Question

Find maximum and minimum values of x³ +3xy² -15x²-15y²+72x.

Answer 1

Given f(x) = x³ +3xy² -15x²-15y²+72x… (1)
STEP 1] for maxima, minima,`(delf)/(delx)=0;(delf)/(dely)=0`
3x²+3y²-30x+72=0 and 6xy – 30y=0
∴ y (6x-30) =0
y=0, x= 5
For x=5; From Equation 3x²+3y²-30x+72=0, we get y2-1=0
Y=±1
Hence (4,0) , (6,0) , (5,1) ,(5,-1) are the stationary points.
STEP 2] Now,`r=(del^2f)/(delx^2)=6x-30;`

`"S"=(del^2f)/(delxdely)=6y;`

`t=(del^2f)/(dely^2)=6x-30`

STEP 3] for (x, y) ≡ (4, 0), r = -6, s = 0, t = -6;
rt –s²=(-6)(-6)-0=36>0 and r< 0.
This shows that the function is maximum at (4, 0)
∴ From Equation (1)
F_max=f (4, 0) =4³+0-15(4²) +0+72(4) =64 – 240 + 288
F_max=112
STEP 4] For (x,y)≡(6,0)
r=6, s=0, t=6
rt-s²=36 but r=6>0
This shows that function is minimum at (6, 0).
From Equation (1),
F_min=f(6,0)=6³+0-15(6)²+0+72(6)=108.
STEP 5] For(x, y) ≡ (5, 1)
r=0, s=6, t=0;
(rt-s²)<0
This shows that at (5, 1) and (5,-1) function is neither maxima nor minima.
These points are saddle points.

Answer 2

Given f(x) = x³ +3xy² -15x²-15y²+72x… (1)
STEP 1] for maxima, minima,`(delf)/(delx)=0;(delf)/(dely)=0`
3x²+3y²-30x+72=0 and 6xy – 30y=0
∴ y (6x-30) =0
y=0, x= 5
For x=5; From Equation 3x²+3y²-30x+72=0, we get y2-1=0
Y=±1
Hence (4,0) , (6,0) , (5,1) ,(5,-1) are the stationary points.
STEP 2] Now,`r=(del^2f)/(delx^2)=6x-30;`

`"S"=(del^2f)/(delxdely)=6y;`

`t=(del^2f)/(dely^2)=6x-30`

STEP 3] for (x, y) ≡ (4, 0), r = -6, s = 0, t = -6;
rt –s²=(-6)(-6)-0=36>0 and r< 0.
This shows that the function is maximum at (4, 0)
∴ From Equation (1)
F_max=f (4, 0) =4³+0-15(4²) +0+72(4) =64 – 240 + 288
F_max=112
STEP 4] For (x,y)≡(6,0)
r=6, s=0, t=6
rt-s²=36 but r=6>0
This shows that function is minimum at (6, 0).
From Equation (1),
F_min=f(6,0)=6³+0-15(6)²+0+72(6)=108.
STEP 5] For(x, y) ≡ (5, 1)
r=0, s=6, t=0;
(rt-s²)<0
This shows that at (5, 1) and (5,-1) function is neither maxima nor minima.
These points are saddle points.