Expert: Alon Mandes - 5/25/2009

Question
So i took a test and I have the opportunity of earning credit back if I redo it.. and I need help with some problems that I didn't get full credit for.. please HELP! =)

3c) s(t)=(e^(-3t))(cos(t/2))
I got after finding the derivative,
s'(t)= e^(-3t)(cos(t/2)-2sin(t/2))

First of all what is the derivative of (t/2)?? I couldn't figure that out but I only got half the credit on this problem.



My next question is:

4. Find the absolute maximum and absolute minimum of
f(x)=x-3x^(2/3) on the interval [-27,27].

I couldn't figure out the derivative here. I hardly got any credit on this problem so this would definitely help if you ran me through the steps on finding this problem with the answer.

also this problem:

5. it is a derivative word problem and I really need help on this also.

Q- You want to plant a rectangular garden agaisnt the side of your house (so you only need to fence three dies of it). You have 120 feet of fencing. What is the largest area you can enclose?

I'd provide my work but I got confused on this problem so it wouldn't explain where i went wrong.

Answer
The derivative of t/2 is 1/2 !
If s(t)=(e^(-3t))(cos(t/2)) then
s'(t)=[e^3t]cos(t/2)+[e^3t][cos(t/2)]'
=[3e^3t]cos(t/2)-[e^3t]*(1/2)*sin(t/2)
=[e^3t][3cos(t/2)-(1/2)sin(t/2)] .
-----------------------------
f(27)=27-3*27^(2/3)=27-3*9=27-27=0.
f(-27)=-27-3(-27)^(2/3)=-27-3*9=-27-27=-54.
f'(x)=1-3*(2/3)*x^(-1/3)
f'(x)=0 -> 1=2/x^(1/3) -> 2=x^(1/3) -> x=8 .
f''(x)=-2*(-1/3)x^(-4/3)
f''(x)=(2/3)/x^(4/3)
f''(8)=(2/3)/8^(4/3) > 0 it's local min .
f(8)=8-3*8^(2/3)=8-3*4=-4 .
Thus,
Absolute max is f(27) & absolute min is f(-27) .
-------------------------------------
The paparel side to the house will be y, & the other 2 will be x.
Thus , y+2x=120 -> y=120-2x.
The areas s=xy=x(120-2x)=120x-2x^2
To find max we derive & solve s'=0 :
s'=120-4x
s'=0 -> 120=4x -> x=30 . & therefore y=120-30*2=120-60=60.
s(30,60)=30*60=1800 ft^2 .

Alon.