Just to clarify: you're asked to classify the stationary points of
Is that correct? If so, what's your partial w.r.t. , and what's your partial w.r.t. ?
I need help solving the following problem. I have already worked it out, but I am getting stuck on the partial derivative relative to y because I am getting a negative square root somewhere along the way...
Here is the problem: classify the stationary points of:
f(x,y) = -2x + 3y + [ (integral) upperbound = y , lowerbound = x ] e^t^2 dt
I hope it makes sense as written...
That is correct. My partial relative to x is: -2 + e^x^2; I was able to get the 0 of fx as : x= sqrt(ln 2)
Relative to y: 3+ e^y^2. I get the 0 of fy as: y = sqrt(ln (-3))
What I attempted to do is get the 0s for the first derivatives relative to x and to y. Then get the second derivatives to form the Hessian and test for the points obtained to classify them as maxima or minima, etc... When getting the derivatives, our prof instructed/suggested that we do not integrate first. I guess that would be the long way, the desperate way...
By the way, I'm new here. How did you get the function input in that form?
Your x partial derivative is incorrect. Because x is the lower limit, there's a minus sign associated with it (just think Fundamental Theorem of the Calculus Part II:
To get equations to look like this, you have to use LaTeX. You can find out more about that here in the stickies.
So, your x partial derivative is
and the partial is, as you've said,
Query: if you set either of these equal to zero, are there any real solutions? Your "solution" unfortunately has you taking the logarithm of a negative number. Negative numbers are not in the domain of the logarithm function, unless you're thinking a different branch of the complex logarithm function. You're going to run into this problem for both derivatives. What does this tell you?