Question

Evaluate the minimum value of x2+y2+z2 subject to x+y+2z=12​

Answer 1

Question :- Evaluate the minimum value of x² + y² + z² subject to x+2y+2z = 12.

Solution :-

Given that, Equation of plane is = x + 2y + 2z = 12.

Now,

Let us Assume that, this plane has a point (x,y,z) .

Now,

Distance from origin (0,0,0) to the point (x,y,z) will be = √{(x - 0)² + (y - 0)² + (z - 0)²}

→ D = √(x² + y² + z²)

→ D²(min.) = (x² + y² + z²)(min.)

Now, This distance will be minimum when they are Perpendicular on the plane.

Therefore,

→ D = | {(Ax1 + By1 + Cz1) - D} / √(A² + B² + C²) |

Putting values now, we get,

→ D = | {(1*0 + 2*0 + 2*0) - 12} /√(1² + 2² + 2²) |

→ D = | (-12) / (√9) |

→ D = | (-12) / 3 |

→ D = | (-4) |

→ D = 4.

Hence,

→ D²(min.) = (x² + y² + z²)(min.)

→ (x² + y² + z²)(min.) = 4² = 16 (Ans.)

Answer 2

Answer:

x²+y²+z² subject to x+y+2z=12​

Step-by-step explanation:

Using Lagrange Multiplier:

Let us have a function F(x,y,z,λ) such that

F(x,y,z,λ)=(x²+y²+z²)-λ(x+y+2z-12)

Partial Derivate the Function with x , y ,z , λ

Partial derivative with x : 2x-λ

Partial derivative with y =2y-λ

Partial derivative with z =2z-2λ

Partial derivative with λ= -x-y-2z+12

Equate the equations i , ii and iii to 0 and find the value of x ,y and z

2x-λ=0   :   x=λ/2

2y-λ=0   :   y=λ/2

z-2λ=0   :   z=λ

Put the values of x ,y ,z in equation iv evaluating it to zero.

-x-y-2z+12=0

(-λ/2)-(λ/2)-2λ+12=0

λ=4

Put values of λ in x , y ,z

∴x=4/2   ,  y=4/2   ,  z=4

So the minimum value of x²+y²+z² is (2)²+(2)²+(4)²=24