2 𝜕𝑤 𝜕𝑥 =0 and 𝜕𝑤 𝜕𝑦 =0 and 𝜕𝑤 𝜕𝑧 =0 Critical (or Stationary) PointsA 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 =0Stationary points for 𝑤=𝑓 𝑥,𝑦,𝑧 satisfy the equations:𝜕𝑤 𝜕𝑥 =0 and 𝜕𝑤 𝜕𝑦 =0 and 𝜕𝑤 𝜕𝑧 =0Extreme PointsAt 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 DownCupped UpNeitherxyzxyzxyzSaddle PointLocal MaxLocal MinCritical Values for 1 VariableFor 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 Down𝑑 2 𝑦 𝑑 𝑥 2 is negativeConcave Up𝑑 2 𝑦 𝑑 𝑥 2 is positiveNeither𝑑 2 𝑦 𝑑 𝑥 2 is neither
3 If D is negative it is neither cupped up or down. If D is positive: The DiscriminateFor 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 𝑧 𝜕 𝑦 2 − 𝜕 2 𝑧 𝜕𝑥𝜕𝑦 2If 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.ExampleFind 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 −12𝑦6𝑥−66𝑦−12𝜕 2 𝑧 𝜕 𝑥 2𝜕 2 𝑧 𝜕 𝑦 2Point𝜕 2 𝑧 𝜕𝑥𝜕𝑦DTypeSet each derivative to zero and solve.3,012-12-144saddle3 𝑥 2 −6𝑥−9=03 𝑥 2 −2𝑥−3 =03 𝑥−3 𝑥+1 =0𝑥=3 and 𝑥=−13 𝑦 2 −12𝑦=03𝑦 𝑦−4 =0𝑦=0 and 𝑦=43,41212144LocalMin−1,0LocalMax-12-12144Since these two equations are independent the stationary points are all the combinations of x and y.−1,4-1212-144saddle
4 Set both equations equal to zero. ExampleFind 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−8Set both equations equal to zero.2𝑥𝑦−8𝑦=0𝑥 2 −8𝑥+6𝑦+12=02𝑦62𝑥−8This 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𝜕𝑦DType2,06-4-16Saddle2𝑥𝑦−8𝑦=02𝑦 𝑥−4 =06,064-16Saddle4, 2 3LocalMin4 368𝑦=0𝑥=4𝑥 2 −8𝑥+12=0𝑥−2 𝑥−6 =0𝑥=2 or 𝑥=616−32+6𝑦+12=06𝑦=4𝑦= 2 3Points:2,0 and 6,0Points:4, 2 3
5 ExampleClassify 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 =0Set to zero and solve.𝜕 2 𝑧 𝜕 𝑦 2𝜕 2 𝑧 𝜕𝑥𝜕𝑦Point𝜕 2 𝑧 𝜕 𝑥 2DType−4 𝑦−5 3 =0𝑦−5=0𝑦=5−2 𝑥+2 =0𝑥=−2-2−2,5?The critical point is: −2,5Because 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 4For 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.