Problem #2: Consider the following function. h(u, v) - 23 +36uv + 18v2 (a) Find

Question

Problem #2: Consider the following function. h(u, v) - 23 +36uv + 18v2 (a) Find the critical points of h. (b) For each critical point in (a), find the value of D(a, b) from the Second Partials test that is used to classify the critical point. Separate your answers with a comma. Your first value of D must correspond to the first critical point in (a) [i.e., the critical point (0, 0)], and your second value of D must correspond with the second critical point (c) Use the Second Partials test to classify each critical point from (a). Note that for each given answer, the first classification corresponds to the first critical point in (a) li.e., the critical point (0, 0)] and the second classification corresponds to the second critical point in (a)

Explanation / Answer

a)

givev h(u,v) =2u3 +36uv +18v2

h/u =hu=6u2 +36v, h/v =hv=36u+36v,

for critical points,hu =0, hv =0

=>6u2 +36v =0,36u+36v=0

=> u2 +6v =0,u=-v

=>(-v)2 +6v =0

=>v2+6v =0

=>v(v+6) =0

=>v =-6 , v =0

v=-6, u=-v =>u=-(-6) =6

v=0, u=-v =>u=0

critical points are (u,v)=(0,0),(6,-6)

(b)

h/u =hu=6u2 +36v, h/v =hv=36u+36v,

2h/u2 =huu=12u, 2h/v2 =hvv=36, 2h/uv =huv=36,D=huuhvv-(huv)2

at (0,0)

2h/u2 =huu=0, 2h/v2 =hvv=0, 2h/uv =huv=36,D=0*0-(36)2

=>D=-1296

at (6,-6)

2h/u2 =huu=72, 2h/v2 =hvv=36, 2h/uv =huv=36,D=72*36 -(36)2

=>D=1296

(c)

D=-1296 <0

=>(0,0) saddle points

D=1296>0,huu=72>0

=>(6,-6) is local minimum

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