Question 14.6.4: Solve the problem min x² + y² + z² s.t.{x + 2y + z = 30 2x −......

The Lagrangian is

The first-order conditions (14.6.11) require that

{\frac{\partial{\mathcal{L}}}{\partial x_{i}}}={\frac{\partial f(\mathbf{x})}{\partial x_{i}}}-\sum\limits_{j=1}^{m}\lambda_{j}{\frac{\partial g_{j}(\mathbf{x})}{\partial x_{i}}}=0     (14.6.11)

{\frac{\partial{\mathcal{L}}}{\partial x}}=2x-\lambda_{1}-2\lambda_{2}=0     (i)

{\frac{\partial{\mathcal{L}}}{\partial v}}=2y-2\lambda_{1}+\lambda_{2}=0         (ii)

{\frac{\partial{\mathcal{L}}}{\partial z}}=2z-\lambda_{1}+3\lambda_{2}=0      (iii)

in addition to the two constraints,

x + 2y + z = 30           (iv)

2x − y − 3z = 10          (v)

So there are five equations, (i) to (v), to determine the five unknowns: x, y, z, λ_{1}, and λ_{2}. Solving (i) and (ii), simultaneously, for λ_{1} and λ_{2} gives λ_{1} = \frac{2}{5}x + \frac{4}{5}y and λ_{2} = \frac{4}{5}x − \frac{2}{5}y. Inserting these expressions for λ_{1} and λ_{2} into (iii) and rearranging yields
x − y + z =0 (vi)

This equation, together with (iv) and (v), constitutes a system of three linear equations in the unknowns x, y, and z. Solving this system by elimination gives (x, y, z) = (10, 10, 0). The corresponding values of the multipliers are λ_{1} = 12 and λ_{2} = 4.
A geometric argument allows us to confirm that we have solved the minimization problem. Each of the two constraints represents a plane in \mathbb{R}^{3}, and the points satisfying both constraints consequently lie on the straight line where the two planes intersect. Now x² + y² + z² measures (the square of) the distance from the origin to a point on this straight line, which we want to make as small as possible by choosing the point on the line that is nearest to the origin. No maximum distance can possibly exist, but it is geometrically obvious that there is a minimum distance, and it must be attained at this nearest point.

An alternative method to solve this particular problem is to reduce it to a one-variable optimization problem by using the two constraints to get y = 20 − x and z = x − 10, the equations of the straight line where the two planes intersect. Then the square of the distance from the origin is

and this function is easily seen to have a minimum when x = 10. See also Exercise 5.