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 .