Sec2Sess30

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Unformatted notes below. Notes on my shorthand method

point where partial derivatives are all 0 is a critical point critical point can be minima, maxima, or saddle point saddle point is neither minima or maxima, looks like middle of a horse saddle shape.

How to find the global mix and max? it could be at a boundry (towards infinity) of the shape, not only min and max

Second derivative test can be used. First consider quadratic example: w = ax^2 + bxy + cy^2 rewritten: (x+y)^2 + 2y^2 because of two square the result will be positive and therefore a minima

  • positive result will typically be minima

in general, if a != 0 w = a(x^2+(b/a)*xy)+cy^2 = a(x+(b/2a)*y)^2 + (c - (b^2/4a))*y^2

2cnd example w=1/4a[4a^2*(x+(b/2a)*y)^2 + (4ac-b^2)*y^2]

  • difference of two squares will typically be a saddle point

3 cases 1) 4ac-b^2 < 0 where one term >= 0 and the other < 0: saddle point occurs 2) 4ac-b^2 = 0 where one square remains: would be equivalent to w = x^2 degenerate critical point occurs.

  • degenerate critical point: a critical point where there are several points at the

extrima same level which are next to each other. 3) 4ac-b^2 > 0 generally speaking: w=1/4a*[+(...)^2+(...)^2] two squares means result is >= 0 if sign of a is positive than it is a minima if sign of a is negative than it is a maxima

  • negative result will typically be maxima

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new example

w = ax^2 + bxy + cy^2 = y^2[a(x/y)^2+b(x/y)+c] if b^2 - 4ac > 0 than: it has solutions (roots to quadratic) and it takes positive and negative values for solutions for example, the two roots can be y-axis intersection points where plot curve travels through both of them along a change in the curge from positive to negative point values or vice versa

w takes both positive and negative values and therefore a saddle point exists.

example: a(x/y)^2+b(x/y)+c is always positive or always negative in its points according to the sign, minima or maxima occurs

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Second derivatives

pd^2(f)/pd^2(x^2) = f_x_x (pd is partial deriv and "_" is subscript) f_x_y is pd(x) with respect to the pd(y)

f_x_y = pd^2(f)/(pd(x)*pd(y)) (where pd^2 is second partial derivative) = pd^2(f)/(pd(y)*pd(x)) = f_y_x

  • Main point: f_x_y = f_y_x

pd^2(f)/pd(y^2) = f_y_y

second derivative test: at a critical point (x_0, y_0) of f let A = f_x_x(x_0,y_0) B = f_x_y(x_0,y_0) C = f_y_y(x_0,y_0)

  • critical point notation x_0, y_0. It seems due to the result being 0 at that point

Cases If AC - B^2 > 0: case 1: A > 0, local minima case 2: A < 0, local maxima if AC - B^2 < 0 case 3: saddle if AC - B^2 = 0 case 4: can not conclude

  • Test the 2cnd derivatives for these cases to find the type of extrima value

The tests are derived from the use of the quadratic formula

verify in special case w=ax^2+bxy+cy^2? w_x = 2ax+by w_x_x = 2a w_x_y = b w_y = bx + 2cy = b = w_y_x w_y_y = 2c

A = 2a, B = b, C = 2c AC-B^2 = 4ac - b^2 See cases above for how extrima can be found once AC - B^2 form is calculated

quadratic approximation: based on tayor series approximation approach: change(f) =~ f_x * (x-x_0)+f_y*(y-y_0) (where =~ is approximately equal) f_x = 0 at critical point f_y = 0 at critical point everything cancels due to equaling zero and therefore linear approximation is no good. non-linear approximation is used when looking for other terms the next terms are quadratic terms change(f) continues with =~ ... +1/2*(f_x_x)*(x-x_0)^2 +1/2*f_x_x*(x-x_0)^2+f_x_y*(x-x_0)(y-y_0)+1/2*f_y_y(y-y_0)^2

the general case reduces to the quadratic case 1/2*f_x_x = 1/2 A = a (small "a") f_x_y = B = b 1/2*f_y_y = 1/2 C = c a degenerate case is not a general case

In degenerate case, what actually happens depends on higher order derivatives.

Notes provided by the class: https://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/2.-partial-derivatives/part-a-functions-of-two-variables-tangent-approximation-and-optimization/session-30-second-derivative-test/MIT18_02SC_MNotes_sd.pdf

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Example problem w = x^3-3xy+y^3

w_x = 3x^2-3y = 0 => y=x^2 w_y = -3x+3y^2= 0 => -x+x^4=0 x = 0 or 1 critical points (0,0) (1,1)

A = w_x_x = 6x B = w_x_y = -3 C = w_y_y = 6y

Study AC-B^2 at crit. pts At (1,1): A = 6 B = -3 C = 6 AC-B^2=27>0 and a > 0 therefore minima At (0,0): A = 0 B = -3 C = 0 AC-B^2 = -9 < 0 -> saddle