Lagrange Multipliers / Height of a Rocket

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Homework Help Overview

The discussion revolves around a problem involving the optimization of a rocket's height using the method of Lagrange multipliers. The problem presents a scenario where the thrust produced by the rocket's engine leads to a constant acceleration, which is to be determined in order to maximize the rocket's height at the moment it runs out of fuel. The height is expressed as a function of time and acceleration, with a constraint relating acceleration and time.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the formulation of the problem and the application of Lagrange multipliers. There are questions about the correct use of the method and the interpretation of the variables involved. Some participants express confusion regarding the steps to take in the optimization process.

Discussion Status

The discussion has seen participants clarifying the mathematical formulation of the problem and confirming the use of Lagrange multipliers. There is acknowledgment of confusion and attempts to align on the problem's requirements, but no explicit consensus has been reached regarding the solution process.

Contextual Notes

Participants note the importance of treating acceleration as a variable during the optimization, despite it being constant while the engine is firing. There are also reminders about physical constraints, such as the requirement that neither acceleration nor time can be zero.

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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
 
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First question is: You titled this thread "Lagrange Multipliers / Height of a Rocket". Where have you used any Lagrange multipliers?
 
defaultusername said:

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:
\text{maximize} \; \frac{1}{2} (a-32) \: t^2 \\<br /> \text{subject to} \; a^2 t = 300\,000
Is that your problem?
 
Ray Vickson said:
Let's leave out all the words and just describe the actual mathematical formulation:
\text{maximize} \; \frac{1}{2} (a-32) \: t^2 \\<br /> \text{subject to} \; a^2 t = 300\,000
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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