What Is the Marginal Profit for 20 Bicycles and the Rocket's Distance Rate?

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1.A Bicycles manufacturer estimates that it can price its bicycles at P=140-0.02x dollars each.
where x is number sold. the cost of producing x bicycles is 900-0.01x^2 dollars. Determine the marginal profit when 20 bicycles are made.

Answer is $59.60

Similarly

2.A rocket is launched straight up. there is an observation stations 7miles from the launching site.a what rate distance between the rocket and station increasing when when rocket is 24 miles high and traveling at 200 miles per hour?

answer is 1920 miles per hour.

Note: it is not homework problems!
just my book exercise:

I try is hardly but i can't solve please help me.

thanks a lot


I following all this but answer coming wrong

revenue = (number sold)(price)

profit = revenue - cost

marginal profit is the derivative of the profit function.


2. let y = height of the rocket , r = distance between the rocket and the station

y^2 + 7^2 = r^2
take the derivative of the above equation w/r to time , then sub in your given/calculated values to determine frac{dr}{dt}

y = 24 miles and \frac{dy}{dt} = 200 mph
 
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neduet said:
1.A Bicycles manufacturer estimates that it can price its bicycles at P=140-0.02x dollars each.
where x is number sold. the cost of producing x bicycles is 900-0.01x^2 dollars. Determine the marginal profit when 20 bicycles are made.

Answer is $59.60

Similarly

2.A rocket is launched straight up. there is an observation stations 7miles from the launching site.a what rate distance between the rocket and station increasing when when rocket is 24 miles high and traveling at 200 miles per hour?

answer is 1920 miles per hour.

Note: it is not homework problems!
just my book exercise:

I try is hardly but i can't solve please help me.

thanks a lot


I following all this but answer coming wrong

revenue = (number sold)(price)

profit = revenue - cost

marginal profit is the derivative of the profit function.
If you sell P=140-0.02x bicycles at x dollars each, you will bring in (140- 0.002x)x= 140x- 0.02x^2 dollars in revenue. Since the cost of those bicycles is 900-0.01x^2 your profit will be 140x- 0.02x^2- 900+ .01x^2= -900+ 140x- .01x^2. Evaluate the derivative of that when x= 20.


2. let y = height of the rocket , r = distance between the rocket and the station

y^2 + 7^2 = r^2
take the derivative of the above equation w/r to time , then sub in your given/calculated values to determine frac{dr}{dt}

y = 24 miles and \frac{dy}{dt} = 200 mph
Is the problem that you can't take the derivative? Both of these use the simple (x^n)'= n x^{n-1}.
Here, 2y dy/dt= 2r dr/dt. You know y= 24, (r= \sqrt{24^2+ 49^2}, and dy/dt= 200 so put those numbers in and solve for dr/dt.
 
you saying Evaluate the derivative of that when x= 20



for first one answer profit derivative is 140 -0.02x and the i put x=20 it is become 139.6
but answer is 59.60
 
Is the problem that you can't take the derivative? Both of these use the simple (x^n)'= n x^{n-1}.
Here, 2y dy/dt= 2r dr/dt. You know y= 24, (r= \sqrt{24^2+ 49^2}, and dy/dt= 200 so put those numbers in and solve for dr/dt.

2(24)(200)=2(54.56)dr/dt
9600/109.12=dr/dt

if i remove 2 on both sides

24(200)=(54.56)dr/dt
4800/54.56=dr/dt

boths answer are coming wrong . please help me
 
neduet said:
1.A Bicycles manufacturer estimates that it can price its bicycles at P=140-0.02x dollars each.
where x is number sold. the cost of producing x bicycles is 900-0.01x^2 dollars. Determine the marginal profit when 20 bicycles are made.

Answer is $59.60
Something doesn't seem right here. Cost of production should increase with x, but the expression actually decreases. Are you sure you copies the expressions correctly?
 
Redbelly98 said:
Something doesn't seem right here. Cost of production should increase with x, but the expression actually decreases. Are you sure you copies the expressions correctly?


yes sir i copy correct .
but both answer coming wrong (not only one)
 
I agree with your answer of $139.6 for the bicycle profit problem. Perhaps the book answer is wrong, it does happen sometimes.
neduet said:
2(24)(200)=2(54.56)dr/dt
9600/109.12=dr/dt

if i remove 2 on both sides

24(200)=(54.56)dr/dt
4800/54.56=dr/dt
The 54.56 is wrong. This number should be the distance from the station to the rocket, when the rocket is 24 miles above the launch site ... just use the pythagorean theorem for that.
 
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