Lagrange Multipliers / Height of a Rocket

  • #1

Homework Statement


I am going to paste the problem word for word, so you can have all the exact information that I have:

You’re part of a team that’s designing a rocket for a specific mission. The thrust (force) produced by the rocket’s engine will give it an acceleration of a feet per second squared. The acceleration a will be constant while the engine is firing, but is undetermined so far. The hope is to design the engine so as to get the value of a that will maximize the rocket’s height at the moment it runs out of fuel.

The height above ground at any time t is f(t, a) = ((a-32) / 2) t2

You’d think you could get more height by carrying more fuel to burn, but that would mean the rocket would be heavier, which would mean you’d need more thrust to accelerate it to the same speed, which would mean you’d need to burn fuel faster. Consequently, carrying more fuel doesn’t help as much as
you’d like.

Without going into details, these considerations lead to the constraint:
a2t = 300,000 for the rocket in question.

Use the method of Lagrange multipliers to find the value of a that maximizes the rocket’s height above ground at the moment the fuel runs out.

Warning:
While the rocket’s engine is firing, a will be a constant. However, during this optimization problem, you have to treat a as a variable.
Hint: Physical considerations require that neither neither a nor t be 0.

Homework Equations


∇f= (∂f/∂t), (∂f/∂a)

The Attempt at a Solution


(∂f/∂a) f(t, a) = ((a-32) / 2) t2 = -31

a2t = 300,000
-312t = 300,000
961t = 3000
t = 312.1
8 ft above ground when fuel runs out

I am trying to follow the pattern in my notes, but I am finding myself extremely confused on what exactly I am supposed to do.

Thanks
 

Answers and Replies

  • #2
stevendaryl
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First question is: You titled this thread "Lagrange Multipliers / Height of a Rocket". Where have you used any Lagrange multipliers?
 
  • #3
Ray Vickson
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Homework Statement


I am going to paste the problem word for word, so you can have all the exact information that I have:

You’re part of a team that’s designing a rocket for a specific mission. The thrust (force) produced by the rocket’s engine will give it an acceleration of a feet per second squared. The acceleration a will be constant while the engine is firing, but is undetermined so far. The hope is to design the engine so as to get the value of a that will maximize the rocket’s height at the moment it runs out of fuel.

The height above ground at any time t is f(t, a) = ((a-32) / 2) t2

You’d think you could get more height by carrying more fuel to burn, but that would mean the rocket would be heavier, which would mean you’d need more thrust to accelerate it to the same speed, which would mean you’d need to burn fuel faster. Consequently, carrying more fuel doesn’t help as much as
you’d like.

Without going into details, these considerations lead to the constraint:
a2t = 300,000 for the rocket in question.

Use the method of Lagrange multipliers to find the value of a that maximizes the rocket’s height above ground at the moment the fuel runs out.

Warning:
While the rocket’s engine is firing, a will be a constant. However, during this optimization problem, you have to treat a as a variable.
Hint: Physical considerations require that neither neither a nor t be 0.

Homework Equations


∇f= (∂f/∂t), (∂f/∂a)

The Attempt at a Solution


(∂f/∂a) f(t, a) = ((a-32) / 2) t2 = -31

a2t = 300,000
-312t = 300,000
961t = 3000
t = 312.1
8 ft above ground when fuel runs out

I am trying to follow the pattern in my notes, but I am finding myself extremely confused on what exactly I am supposed to do.

Thanks
Let's leave out all the words and just describe the actual mathematical formulation:
[tex] \text{maximize} \; \frac{1}{2} (a-32) \: t^2 \\
\text{subject to} \; a^2 t = 300\,000 [/tex]
Is that your problem?
 
  • #4
Let's leave out all the words and just describe the actual mathematical formulation:
[tex] \text{maximize} \; \frac{1}{2} (a-32) \: t^2 \\
\text{subject to} \; a^2 t = 300\,000 [/tex]
Is that your problem?
Thank you for clarifying!!
Yes, that is exactly what I was trying to figure out.

Sorry for the late response...unexpected family issues arose.
But yes, from there I was able to get it solved.
 

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