# Homework Help: Optimizing with derivatives I think

1. Sep 17, 2007

### michaeldw

Optimizing with derivatives Question

1. The problem statement, all variables and given/known data
You're on the moon and you need a bit from the base 900 feet away so you can continue your experiment. The base has a roover/launcher used for retrieving the bit. The roover can move a max dist of 600 feet. The launcher is controled by angle. At this time the angle is inclined at 70 degrees. It cost $375 per foot to move the roover, and$20,000 per degree to change the launch angle. What are the optimal settings (what angle should I launch from where) to minimize cost?

acceleration due to grav on moon = 5.31 ft/s^2
initial velocity of bit leaving launcher = 75 ft/s

2. Relevant equations

R = V^2(sin(2x))/g ; where R = total distance, V= Velocity, x= theta (some angle) and g= gravitational acceleration.

3. The attempt at a solution
I made an equation for constraint and pne for the optimaization.

Constraint:
900 = [[(75^2)*sin(2x)]/5.31]+ y ; where y = distance roover moves and x = theta

optimization
minCost = [20000*|70-x|]+2y(375) ; |...| = absolute value, 2y because the roover has to return to base.

I remember doing problems like this in HS but I need help with this one. I solved for y in the constraint equation, then substiituted it in for y in the optimization equation. Here's what I got:

minCost= [-7.945*(10^5)*sin(2x)+675000+20000*|x-70|]

Now if im correct I get the 1st derivative of this and I will get the extrema or something like that which will be the optimal angle. Then I plug that into the first equation and get y....

But how do I find the first derivative. This might not even be the right method so please help thatnks.

Last edited: Sep 17, 2007
2. Sep 17, 2007

### michaeldw

anyone?

3. Sep 17, 2007

### Dick

You are probably having a hard time getting answers because the problem is nasty and not terribly interesting. I think I agree with your minCost function. Unfortunately, life is not so easy as to just take a derivative and set equal to zero. For one thing you have the absolute value function which makes the form different for x>70 and x<70. Write down these two forms, take derivative and set equal to zero (be careful since you are working in degrees - if x is in degrees, sin(2x) should be sin(2*(pi/180)*x). For the record I get an answer just a few degrees below 70 degrees (as you might expect, since it is so expensive to adjust the angle).

4. Sep 18, 2007

### michaeldw

i understand what you are saying. how do I set up the two forms though? my math skills are not that great. and is d/dx sin(2x) = to cos(2x)?

5. Sep 18, 2007

### Dick

d/dx sin(k*x)=k*cos(k*x) (that's the chain rule). But that assumes k*x is in radians. You are working in degrees, don't forget the conversion. And for the two forms |x-70| is (x-70) if x>70 and (70-x) if x<70.

6. Sep 18, 2007

### michaeldw

is this the correct derivative:
(if x > 70)
d/dx= 20000+7.945^5(cos(x(pi/90)))(pi/90)

7. Sep 18, 2007

### Dick

Yes, I'll go along with that. You'll find x>70 is not very interesting though. Above 70 y increases AND (70-x) increases. x<70 is the place to be.

8. Sep 18, 2007

### michaeldw

yAY completed and turned in. Thanks for All your help!! this is an awesme forum.