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Q 3. If f(x,y) = x^3/3+4/3y^3-x^2-3x-4y-3
Find relative extrema and saddle points of f.
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Q 3. If f(x,y) = x^3/3+4/3y^3-x^2-3x-4y-3
Find relative extrema and saddle points of f.
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To find relative extrema and saddle points you need to find the partial derivatives Fx and Fy. Set Fx and Fy equal to zero and find the values of x and y that make the partial derivatives equal to zero, which will give you the critical points of the surface.
Now that you have your critical points you can apply this test:
If D > 0 and Fxx > 0 then f has a min at the critical point.
If D > 0 and Fxx < 0 then f has a max at the critical point.
If D < 0 then it is a saddle point.
If D = 0 then you cannot make any conclusions.
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