Why isn't my rearranging correct? How would you correctly rearrange this to find t?
We have the equation but we don't know the value of yet (since we're looking for it...) hence we cannot divide the equation by without considering two cases :
- First case : . We can safely divide the equation by (since it does not equal 0) and it gives us . That's one of the solutions.
- Second case : . For , the equation is true since hence is the other solution.
Is it clearer ?
OH you're right! I had a derivative already, so I ignored the fact that I needed to find the derivative of THAT to get the maximum.
I still don't see how we got the minimums though. Unless the minimums happened to be the minimums/maximums of the original formula as well, I guess they were.
Thanks very much for your help
Just a query...I differentiated to find the maximum, and came out with t=0 or t=60. Is this right? t shouldn't be 0.
I differentiated to:
1/100(180t-3t^2)
Then, put that equal to 0, and rearranged:
0=1.8t-0.03t^2
0=t(1.8-0.03t)
So, due to the law "If A*B=0, either A or B=0", I decided that t=0 or (1.8-0.03t)=0. This gives t=0 or t=60.
Am I wrong?
No, that's it. The thing is that the points such that can be local maximums or local minimums.
To show that is a local maximum, you can, for example, show that increases on and decreases on . ( on and on )
The same idea applies to show that is a local minimum : simply show that the derivative is negative for and positive for .