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Section 16.5 Local Extreme Values

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

2 πœ•π‘€ πœ•π‘₯ =0 and πœ•π‘€ πœ•π‘¦ =0 and πœ•π‘€ πœ•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 𝑧=𝑓 π‘₯,𝑦 satisfy equations: πœ•π‘§ πœ•π‘₯ =0 and πœ•π‘§ πœ•π‘¦ =0 Stationary points for 𝑀=𝑓 π‘₯,𝑦,𝑧 satisfy the equations: πœ•π‘€ πœ•π‘₯ =0 and πœ•π‘€ πœ•π‘¦ =0 and πœ•π‘€ πœ•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 𝑧=𝑓 π‘₯,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 π‘₯ 2 is negative Concave Up 𝑑 2 𝑦 𝑑 π‘₯ 2 is positive Neither 𝑑 2 𝑦 𝑑 π‘₯ 2 is neither

3 If D is negative it is neither cupped up or down. If D is positive:
The Discriminate For a function 𝑧=𝑓 π‘₯,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 π‘₯,𝑦 : 𝐷= πœ• 2 𝑧 πœ• π‘₯ πœ• 2 𝑧 πœ• 𝑦 2 βˆ’ πœ• 2 𝑧 πœ•π‘₯πœ•π‘¦ 2 If D is negative it is neither cupped up or down. If D is positive: a. If πœ• 2 𝑧 πœ• π‘₯ 2 is positive it is cupped up. b. If πœ• 2 𝑧 πœ• 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. 𝑧= π‘₯ 3 βˆ’3 π‘₯ 2 βˆ’9π‘₯+ 𝑦 3 βˆ’6 𝑦 2 πœ• 2 𝑧 πœ• π‘₯ 2 =6π‘₯βˆ’6 πœ• 2 𝑧 πœ• 𝑦 2 =6π‘¦βˆ’12 πœ• 2 𝑧 πœ•π‘₯πœ•π‘¦ =0 πœ•π‘§ πœ•π‘₯ =3 π‘₯ 2 βˆ’6π‘₯βˆ’9 πœ•π‘§ πœ•π‘¦ =3 𝑦 2 βˆ’12𝑦 6π‘₯βˆ’6 6π‘¦βˆ’12 πœ• 2 𝑧 πœ• π‘₯ 2 πœ• 2 𝑧 πœ• 𝑦 2 Point πœ• 2 𝑧 πœ•π‘₯πœ•π‘¦ D Type Set each derivative to zero and solve. 3,0 12 -12 -144 saddle 3 π‘₯ 2 βˆ’6π‘₯βˆ’9=0 3 π‘₯ 2 βˆ’2π‘₯βˆ’3 =0 3 π‘₯βˆ’3 π‘₯+1 =0 π‘₯=3 and π‘₯=βˆ’1 3 𝑦 2 βˆ’12𝑦=0 3𝑦 π‘¦βˆ’4 =0 𝑦=0 and 𝑦=4 3,4 12 12 144 Local Min βˆ’1,0 Local Max -12 -12 144 Since these two equations are independent the stationary points are all the combinations of x and y. βˆ’1,4 -12 12 -144 saddle

4 Set both equations equal to zero.
Example Find the critical (stationary) points of the surface to the right and classify them. 𝑧= π‘₯ 2 π‘¦βˆ’8π‘₯𝑦+ 3𝑦 2 +12𝑦 πœ•π‘§ πœ•π‘₯ =2π‘₯π‘¦βˆ’8𝑦 πœ•π‘§ πœ•π‘¦ = π‘₯ 2 βˆ’8π‘₯+6𝑦+12 πœ• 2 𝑧 πœ• π‘₯ 2 =2𝑦 πœ• 2 𝑧 πœ• 𝑦 2 =6 πœ• 2 𝑧 πœ•π‘₯πœ•π‘¦ =2π‘₯βˆ’8 Set both equations equal to zero. 2π‘₯π‘¦βˆ’8𝑦=0 π‘₯ 2 βˆ’8π‘₯+6𝑦+12=0 2𝑦 6 2π‘₯βˆ’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 πœ• 2 𝑧 πœ• π‘₯ 2 πœ• 2 𝑧 πœ• 𝑦 2 πœ• 2 𝑧 πœ•π‘₯πœ•π‘¦ D Type 2,0 6 -4 -16 Saddle 2π‘₯π‘¦βˆ’8𝑦=0 2𝑦 π‘₯βˆ’4 =0 6,0 6 4 -16 Saddle 4, 2 3 Local Min 4 3 6 8 𝑦=0 π‘₯=4 π‘₯ 2 βˆ’8π‘₯+12=0 π‘₯βˆ’2 π‘₯βˆ’6 =0 π‘₯=2 or π‘₯=6 16βˆ’32+6𝑦+12=0 6𝑦=4 𝑦= 2 3 Points: 2,0 and 6,0 Points: 4, 2 3

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


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