# [SOLVED] Optimization in Several Variables

• Dec 9th 2008, 07:25 PM
coe236
[SOLVED] Optimization in Several Variables
F(x,y) = (2*y+1)*e^(x^2-y)
Find critical point and prove there is only one.
Use second derivative test to determine nature of crit. pt.

I know the procedure in solving it: set partial derivatives to zero and solve resulting equations. And by second derivative test, if D>0, f(a,b) is local min/max; D<0, (a,b) is saddle point. if f_xx(a,b)>0, f(a,b) is min
where D=D(a,b)=f_xx(a,b)f_yy(a,b)-f_xy(a,b)^2

I have no idea how to get the partial derivatives and start the problem. Any help will be appreciated, thanks.
• Dec 11th 2008, 12:26 PM
Mush
Quote:

Originally Posted by coe236
F(x,y) = (2*y+1)*e^(x^2-y)
Find critical point and prove there is only one.
Use second derivative test to determine nature of crit. pt.

I know the procedure in solving it: set partial derivatives to zero and solve resulting equations. And by second derivative test, if D>0, f(a,b) is local min/max; D<0, (a,b) is saddle point. if f_xx(a,b)>0, f(a,b) is min
where D=D(a,b)=f_xx(a,b)f_yy(a,b)-f_xy(a,b)^2

I have no idea how to get the partial derivatives and start the problem. Any help will be appreciated, thanks.

$f(x,y) = 2y.exp(x^2-y) + exp(x^2-y)$

$\frac{\partial f}{\partial x} = 2x.2y.exp(x^2-y) + 2x.exp(x^2-y) = exp(x^2-y)(4xy+2x) = 0$

The exponential function is never zero, hence $4xy+2x = 0$

$\frac{\partial f}{\partial y} = 2exp(x^2-y) + 2y(-1)exp(x^2-y) -exp(x^2-y) = exp(x^2-y)(2-2y-1) = 0$

Again exponential never zero, so: $2-2y-1 = 0$

Surely ou can continue from there?
• Dec 11th 2008, 06:57 PM
coe236
yup, that simplified things a lot. Thank you(Nod)