# Thread: stationary points(an unholy mess)

1. ## stationary points(an unholy mess)

Determine the stationary points of the function

f(x,y)=(x^2+(1/2)xy+y^2)e^(x+y)

after differentiating i get

df/dx = x^2 + 2x +(1/2)xy + (1/2)y +y^2 = 0 ...eq1

and

df/dy = y^2 + (1/2)x +(1/2)xy + 2y +y^2 = 0 ...eq2

i then solved these 2 eqs simultanously and got

y= (-4/3)x^2 - x

which i subed back into eq 1 which turnrf into a unholy mess

2. Originally Posted by fitzgilbert
Determine the stationary points of the function

f(x,y)=(x^2+(1/2)xy+y^2)e^(x+y)

after differentiating i get

df/dx = x^2 + 2x +(1/2)xy + (1/2)y +y^2 = 0 ...eq1

and

df/dy = y^2 + (1/2)x +(1/2)xy + 2y + y^2 = 0 ...eq2 Mr F says: This is wrong. The term in red should be x^2. Is it a typo or were you using it when attempting the calculations?

i then solved these 2 eqs simultanously and got

y= (-4/3)x^2 - x

which i subed back into eq 1 which turnrf into a unholy mess
Since f(x, y) is symmetric in x and y you can easily get $\frac{\partial f}{\partial y}$ simply by swapping the x's and y's around in your expression for $\frac{\partial f}{\partial x}$. No need to waste time partial differentiating again ......