Find the critical points of f(x, y) = x^3 -y^3 - 3xy and test for local max or m

Question

Find the critical points of f(x, y) = x^3 -y^3 - 3xy and test for local max or min or saddle. Contour lines are drawn at the right at intervals of z = 1.] Your task now is to find the global MAX and MIN points of the function f(x, y) on the circle x^2 +y^2 lessthanorequalto 4 Use the contour diagram to identify (graphically the maximum and minimum points. Mark them on the contour graph provided. Parameterize the circle with the functions x = 2cost y = 2sint This parameterization defines z as a function g of t g(t) = f(2cos(t), 2sin(t)). Using only the contour lines as an aid, draw a rough graph of g(t) for 0

Explanation / Answer

Solution: (a)

F(x,y) = x3 - y3 - 3xy

F = 3x2 - 3y
F = - 3y2 - 3x

To find the stationary points we set these to zero and solve the equations for x & y

=> 3x2 - 3y = 0 => 3x2 = 3y => x2 = y .....(i)
=> -3y2 - 3x = 0 => y2 + x = 0 => (x2)2 + x = 0 => x4 + x = 0 => x(x3 + 1) = 0 ......(ii)

So from the first equation we can see that if x=0 then y=0 and if x = -1 then y = 1 so you have a critical point at (-1,1) and (0,0)

To find the type of stationary point first we find the partial second derivatives
F = 6x, F = -6y, F = -3 and compute det(H) = FF - F2

For each stationary point, we examine these values

( 0, 0 ) : F = 0, F = 0, F = -3, det(H) = -9

Since det(H) < 0 this is a saddle point with F(0, 0)

( -1, 1 ) : F = -6, F = -6 , F = -3, det(H) = (-6)(-6) - (-3)^2 = 36 - 9 = 27

Since det(H) > 0 and F < 0 this is a maximum with F(-1,1) = 1

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