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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) t*

^{2}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:

*a*

^{2}

*t*= 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.

**While the rocket’s engine is firing,**

Warning:

Warning:

*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)

*8 ft above ground when fuel runs out*

*f*(*t, a*) = (*(a-32) / 2) t*

-31

961t = 3000

t = 312.1^{2}= -31*a*^{2}*t*= 300*,*000-31

^{2}*t*= 300*,*000961t = 3000

t = 312.1

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