Question 14.6.2: Solve the problem min f (x, y, z) = (x − 4)² + (y − 4)² + (z......
Solve the problem
\min f (x, y, z) = (x − 4)² + (y − 4)² + (z − \frac{1}{2})² s.t. x² + y² = z
Can you give a geometric interpretation of the problem?
The Lagrangian is
\mathcal{L}(x,y,z)=(x-4)^{2}+(y-4)^{2}+\left(z-{\textstyle\frac{1}{2}}\right)^{2}-\lambda(x^{2}+y^{2}-z)and the first-order conditions are:
{\mathcal L}_{1}^{\prime}(x,y,z)=2(x-4)-2\lambda x=0 (i)
{\mathcal L}_{2}^{\prime}(x,y,z)=2(y-4)-2\lambda y=0 (ii)
{\mathcal L}_{3}^{\prime}(x,y,z)=2\left(z-{\frac{1}{2}}\right)+\lambda\ \ =0 (iii)
x^{2}+y^{2}=z (iv)
From (i) we see that x = 0 is impossible. Equation (i) thus gives λ = 1 − 4/x. Inserting this into (ii) and (iii) gives y = x and z = 2/x. Using these results, Eq. (iv) reduces to 2x² = 2/x, that is, x³ = 1, so x = 1. It follows that (x, y, z) = (1, 1, 2) is the only solution candidate for the problem.
The expression (x − 4)² + (y − 4)² + (z − 1/2)² measures the square of the distance from the point (4, 4, 1/2) to the point (x, y, z). The set of points (x, y, z) that satisfy z = x² + y² is a surface known as a paraboloid, part of which is shown in Fig. 14.6.1. The minimization problem is therefore to find that point on the paraboloid which has the smallest (square) distance from (4, 4, 1/2). It is “geometrically obvious” that this problem has a solution. On the other hand, the problem of finding the largest distance from (4, 4, 1/2) to a point on the paraboloid does not have a solution, because the distance can be made as large as we like.
