Recent content by Pulty

So the circular boundary: g(1,0) = 3 linear boundary: g(0,0) = 3 (?) g(0,-1) = 2 (?) g(0, 1) = 2 (?) Interior (2/3, 0) = saddle Can there be this many points with the same values and still be extrema?

so, x = cos(t) = cos(0) = 1 y = sin(t) = sin(0) = 0 (1.0) for the boundary?

is that just substituting t = 0 back into the function g(t) g(t) = 2

g'(t) = -3sin(x) cos^2(x) = 0 x = 0 in the given range?

So what I've done so far is: g(x,y) = 3 + x^3 - x^2 - y^2 x^2+y^2 <= 1, x>=0 gx(x,y) = 3x^2 - 2x = x(3x - 2) = 0 x = 0, or x = 2/3 gy(x,y) = -2y = 0 y = 0 (2/3, 0) interior: saddle point using second partials x [0, 1], y [-1, 1] (0, 0) boundary: 3 (1, 0) boundary: 3 (0, -1) boundary...

Do i take the derivative of the single variable function and solve it at 0? y=0?

second partials: gxx(x,y) = 6x - 2 gxy(x,y) = 0 gyy(x,y) = -2 gxx(x,y)gyy(x,y) - [gxy(x,y)]^2 = (2)(-2) - (0)^2 = -4 So the stationary point is a saddle point? What does finding the global extreme points of g in D mean exactly? Thanks!

Taking the partials i get; gx(x,y) = 3x^2 - 2x = 0 solving this I get x = 0, or x = 2/3 gy(x,y) = -2y = 0 solving this I get y = 0 The only point that falls in the domain is (2/3, 0)

The function g is defined by g(x,y) = 3 + x^3 - x^2 - y^2 on the domain D given by points in the xy-plane satisfying x^2 + y^2 <=1 and x >= 0. So I need to find and classify the stationary points of g, and find the global extreme points of g in D. Do I start with taking partial derivatives and...

Thanks Mark! I have a similar question I will ask in a new thread.

fxx(x,y) = e^y - 2x fyy(x,y) = -9e^(3y)y - 6e^(3y) + (e^(y)+x^(2))/2 fyx(x,y) = e^(y)x At point (0,-1/3): fxx(x,y)fyy(x,y) - fxy(x,y)^2 = -0.79079 (Saddle point) At point (e^1/6, -1/6) fxx(x,y)fyy(x,y) - fxy(x,y)^2 = 3.08049 (Maximum point) How does this look? Also, I am confused as to...

$$f(x,y) = \frac{1}{2}{x}^{2}{e}^{y}-\frac{1}{3}{x}^{3}-y{e}^{3y}$$ To start off I found the partial derivatives of $$x: {e}^{y}x - {x}^{2}$$ $$y: \frac{{e}^{y}{x}^{2}}{2}-{e}^{3y}-3{e}^{3y}y$$ Then solved simultaneously for each equation equal to 0. $$y = \ln(x) = -\frac{1}{6}$$ $$x =...