Section 16.5 Local Extreme Values Chapter 16 Section 16.5 Local Extreme Values

𝜕𝑤 𝜕𝑥 =0 and 𝜕𝑤 𝜕𝑦 =0 and 𝜕𝑤 𝜕𝑧 =0 Critical (or Stationary) Points A critical or stationary point is a point (i.e. values for the independent variables) that give a zero gradient. On a surface this will make the tangent plane horizontal. Stationary points for 𝑧=𝑓 𝑥,𝑦 satisfy equations: 𝜕𝑧 𝜕𝑥 =0 and 𝜕𝑧 𝜕𝑦 =0 Stationary points for 𝑤=𝑓 𝑥,𝑦,𝑧 satisfy the equations: 𝜕𝑤 𝜕𝑥 =0 and 𝜕𝑤 𝜕𝑦 =0 and 𝜕𝑤 𝜕𝑧 =0 Extreme Points At a stationary point a surfaced can be cupped up, cupped down or neither, this determines if the point is a local max, local min or saddle point. Cupped Down Cupped Up Neither x y z x y z x y z Saddle Point Local Max Local Min Critical Values for 1 Variable For one variable functions a local max or min could be determined by looking at the sign of the second derivative. For a surface 𝑧=𝑓 𝑥,𝑦 there are 3 different second derivatives. How do they combine to tell you if it is cupped up or down? Concave Down 𝑑 2 𝑦 𝑑 𝑥 2 is negative Concave Up 𝑑 2 𝑦 𝑑 𝑥 2 is positive Neither 𝑑 2 𝑦 𝑑 𝑥 2 is neither

If D is negative it is neither cupped up or down. If D is positive: The Discriminate For a function 𝑧=𝑓 𝑥,𝑦 the discriminate is a combination of all second derivatives that determine if the function is cupped up, down or neither. It is abbreviated with 𝐷. If the point you are evaluating this at is a stationary point it determines if it is a local max, min or saddle point. For a surface 𝑧=𝑓 𝑥,𝑦 : 𝐷= 𝜕 2 𝑧 𝜕 𝑥 2 𝜕 2 𝑧 𝜕 𝑦 2 − 𝜕 2 𝑧 𝜕𝑥𝜕𝑦 2 If D is negative it is neither cupped up or down. If D is positive: a. If 𝜕 2 𝑧 𝜕 𝑥 2 is positive it is cupped up. b. If 𝜕 2 𝑧 𝜕 𝑥 2 is negative it is cupped down. 3. If D is zero can not tell might be up down or neither. Example Find the critical (stationary) points of the surface to the right and classify them. 𝑧= 𝑥 3 −3 𝑥 2 −9𝑥+ 𝑦 3 −6 𝑦 2 𝜕 2 𝑧 𝜕 𝑥 2 =6𝑥−6 𝜕 2 𝑧 𝜕 𝑦 2 =6𝑦−12 𝜕 2 𝑧 𝜕𝑥𝜕𝑦 =0 𝜕𝑧 𝜕𝑥 =3 𝑥 2 −6𝑥−9 𝜕𝑧 𝜕𝑦 =3 𝑦 2 −12𝑦 6𝑥−6 6𝑦−12 𝜕 2 𝑧 𝜕 𝑥 2 𝜕 2 𝑧 𝜕 𝑦 2 Point 𝜕 2 𝑧 𝜕𝑥𝜕𝑦 D Type Set each derivative to zero and solve. 3,0 12 -12 -144 saddle 3 𝑥 2 −6𝑥−9=0 3 𝑥 2 −2𝑥−3 =0 3 𝑥−3 𝑥+1 =0 𝑥=3 and 𝑥=−1 3 𝑦 2 −12𝑦=0 3𝑦 𝑦−4 =0 𝑦=0 and 𝑦=4 3,4 12 12 144 Local Min −1,0 Local Max -12 -12 144 Since these two equations are independent the stationary points are all the combinations of x and y. −1,4 -12 12 -144 saddle

