Respuesta :
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 [tex]\frac{dt}{dx} \\[/tex]
and [tex]\frac{dt}{dy} \\[/tex].
therefore, [tex]\frac{dt}{dx} \\[/tex] = 36-36x² -6y² ...................................eqn1
[tex]\frac{dt}{dy} \\[/tex] = -12xy .....................................eqn2
(2) for stationary points,
for eqn1, 36-36x² -6y² = 0
36x² + 6y² = 36
6x² + y² = 6
y= [tex]\sqrt{6-6x^{2} }[/tex] ...................................eqn3
then, substitute y in eqn2
then for, -12xy = 0
-12x[tex]\sqrt{6-6x^{2} }[/tex] = 0
solving this, give x=0 ...................................eqn4
substitute eqn 4 in eqn3, gives
y= [tex]\sqrt{6}[/tex]
y= 2.45
(3) therefore, for eqn 1 and 2, the stationary points occur at x=0 and y=2.45.
(4)determine [tex]\frac{d²t}{dx²} \\[/tex]
since, [tex]\frac{dt}{dx} \\[/tex] = 36-36x² -6y²
[tex]\frac{d²t}{dx²} \\[/tex] = 36x ...................................eqn5
(5)determine [tex]\frac{d²t}{dy²} \\[/tex]
since, [tex]\frac{dt}{dy} \\[/tex] = -12xy
then [tex]\frac{d²t}{dy²} \\[/tex] = -12x ...................................eqn6
(6)determine [tex]\frac{d²t}{dxdy} \\[/tex] = [tex]\frac{d}{dx} \\[/tex].[tex]\frac{dt}{dy} \\[/tex]
= 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= ([tex]\frac{d²t}{dxdy} \\[/tex])² - [[tex]\frac{d²t}{dy²} \\[/tex] * [tex]\frac{d²t}{dx²} \\[/tex]]
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 [tex]\frac{dt}{dx} \\[/tex]
and [tex]\frac{dt}{dy} \\[/tex].
therefore, [tex]\frac{dt}{dx} \\[/tex] = 3x² -6y-6 ...................................eqn1
[tex]\frac{dt}{dy} \\[/tex] = 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 [tex]\frac{d²t}{dx²} \\[/tex]
since, [tex]\frac{dt}{dx} \\[/tex] = 3x² -6y-6
[tex]\frac{d²t}{dx²} \\[/tex] = 6x ...................................eqn5
(5)determine [tex]\frac{d²t}{dy²} \\[/tex]
since, [tex]\frac{dt}{dy} \\[/tex] = 6x - 12y
then [tex]\frac{d²t}{dy²} \\[/tex] = -12y ...................................eqn6
(6)determine [tex]\frac{d²t}{dxdy} \\[/tex] = [tex]\frac{d}{dx} \\[/tex].[tex]\frac{dt}{dy} \\[/tex]
= 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= ([tex]\frac{d²t}{dxdy} \\[/tex])² - [[tex][\frac{d²t}{dy²}[/tex]* [tex]\frac{d²t}{dx²} \\[/tex]]
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.
