Find the maximum value of the function?

Question

Find the maximum value of the function f(x,y,z)= x+2y-3z subject to the constraint z=4x^2+y^2

Answer 1

Substitute the z constraint into f(x,y,z).
f(x,y) = x+2y - 12x^2 -3y^2
Take partial derivatives with respect to each variable and set them to zero.
f'(x,y)(x) = 1-24x = 0
f'(x,y)(y) = 2 - 6y = 0

f'(x,y)(x) => x=1/24
f'(x,y)(y) => y = 1/3

Evaluate the original function at this (x,y) coordinate and you'll have your max value.

Answer 2

Let's see. When z = 4x^2 + y^2,
f(x,y,z) = f(x,y) = x + 2y - 3 ( 4x^2 + y^2 )
= x + 2y - 12 x^2 - 3 y^2

One can find extreme points by taking partial derivatives with respect to x and with respect to y and setting them equal to zero:

∂f / ∂x = 1 - 24 x = 0
∂f / ∂y = 2 - 6y = 1 - 3y = 0

Simultaneous solution is not necessary. This condition is reached when x = 1/24 and y = 1/3. At this point,

f(x,y) = x + 2y - 12 x^2 - 3 y^2
= 1/24 + 2(1/3) - 12 (1/24)^2 - 3(1/3)^2
= 1/24 + 2/3 - 1/48 - 1/3
= ( 2 + 32 - 1 - 16) / 48
= 17 / 48


To verify that this is a maximum, it is sufficient that the second partial derivatives are negative.

∂^2 f / ∂x^2 = -24 ok
∂^2 f / ∂y^2 = -3 ok

Answer 3

you can use lagrangian multipliers

F(x,y,z,k) = x+2y-3z+k(4x^2+y^2-z)
Fx=1+8kx=0
Fy=2+2ky=0
Fz= -3-k=0 so k=-3
1-24x=0 x= 1/24
2-6y= 0 so y=1/3 and z= 4(1/24)^2 +1/9

Answer 4

we will use Lagrange multiplier method
let g(x,y,z) = z-4x^2-y^2
now consider function V(x,y,z,λ) = f(x,y,z)+λg(x,y,z) where λ is a variable.
now for maximum value,(let me denote partial differentiation by p, it is due to lack of the symbol del )
pV/px =0;
pV/py =0;
pV/pz =0;
pV/px = 1-8λx=0;
pV/py = 2-2λy=0;
pV/pz = λ-3=0;
so ,we get value of x= 1/24 and y = 1/3
putting these value we get z = 17/144

so, f(x,y,z) = 1/24 + 2/3 -3*17/144 =17/48