f(x,y)=x^2y on the ellipse 4x^2+9y^2=36 find the min and max values. Also explain the geometery of the solution by plotting ellipse and comparing with contours of function f
the ellipse is on the xy plane; and it has a cover of z = x^2 y? or x^(2y) ?
when df/dx, df/dy = (0,0) we have a critical point; since the normal is pointing straight up or down
ill assume the x^2 y version df/dx = 2xy ; and df/dy = x^2 (2xy , x^2) = (0,0) you got a critical point
2xy = 0 when x or y or both = 0 x^2 = 0 when x = 0; so at some points (0,y) we should have a normal that is pointing up .. if i see this right
could you expand on how to find the min and maxes ? I think I understand the critical points but am unsure how I plug them back in
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