# Optimization in Several Variables

1. Dec 10, 2008

### soe236

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.

2. Dec 10, 2008

### Office_Shredder

Staff Emeritus
To take a partial derivative, you just treat one of either x or y as a constant, and differentiate with respect to the other one. So for example if the function was f(x,y) = x2y2 + x3 the partial derivative with respect to x is

2xy2 + 3x2

and with respect to y

2yx2

3. Dec 10, 2008

### soe236

Well yea I know that.. I meant I did not know how to get the partials specifically for the function I posted above : F(x,y) = (2*y+1)*e^(x^2-y).
Is dF/dx = (2*y+1)*e^(x^2-y)*2*x ? and dF/dy = ?

4. Dec 10, 2008

### HallsofIvy

Staff Emeritus
Yes. To find dF/dy, use the product rule. What are d(2y+1)/dy and de^(x^2-y)/y separately?

5. Dec 10, 2008

### soe236

d(2y+1)/dy= 2
de^(x^2-y)/dy = -e^(x^2-y)
so by product rule, dF/dy= 2*e^(x^2-y) ?

6. Dec 10, 2008

### Office_Shredder

Staff Emeritus
The product rule is d(f*g)/dy = df/dy*g + dg/dy*f not df/dy*dg/dy

7. Dec 10, 2008

### soe236

So dF/dy= (2y+1)*(-e^(x^2-y) + 2*(e^(x^2-y)) ?

8. Dec 10, 2008

### Staff: Mentor

Technically speaking, the two first derivatives are partial derivatives $$\frac{\partial{F}}{\partial{x}}$$ and $$\frac{\partial{F}}{\partial{y}}$$, also written as $$F_x$$ and $$F_y$$.