Question

1. Find the critical points of the function. Then use the Second Derivative Test to determine whether they are local minima, local maxima, or saddle points (or state that the test fails.) (Order your answers from smallest to largest x, then from smallest to largest y.)f(x, y) = 36x − 12x3 − 6xy22. Find the critical points of the function. Then use the Second Derivative Test to determine whether they are local minima, local maxima, or saddle points (or state that the test fails.) (Order your answers from smallest to largest x, then from smallest to largest y.)f(x, y) = x3 + 6xy − 6y2 − 6x

Answer 1

Answer:

1. the critical point are at x=0 and y=2.45, and the critical points occur at the saddle points.

2. the critical point are at x=2 and y=1, and the critical points occur at the saddle points.

Step-by-step explanation:

1.

let  t= F(x,y)

Given, t= F(x,y)= 36x - 12x³ -6xy²

(1) determine $\frac{dt}{dx}$

and  $\frac{dt}{dy}$.

therefore, $\frac{dt}{dx}$ = 36-36x² -6y²     ...................................eqn1

$\frac{dt}{dy}$ = -12xy                                 .....................................eqn2

(2) for stationary points,

for eqn1, 36-36x² -6y² = 0

36x² + 6y² = 36

6x² + y² = 6

y= $\sqrt{6-6x^{2} }$               ...................................eqn3                

then, substitute y in eqn2

then for, -12xy = 0

-12x$\sqrt{6-6x^{2} }$ = 0

solving this, give x=0                      ...................................eqn4

substitute eqn 4 in eqn3, gives

y= $\sqrt{6}$

y= 2.45

(3) therefore, for eqn 1 and 2, the stationary points occur at x=0 and y=2.45.

(4)determine  $\frac{d²t}{dx²}$

since, $\frac{dt}{dx}$ = 36-36x² -6y²    

$\frac{d²t}{dx²}$ = 36x                                  ...................................eqn5

(5)determine  $\frac{d²t}{dy²}$

since,  $\frac{dt}{dy}$ = -12xy

then $\frac{d²t}{dy²}$ = -12x                       ...................................eqn6      

(6)determine  $\frac{d²t}{dxdy}$ =  $\frac{d}{dx}$.$\frac{dt}{dy}$

= d/dx (-12xy) = -12y                                            ...................................eqn7

substitute values of x and y (0, 2.45) in eqn 5,6,7

eqn 5 =0

eqn 6 =0

also,

eqn7 =-29.4

also for, sqr eqn7 = (-12y)² = 144y² = 864.4

(7) finally, determine whether it is maxima, minima or saddle point by:

Δ=  ($\frac{d²t}{dxdy}$)² - [$\frac{d²t}{dy²}$ * $\frac{d²t}{dx²}$]

for (x=0, y=2.45)

Δ = 864.4 - {(0)(0)} = 864.4

since, Δ > 0, therefore the stationary points (0,2.45) is a saddle point.

2.

let  t= F(x,y)

Given, t= F(x,y)= x³ + 6xy - 6y² - 6x

(1) determine $\frac{dt}{dx}$  

and  $\frac{dt}{dy}$.

therefore, $\frac{dt}{dx}$ = 3x² -6y-6     ...................................eqn1

$\frac{dt}{dy}$ = 6x - 12y                        .....................................eqn2

(2) for stationary points,

for eqn1, 3x² -6y-6 = 0

x² -2y -2 = 0

x²  =2+2y

x =sqrt(2+2y)                                                                            ...................eqn3                

then, substitute x in eqn2

then for, 6x - 12y= 0

-6(sqrt(2+2y) - 12y= 0

solving this, give y= 1 and -0.5                      ...................................eqn4

substitute eqn 4 in eqn3, gives

x = 2 and 1

(3) therefore, for eqn 1 and 2, the stationary points occur at x=2, y=1 and x=1, y=-0.5.

thus, the stationary points occur at (2,1) and (1, -0.5)

(4)determine  $\frac{d²t}{dx²}$

since, $\frac{dt}{dx}$ = 3x² -6y-6    

$\frac{d²t}{dx²}$ = 6x                                ...................................eqn5

(5)determine  $\frac{d²t}{dy²}$

since,  $\frac{dt}{dy}$ = 6x - 12y

then $\frac{d²t}{dy²}$ = -12y                      ...................................eqn6      

(6)determine  $\frac{d²t}{dxdy}$ =  $\frac{d}{dx}$.$\frac{dt}{dy}$

= d/dx (6x - 12y) = 6                                            ...................................eqn7

substitute values of x and y as (2,1) and (1, -0.5) in eqn 5,6,7

at (2,1) eqn 5 =12, eqn 6 =-12 also, eqn7 =6

at (1,-0.5) eqn 5 =6, eqn 6 =6 also, eqn7 =6

also for, sqr eqn7 = (6)² = 36

(7) finally, determine whether it is maxima, minima or saddle point by:

Δ=  ($\frac{d²t}{dxdy}$)² - [$[\frac{d²t}{dy²}$* $\frac{d²t}{dx²}$]

for (x=2, y=1)

Δ = 36 - {(12)(-12)} = 36+144 = 180

for (x=1, y=-0.5)

Δ = 36 - {(6)(6)} = 36-36 = 0

since, Δ > 0, for (2,1) therefore the stationary points (2,1) is a saddle point.