Find the stationary points of the graph x^3 + y^3 = 3xy +35
Differentiate wrt x to get 3x2 + 3y2dy/dx = 3y + 3x dy/dx Rearrange to get dy/dx = (3x2 - 3y)/(3x-3y2). Set dy/dx =0 and infer y=x2. Substitute in for y into original equation and rearrange to get x6 -2x3 -35 =0. Let p= x3 . Equation in terms of p becomes p2 -2p -35 =0. p=7 or -5. Therefore stationary points are (7(1/3), 7(2/3)) and (-5(1/3), -5(2/3)).
JB
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