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zalisa [80]
3 years ago
14

A manufacturer sells two goods, one at a price of $1000 a unit and the other at a price of $16,500 a unit. A quantity q1 of the

first good and q2 of the second good are sold at a total cost of $5000 to the manufacturer. (a) Express the manufacturer's profit, π (in thousands of dollars), as a function of q1 and q2.
Physics
1 answer:
vaieri [72.5K]3 years ago
4 0

Answer:

manufacturing cost = $16.5 q₂ + $1 q₁ - $ 5

Explanation:

given,

one unit price of  first good is = $ 1000

one unit price of second good is = $ 16500

quantity of the first good = q₁

quantity of the second good = q₂                      

total cost to the manufacturer =  $ 5000

revenue from  good one = $1000 q₁                    

revenue from second good = $16500 q₂                      

manufacturing cost = total revenue - total manufacturing cost

manufacturing cost = $16500 q₂ + $1000 q₁ - $ 5000        

but in the question we have to answer in thousands of dollar

so, divide the result by 1000                  

manufacturing cost = $16.5 q₂ + $1 q₁ - $ 5                

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2 years ago
A seagull flies at a velocity of 9.00 m/s straight into the wind.
RideAnS [48]

a)If it takes the bird 18.0 minutes to fly 6 km away from the earth, the wind's speed will be 4 m/s.

b) The bird would need 7 minutes and 42 seconds to fly back 6 kilometers if he turned around and flew with the wind.

c)Compared to the 133.33 seconds it would take without the wind, the overall round-trip time is affected by the wind.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

A seagull flies at a velocity,\rm v_{SA}  = 9 \ m/sec

The time the bird takes,t=18.0 min

The distance traveled relative to the earth = 6.00 km

a)

The seagull's relative velocity with reference to the ground as;

\rm v_{sg} = \frac{6.00 \times 10^3 \ m }{(20 min) \times \frac{60 s }{1 \ min}} \\\\ v_{sg}= 5.00 \ m/sec

Air velocity with reference to the ground is;

\rm v_{AG}= v_{SG}-v_{SA} \\\\ v_{AG} = 5.00 \ m/sec - 9.00 \ m/sec \\\\ v_{AG} = -4.00 \ m/sec

b)

If the bird turns around and flies with the wind, The time will he take to return 6.00 km is;

\rm v_{SG}=v_{SA}+v_{AG} \\\\ v_{SG}=-900 \ m/sec +(-4.00 \ m/sec) \\\\ v_{SG}= -13.00 \ m/sec

The time the bird takes;

\rm t = \frac{x_{SG}}{v_{SG}} \\\\ t = \frac{6.00 \times 10^3 \ m }{13.00 \ m/sec } \\\\ t = 462 m/sec \\\\ t = 7  \ min \  and  \ 42  \ sec

c)\

The total round-trip time compared to what it would be with no wind. is;

\rm  t = 20 \ min( \frac{60 \ sec }{1 \ min} )+ 462 \ sec \\\\ t = 1200 \ sec +6 462 \ ec \\\\ t= 1662 \ sec

The time for the round trip is;

\rm  t = \frac{12 \times 10^ 3 }{ 9 \ m/sec }  \\\\ t  = 1333.33 \ sec

Hence the wind's speed, the time bird would need to fly back the total round-trip time will be  4 m/s, 7 minutes and 42 seconds and 1333.33 sec.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

#SPJ1

4 0
2 years ago
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