# Section 16.5 Local Extreme Values

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

ππ€ ππ₯ =0 and ππ€ ππ¦ =0 and ππ€ ππ§ =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 π§=π π₯,π¦ satisfy equations: ππ§ ππ₯ =0 and ππ§ ππ¦ =0 Stationary points for π€=π π₯,π¦,π§ satisfy the equations: ππ€ ππ₯ =0 and ππ€ ππ¦ =0 and ππ€ ππ§ =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 π§=π π₯,π¦ there are 3 different second derivatives. How do they combine to tell you if it is cupped up or down? Concave Down π 2 π¦ π π₯ 2 is negative Concave Up π 2 π¦ π π₯ 2 is positive Neither π 2 π¦ π π₯ 2 is neither

If D is negative it is neither cupped up or down. If D is positive:
The Discriminate For a function π§=π π₯,π¦ the discriminate is a combination of all second derivatives that determine if the function is cupped up, down or neither. It is abbreviated with π·. 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 π§=π π₯,π¦ : π·= π 2 π§ π π₯ π 2 π§ π π¦ 2 β π 2 π§ ππ₯ππ¦ 2 If D is negative it is neither cupped up or down. If D is positive: a. If π 2 π§ π π₯ 2 is positive it is cupped up. b. If π 2 π§ π π₯ 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. π§= π₯ 3 β3 π₯ 2 β9π₯+ π¦ 3 β6 π¦ 2 π 2 π§ π π₯ 2 =6π₯β6 π 2 π§ π π¦ 2 =6π¦β12 π 2 π§ ππ₯ππ¦ =0 ππ§ ππ₯ =3 π₯ 2 β6π₯β9 ππ§ ππ¦ =3 π¦ 2 β12π¦ 6π₯β6 6π¦β12 π 2 π§ π π₯ 2 π 2 π§ π π¦ 2 Point π 2 π§ ππ₯ππ¦ D Type Set each derivative to zero and solve. 3,0 12 -12 -144 saddle 3 π₯ 2 β6π₯β9=0 3 π₯ 2 β2π₯β3 =0 3 π₯β3 π₯+1 =0 π₯=3 and π₯=β1 3 π¦ 2 β12π¦=0 3π¦ π¦β4 =0 π¦=0 and π¦=4 3,4 12 12 144 Local Min β1,0 Local Max -12 -12 144 Since these two equations are independent the stationary points are all the combinations of x and y. β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. π§= π₯ 2 π¦β8π₯π¦+ 3π¦ 2 +12π¦ ππ§ ππ₯ =2π₯π¦β8π¦ ππ§ ππ¦ = π₯ 2 β8π₯+6π¦+12 π 2 π§ π π₯ 2 =2π¦ π 2 π§ π π¦ 2 =6 π 2 π§ ππ₯ππ¦ =2π₯β8 Set both equations equal to zero. 2π₯π¦β8π¦=0 π₯ 2 β8π₯+6π¦+12=0 2π¦ 6 2π₯β8 This system of equations is not independent (i.e. there are xβs and yβs in both). We need to solve one and substitute into the other. Point π 2 π§ π π₯ 2 π 2 π§ π π¦ 2 π 2 π§ ππ₯ππ¦ D Type 2,0 6 -4 -16 Saddle 2π₯π¦β8π¦=0 2π¦ π₯β4 =0 6,0 6 4 -16 Saddle 4, 2 3 Local Min 4 3 6 8 π¦=0 π₯=4 π₯ 2 β8π₯+12=0 π₯β2 π₯β6 =0 π₯=2 or π₯=6 16β32+6π¦+12=0 6π¦=4 π¦= 2 3 Points: 2,0 and 6,0 Points: 4, 2 3

Example Classify the critical (stationary) points of the surface: π§=16β π₯+2 2 β π¦β5 4 . ππ§ ππ₯ =β2 π₯+2 ππ§ ππ¦ =β4 π¦β5 3 π 2 π§ π π₯ 2 =β2 π 2 π§ π π¦ 2 =β12 π¦β5 2 π 2 π§ ππ₯ππ¦ =0 Set to zero and solve. π 2 π§ π π¦ 2 π 2 π§ ππ₯ππ¦ Point π 2 π§ π π₯ 2 D Type β4 π¦β5 3 =0 π¦β5=0 π¦=5 β2 π₯+2 =0 π₯=β2 -2 β2,5 ? The critical point is: β2,5 Because π·=0 can not draw a conclusion! How do you tell what this critical point is? Form π β2+β,5+π βπ β2,5 and look at this for small values of h and k. π β2+β,5+π βπ β2,5 =16β β2+β+2 2 β 5+πβ5 4 β16=β β 2 β π 4 For all small values of h and k this expression will always be negative that means that the point β2,5 is a local max since the functionβs value is zero there. If the point π₯ 0 , π¦ 0 is a stationary point and the discriminate π· π₯ 0 , π¦ 0 =0, for small values of h and k consider: If π π₯ 0 +β, π¦ 0 +π βπ π₯ 0 , π¦ 0 is negative then the point π₯ 0 , π¦ 0 is a local max. If π π₯ 0 +β, π¦ 0 +π βπ π₯ 0 , π¦ 0 is positive then the point π₯ 0 , π¦ 0 is a local min. If π π₯ 0 +β, π¦ 0 +π βπ π₯ 0 , π¦ 0 is neither then the point π₯ 0 , π¦ 0 is a saddle.