Kneeling for the solution of the extreme value of multivariate function under the constraint condition, 20 10 postgraduate entrance examination three mathematics 17 questions, can not be solved after listing the equations. The symbols are not easy to type, please match the words. Please forgive me.

The first equation: y+lambda*2x = 0.

The second equation: x+2z+lambda*2y = 0.

The third equation: 2y+lambda*2z = 0.

The fourth equation: x 2+y 2+z 2- 10 = 0.

According to equation 1 and equation 3, we can get the relationship: z=2x, and if we substitute it into equation 2, we can get 5x+λ* 2y = 0.

Substituting the above formula into the equation 1, you can get y =-λ* 2x =-λ* 2 *(λ* 2y)/5.

That is y = y* 2/5 * lambda^2 2 2.

Discuss the condition of this formula: (1) y = 0, then it can be deduced that x = z = 0 does not satisfy Equation 4, and this condition does not hold;

(2) If y is not equal to zero, you can omit y and get 2/5 *λ2 = 1.

Solution, λ = (plus or minus) 2 radical number 5,

The back is easy. Forget it. Replace both Y and Z in Equation 4 with X to get X 2+5x 2+4x 2 =10+00, and x = plus or minus1;

Y = plus or minus root number 5;

Z = plus or minus 2.

Pay attention to the symbolic pairing of x y z and lambda, there should be four groups of solutions.