conic sectionslagrange multipliermultivariable-calculusoptimization
I've solved the problem without the use of Lagrange multipliers, but am unsure on how to solve this problem using lagrange multipliers like was intended?
Is there some way to form my own constraint? I wanted to use distance as a formula and have the ellipse formula act as the constraint, but was unsure on how to go about it?
Does anyone have any insight on how to do this problem? I've scoured the internet for tips, but was unsuccessful.
Warning: This method is cute, but does not use Lagrange multipliers.
Let $x=3w$ and $y=2u$. Then we have $xyz=6\implies uwz=1$ and we want to minimize $$xy+2xz+3yz=6uw+6wz+6uz=6(uw+wz+uz)$$ Now by the AM-GM inequality, we have $$uw+wz+uz\ge3\sqrt[3]{uwz}=3,$$ with equality when $uw=wz=uz$ which implies $u=w=z=1$, so we have $$xy+2xz+3yz\ge6\times3=18$$ and we can see $18$ is a true minimum because it is achieved at $x=3$, $y=2$, and $z=1$.
The constraint in problems like these is usually given as a level curve of some function. We’ll proceed assuming that this is the case here, i.e., that the constraint is of the form $g(x,y)=c^2$, with $c>0$.
After differentiating and introducing a Langrange multiplier, you have a system of three equations in three unknowns:$$\begin{equation} 2x = λ (34x+12y) \\ 2y = λ (12x +16y) \\ 17x^2 + 12xy + 8y^2 = c^2 \end{equation}$$ Solve for $\lambda$ in one of them, say, the first, and substitute into the other one:$$ \lambda = \frac x {17x+6y} \\ 2y=\frac{12x^2+16xy}{17x+6y}. $$ Solve this new equation for $y$:$$ 34xy+12y^2=12x^2+16xy \\ y = \frac x2, -2x. $$ Plugging either of these values into $g(x,y)=c^2$ yields $x^2=\frac{c^2}{25}$. So, you have the following two pairs of points as the candidates for extrema: $\left(\frac15c,\frac1{10}c\right)$, $\left(-\frac15c,-\frac1{10}c\right)$ and $\left(\frac15c,-\frac25c\right)$, $\left(-\frac15c,\frac25c\right)$, with distances from the origin of $\frac1{2\sqrt5}c$ and $\frac1{\sqrt5}c$, respectively. I’ll leave verifying that these are minima and maxima to you.
Solving for $\lambda$ in the second equation and substituting into the first yields $x=2y,-\frac y2$, which doesn’t produce any additional solutions to the system.
To check the solution, note that the level curves of $g$ are a family of ellipses centered on the origin, so the given problem is equivalent to determining the semimajor and semiminor axes of these ellipses. As a quadratic form, $g$ is represented by the matrix $$ \pmatrix{17 & 6 \\ 6 & 8} $$ which has eigenvalues $20$ and $5$, with eigenvectors $\langle2,1\rangle$ and $\langle-1,2\rangle$, respectively. The semiminor axis is in the direction of the eigenvector corresponding to the larger eigenvalue, and the semimajor axis is orthogonal to it, in the direction of the other eigenvector. The ratio of the eigenvalues is the square of the ratio of the ellipse’s axis lengths, all of which matches the solution above.
Curiously, if you compute the numerical values of $\lambda$ above, they turn out to be the reciprocals of the eigenvalues.
Best Answer
You can formulate problem (a) as
\begin{align} \operatorname{maximize} & \quad x \\ \text{subject to} & \quad 4x^2 + 7xy + 8y^2 = 60. \end{align} The optimization variables are $x, y \in \mathbb R$.
You can formulate problem (b) as \begin{align} \operatorname{minimize} & \quad y \\ \text{subject to} & \quad 4x^2 + 7xy + 8y^2 = 60. \end{align} The optimization variables are again $x, y \in \mathbb R$.