(x*z=36) what two numbers make x+4z minimum?
So ....using the first equation...we have
z = 36/x
And substituting this into the second equation, we have
x + 4(36/x)
So...taking the derivative of this and setting it to 0, we have
1 - 144/x^2 = 0
1 = 144/x^2
x^2 = 144
x =12 or -12
And putting both of these values into the first equation, we have
z = 36/12 = 3 or z = 36/-12 = -3
And using the second equation, we have
12 + 4(3) = 24 or
-12 + 4(-3) = -24
So x = -3 and z = -12 minimize x + 4z
P.S. - Could another mathematician check this??? Thanks!!
Hi Chris, your answer is correct right down to the last statement. Then you have drawn an incorrect conclusion.
In NSW, Australia, the dept of Education demand that the second derivative used to determine the nature of the stat point. In this case there could even be a good reason.
I let T=x+4z=x+144x-1
What sould be noted is that T=x+144x-1 has a discontinuity at x=0
This is why your 'obvious' conclusion is incorrect. (Look at the graph)
When x>0, T''>0 therefore any stat pt will be a minimum
When x<0, T''
Therefore there is a local minimum at x=12 and z=3 but this is not an absolute minimum(not sure about the wording)
All negative x values will make x+4z smaller. There is no absolute minimum value.
Thanks, Melody....I wasn't too sure about this one!!! I see my error, now!!
Glad you checked it !!! "Thumbs Up" and a green check from me!!
Chris and I take turns at getting things wrong. I'm on a roll at the moment but a few days back I was starting to feel like everything I posted was wrong!
Maybe they are still wrong and the errors just haven't been 'pick up' yet. lol