# Using Lagrange multipliers without a given constraint? 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.

The function $$4x^2+7xy+8y^2=60$$ is in fact your constraint. The function you're maximizing is $$f(x,y)=x$$, and the function you're minimizing is $$g(x,y)=y$$.
Note that $$f(x,y)=x$$ is a vertical line somewhere in the $$xy$$-plane. So essentially what you're doing is finding the equation for a vertical line that is "furthest to the right" while intersecting the ellipse. The analogous is true for $$g(x,y)=y$$; you're finding the equation for a horizontal line that is "as low as possible" while intersecting the ellipse.
\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 $$x, y \in \mathbb R$$.
For the first case, consider $$F=x+\lambda(4x^2+7xy+8y^2-60)$$ $$\frac{\partial F}{\partial x}=1+\lambda (8 x+7 y)\qquad \frac{\partial F}{\partial y}=\lambda (7 x+16 y)\qquad \frac{\partial F}{\partial \lambda}=4x^2+7xy+8y^2-60$$ From $$\frac{\partial F}{\partial y}=0$$ we have $$y=-\frac 7 {16}x$$. Plug it in $$\frac{\partial F}{\partial \lambda}$$ to get $$\frac{79 }{32}x^2=60\implies x=8 \sqrt{\frac{30}{79}}$$