Given f(x,y)= -x^3+4xy-2y^2+3 Find all critical points and determine if the poin

Question

Given f(x,y)= -x^3+4xy-2y^2+3 Find all critical points and determine if the points are relative extrema or saddle points. Can you please show me how this is done and what you get for a final answer

Explanation / Answer

f(x,y) = x³ + y³ -3xy + 10 So, we need to find all the critical points of this. First find Fx, which is the partial derivative of f with respect to x. That's just Fx = 3x² - 3y since y³ and 10 are constants, they go to 0, whereas 3y is a constant, so 3xy has a derivative with respect to x of 3y. Now Fy by the same arguments is just Fy = 3y² -3x So we have the system Fx = 3x² - 3y Fy = 3y² -3x Critical points are at such that 3x² - 3y = 0 3y² -3x = 0 Let's use substitution to solve this. Look at the first equation and solve for y: 3x² = 3y y = x² Substitute that for y in the second equation to get 3(x²)² - 3x = 0 3x^4 - 3x = 0 Factor out 3x, 3x [x³ - 1] = 0 Clearly, either x = 0 or x = 1. So our critical points are at x = 0 or x = 1. Recall that y = x². Therefore given that x = 0, y = 0 and given that x = 1, y = 1. So the critical points are at (0,0) and (1,1) In the computer, I suppose you would input [p(0,0), p(1,1)] To determine if the critical points are relative extrema, a saddle point, or indeterminate, we need to use the following definition: D(a,b) = Fxx(a,b) * Fyy(a,b) - [Fxy(a,b)]² Basically, (a,b) is some critical point of f(x,y) and you use the second-order partial derivatives Fxx, Fyy, and Fxy. We can then say the following: if D(a,b) > 0, and Fxx(a,b) > 0, then (a,b) is a relative minimum if D(a.b) > 0, (a,b) and Fxx(a,b) < 0, then (a,b) is a relative maximum if D(a,b) = 0, this test is inconclusive if D (a,b) < 0, then then (a, b) must be a saddle point Let's compute Fxx, Fyy, and Fxy first. Given that Fx = 3x² - 3y, then Fxx = 6x. 3y is a constant and goes to 0. Given that Fy = 3y² -3x, then Fyy = 6y. Same arguments apply. Given that Fx = 3x² - 3y, Fxy = -3, since 3x² is a constant with respect to y. Therefore, Fxx = 6x Fyy = 6y Fxy = -3 Let's look at D in general first and get that D = Fxx * Fyy - [Fxy]² D = (6x)(6y) - (-3)² D = 36xy - 9 So, let's look at each critical point. D(0,0) = 36(0)(0) - 9 = -9 < 0 D is clearly less than 0, so (0,0) is a saddle point per the test. D(1,1) = 36(1)(1) - 9 = 36 - 9 = 27 > 0 Since D > 0 and Fxx = 1 > 0, (1,1) is a relative minimum. Therefore (0,0) is a saddle point and (1,1) is a relative minimum.

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