Problem 1
1. Minimize \(f(x,y,z) = x^2+y^2+z^2\) subject to the constraint \(x+y+z=1\). (5 points)
Solution. We solve using Lagrange multipliers. We have \(\nabla f = (2x,2y,2z)\) and \(\nabla g = (1,1,1)\), giving us the system of equations
\[ \begin{cases} 2x = \lambda \\ 2y = \lambda \\ 2z = \lambda \\ x+y+z = 1. \end{cases} \]
Combining the first three equations gives us \(x=y=z\), and upon substitution into the last equation, we get \(x=1/3\) and so on. This gives us a value of \(f(1/3,1/3,1/3) = 1/3\). This is a minimum: there is no max, as something with negative values can grow much larger, such as \(f(100,-100,0) = 20000\).Problem 2
2. Find the absolute maximum and minimum values of \(4xy\) subject to \(\dfrac{x^2}{9}+\dfrac{y^2}{16}=1\). (5 points)
Solution. We again use Lagrange multipliers. We get the system:
\[ \begin{cases} 4y = 2\lambda x/9 \\ 4x = \lambda y/8 \\ \dfrac{x^2}{9}+\dfrac{y^2}{16} = 1. \end{cases} \]
We can multiply the first equation by \(x\) and the second equation by \(y\) to get \[ 4xy = \lambda x^2/18 \quad \text{and} \quad 4xy = \lambda y^2/32. \]
Setting them equal gives us \(\lambda x^2/18 = \lambda y^2/32\) or equivalently \(y^2/16 = x^2/9\). We can substitute these into the third equation to get \(2x^2/9 = 1\) and \(y^2/8 = 1\), giving us the critical points \(x = \pm 3/\sqrt{2}\) and \(y = \pm 2\sqrt{2}\). Just looking at the positive points, we get \(f(3/\sqrt{2}, 2\sqrt{2}) = 24\). The signs in the critical point will only change the sign of the answer, so we conclude \((3/\sqrt{2}, 2\sqrt{2})\) and \((-3/\sqrt{2}, -2\sqrt{2})\) are the places we hit a maximum, with a value of 24. On the other hand, at \((3/\sqrt{2}, -2\sqrt{2})\) and \((-3/\sqrt{2}, 2\sqrt{2})\) we hit an absolute minimum of \(-24\).