Lagrange Multipliers / Height of a Rocket

In summary: Thank you!In summary, the problem is to design a rocket's engine in order to maximize its height at the moment it runs out of fuel. The height above ground at any time t is given by f(t,a) = (a-32)/2 * t^2. The constraints for this problem are that a and t cannot be equal to 0, and a^2 * t = 300,000. Using the method of Lagrange multipliers, the maximum height is achieved when a = 64 and t = 7500.
  • #1
defaultusername
17
0

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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  • #2
First question is: You titled this thread "Lagrange Multipliers / Height of a Rocket". Where have you used any Lagrange multipliers?
 
  • #3
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:
[tex] \text{maximize} \; \frac{1}{2} (a-32) \: t^2 \\
\text{subject to} \; a^2 t = 300\,000 [/tex]
Is that your problem?
 
  • #4
Ray Vickson said:
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.
 

1. What are Lagrange multipliers?

Lagrange multipliers are a mathematical tool used to optimize a function subject to one or more constraints. They allow us to find the maximum or minimum value of a function while satisfying the given constraints.

2. How are Lagrange multipliers used in rocket design?

In rocket design, Lagrange multipliers are used to optimize the height of a rocket while considering various constraints such as fuel consumption, payload weight, and atmospheric conditions. By using Lagrange multipliers, engineers can find the optimal height for a rocket that maximizes performance while satisfying these constraints.

3. Can Lagrange multipliers be used for other applications besides rocket design?

Yes, Lagrange multipliers can be used for a variety of applications in mathematics, physics, engineering, and economics. They are a powerful tool for solving optimization problems with constraints, making them applicable in many fields.

4. What is the height of a rocket?

The height of a rocket refers to the distance between the ground and the highest point reached by the rocket during its flight. This height can vary depending on the rocket's design, intended purpose, and external factors such as atmospheric conditions.

5. How do Lagrange multipliers help optimize the height of a rocket?

Lagrange multipliers help optimize the height of a rocket by finding the maximum or minimum value of a function (in this case, the height of the rocket) while satisfying various constraints. By using this method, engineers can determine the best height for a rocket that meets performance requirements while considering limitations such as weight, fuel, and atmospheric conditions.

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