have:

$F_{out}=5880 N$ is the output force (half the weight of the car)

$d_{in} = 40.0 cm = 0.40 m$ is the arm of the input force

$d_{out} = 5.0 mm = 0.005 m$ is the arm of the output force

Here we are told that the input force this time is

$F_{in}=90.0N$

Substituting into the equation, we find the new efficiency of the jack:

$\eta = \frac{(5880)(0.005)}{(90.0)(0.40)}=0.82$

Which means an efficiency of 82%.

Learn more about levers:

brainly.com/question/5352966

#LearnwithBrainly

8 0

4 years ago