Set both equations equal to zero. Example Find the critical (stationary) points of the surface to the right and classify them. 𝑧= 𝑥 2 𝑦−8𝑥𝑦+ 3𝑦 2 +12𝑦 𝜕𝑧 𝜕𝑥 =2𝑥𝑦−8𝑦 𝜕𝑧 𝜕𝑦 = 𝑥 2 −8𝑥+6𝑦+12 𝜕 2 𝑧 𝜕 𝑥 2 =2𝑦 𝜕 2 𝑧 𝜕 𝑦 2 =6 𝜕 2 𝑧 𝜕𝑥𝜕𝑦 =2𝑥−8 Set both equations equal to zero. 2𝑥𝑦−8𝑦=0 𝑥 2 −8𝑥+6𝑦+12=0 2𝑦 6 2𝑥−8 This system of equations is not independent (i.e. there are x’s and y’s in both). We need to solve one and substitute into the other. Point 𝜕 2 𝑧 𝜕 𝑥 2 𝜕 2 𝑧 𝜕 𝑦 2 𝜕 2 𝑧 𝜕𝑥𝜕𝑦 D Type 2,0 6 -4 -16 Saddle 2𝑥𝑦−8𝑦=0 2𝑦 𝑥−4 =0 6,0 6 4 -16 Saddle 4, 2 3 Local Min 4 3 6 8 𝑦=0 𝑥=4 𝑥 2 −8𝑥+12=0 𝑥−2 𝑥−6 =0 𝑥=2 or 𝑥=6 16−32+6𝑦+12=0 6𝑦=4 𝑦= 2 3 Points: 2,0 and 6,0 Points: 4, 2 3

Example Classify the critical (stationary) points of the surface: 𝑧=16− 𝑥+2 2 − 𝑦−5 4 . 𝜕𝑧 𝜕𝑥 =−2 𝑥+2 𝜕𝑧 𝜕𝑦 =−4 𝑦−5 3 𝜕 2 𝑧 𝜕 𝑥 2 =−2 𝜕 2 𝑧 𝜕 𝑦 2 =−12 𝑦−5 2 𝜕 2 𝑧 𝜕𝑥𝜕𝑦 =0 Set to zero and solve. 𝜕 2 𝑧 𝜕 𝑦 2 𝜕 2 𝑧 𝜕𝑥𝜕𝑦 Point 𝜕 2 𝑧 𝜕 𝑥 2 D Type −4 𝑦−5 3 =0 𝑦−5=0 𝑦=5 −2 𝑥+2 =0 𝑥=−2 -2 −2,5 ? The critical point is: −2,5 Because 𝐷=0 can not draw a conclusion! How do you tell what this critical point is? Form 𝑓 −2+ℎ,5+𝑘 −𝑓 −2,5 and look at this for small values of h and k. 𝑓 −2+ℎ,5+𝑘 −𝑓 −2,5 =16− −2+ℎ+2 2 − 5+𝑘−5 4 −16=− ℎ 2 − 𝑘 4 For all small values of h and k this expression will always be negative that means that the point −2,5 is a local max since the function’s value is zero there. If the point 𝑥 0 , 𝑦 0 is a stationary point and the discriminate 𝐷 𝑥 0 , 𝑦 0 =0, for small values of h and k consider: If 𝑓 𝑥 0 +ℎ, 𝑦 0 +𝑘 −𝑓 𝑥 0 , 𝑦 0 is negative then the point 𝑥 0 , 𝑦 0 is a local max. If 𝑓 𝑥 0 +ℎ, 𝑦 0 +𝑘 −𝑓 𝑥 0 , 𝑦 0 is positive then the point 𝑥 0 , 𝑦 0 is a local min. If 𝑓 𝑥 0 +ℎ, 𝑦 0 +𝑘 −𝑓 𝑥 0 , 𝑦 0 is neither then the point 𝑥 0 , 𝑦 0 is a saddle.