16. Max-Min Problems

d. Multiple Constraint Problems (Optional)

It is also possible to solve max-min problems with \(2\) or more constraints. Again there are \(3\) methods of solution:

Each constraint allows you to eliminate \(1\) variable.

For each constraint, you add a Lagrange multiplier times the gradient of the constraint to the Lagrange equations. For example, if there are \(2\) constraints \(g=C\) and \(h=D\), then the Lagrange equations are: \[ \vec\nabla f=\lambda\vec\nabla g+\mu\vec\nabla h \] where \(\lambda\) and \(\mu\) are Lagrange multipliers. We solve the Lagrange equations and the constraints for the coordinates and \(\lambda\) and \(\mu\).

If you can parametrize the intersection of the constraints, then this method can be very powerful. However, parametrizing the intersection can be very hard.

We need to minimize the distance \(D=\sqrt{x^2+y^2+z^2}\). However, it is equivalent and easier to minimize the square of the distance: \[ f=D^2=x^2+y^2+z^2 \] We just need to remember to take the square root of \(f\) when we compute the final distance.

We define the constraint functions: \[ g=3x+4y+5z \qquad \text{and} \qquad h=z^2-x^2-y^2 \] Then we need to minimize \(f\) subject to the constrtaints \(g=1\) and \(h=0\). The gradients are \[\begin{aligned} \vec\nabla f&=\langle 2x,2y,2z\rangle \\ \vec\nabla g&=\langle 3,4,5\rangle \\ \vec\nabla h&=\langle -2x,-2y,2z\rangle \\ \end{aligned}\] The components of the Lagrange equations \(\vec\nabla f=\lambda\vec\nabla g+\mu\vec\nabla h\) are: \[\begin{aligned} 2x&=\lambda 3-\mu2x \qquad \text{(1)} \\ 2y&=\lambda 4-\mu2y \qquad \text{(2)} \\ 2z&=\lambda 5+\mu2z \qquad \text{(3)} \end{aligned}\] We solve these together with the constraints: \[\begin{aligned} 3x+4y+5z&=1 \qquad \text{(4)} \\ z^2-x^2-y^2&=0 \qquad \text{(5)} \end{aligned}\]

For a single constraint, a useful technique is to eliminate \(\lambda\). Here, we try to eliminate \(\lambda\) and \(\mu\). To eliminate \(\lambda\), we compute \(3\) times equation (2) minus \(4\) times equation (1) and \(3\) times equation (3) minus \(5\) times equation (1). These give: \[\begin{aligned} 6y-8x&=-\mu6y+\mu8x \\ 6z-10x&=\mu6z+\mu10x \end{aligned}\] These simplify to: \[\begin{aligned} 3y-4x&=-\mu(3y-4x) \qquad \text{(6)} \\ 3z-5x&=\mu(3z+5x) \qquad \text{(7)} \end{aligned}\] Equation (6) says \(\mu=-1\) or \(3y=4x\). We look at these two cases:

• Case 1: \(\mu=-1\): Equation (7) says \(3z-5x=-(3z+5x)\) or \(z=0\). Then the constraint (5) says \(-x^2-y^2=0\) or \(x=y=0\). This contradicts constraint (4). So this case does not leads to any critical points.
• Case 2: \(3y=4x\): We need to solve this along with equations (4) and (5). These become \(3x+\dfrac{16}{3}x+5z=1\) and \(z^2-x^2-\,\dfrac{16}{9}x^2=0\), or \[\begin{aligned} 25x+15z&=3 \qquad \text{(8)} \\ 9z^2-25x^2&=0 \qquad \text{(9)} \end{aligned}\] Equation (8) says: \[ z=\dfrac{1}{5}-\,\dfrac{5}{3}x \qquad \text{(15)} \] Then (9) becomes: \[ 9\left(\dfrac{1}{5}-\,\dfrac{5}{3}x\right)^2-25x^2=0 \] This is a quadratic equation to solve for \(x\). We clear the denominator by multiplying by \(25\): \[ (3-25x)^2-25^2x^2=0 \] This expands to \[ 9-150x=0 \] So \(x=\dfrac{3}{50}\). Substituting back, we have: \[ y=\dfrac{4}{3}x=\dfrac{4}{50}=\dfrac{2}{25} \] \[ z=\dfrac{1}{5}-\,\dfrac{5}{3}x=\dfrac{1}{5}-\,\dfrac{5}{50}=\dfrac{5}{50}=\dfrac{1}{10} \]

So the critical point is \[ (x,y,z)=\left(\dfrac{3}{50},\dfrac{4}{50},\dfrac{5}{50}\right) =\left(\dfrac{3}{50},\dfrac{2}{25},\dfrac{1}{10}\right) \] and the function value is \[ f=\left(\dfrac{3}{50}\right)^2+\left(\dfrac{4}{50}\right)^2+\left(\dfrac{5}{50}\right)^2 =\dfrac{3^2+4^2+5^2}{50^2} =\dfrac{1}{50} \] and the smallest distance is the square root of this \(D=\sqrt{f}=\dfrac{1}{\sqrt{50}}\).