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.
On
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 $x, y \in \mathbb R$.
On
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 $$\frac{79 }{32}x^2=60\implies x=8 \sqrt{\frac{30}{79}}$$
On
This looked like a good problem to illustrate how the method from first-semester calculus is related to the use of Lagrange multipliers for answering these questions. The curve $ \ 4x^2 \ + \ 7xy \ + \ 8y^2 \ = \ 60 \ $ is a rotated ellipse centered on the origin, so it has symmetry about the origin. This means that the point $ \ (X \ , \ Y) \ $ on the ellipse for which the $ \ x-$ coordinate is largest is "mirrored" in the opposite quadrant by the point $ \ (-X \ , \ -Y) \ , $ which is the point with the smallest $ \ x \ \ ; $ a similar statement can be made for the points with the largest and smallest $ \ y-$ values.
Determining these points is equivalent to finding the tangent points on the vertical and horizontal tangent lines to the ellipse (as mentioned by RyRy the Fly Guy), a problem commonly set in single-variable calculus. We apply implicit differentiation to the curve equation to produce
$$ \frac{d}{dy}[ \ 4x^2 \ + \ 7xy \ + \ 8y^2 \ ] \ = \ \frac{d}{dy}[60] \ \ \Rightarrow \ \ 8x \ + \ 7y \ + \ 7x·\frac{dy}{dx} \ + \ 16y·\frac{dy}{dx} \ \ = \ \ 0 $$ $$ \Rightarrow \ \ \frac{dy}{dx} \ \ = \ \ -\frac{8x \ + \ 7y}{7x \ + \ 16y} \ \ . $$
The expressions in the numerator and denominator of this implicit derivative are the same as those which appear in the calculation Claude Leibovici presents.
The horizontal tangent lines are those for which this expression is zero, and so for which the numerator is zero (since the denominator is not also zero for those coordinates). We learn to insert $ \ 8x \ + \ 7y \ = \ 0 \ \ \Rightarrow \ \ y \ = \ -\frac87 \ x \ $ into the curve equation, which gives us the tangent points $ \left( \pm \ 7\sqrt{\frac{15}{79}} \ , \ \mp \ 8\sqrt{\frac{15}{79}} \right) \ \approx \ (\pm \ 3.05 \ , \ \mp \ 3.49) \ \ . $ This line through the origin (in orange) and the horizontal tangent points are seen in the graph below. [In the Lagrangian analysis, this will come from $$ F \ = \ \mathbf{y} \ + \ \lambda(4x^2+7xy+8y^2-60) \ \ \Rightarrow \ \ \frac{\partial F}{\partial x} \ = \ \lambda (8 x+7 y) \ = \ 0 \ \ . ] $$
By the same token, we can differentiate the curve equation implicitly with respect to $ \ x \ $ (or simply use the relation $ \frac{dx}{dy} \ \ = \frac{1}{\frac{dy}{dx}} \ $ ) to obtain $ \ \frac{dx}{dy} \ = \ -\frac{7x \ + \ 16y}{8x \ + \ 7y} \ . $ Setting this ratio equal to zero permits us to locate the vertical tangent points from the condition $ \ 7x \ + \ 16y \ = \ 0 \ \ \Rightarrow \ \ y \ = \ -\frac{7}{16} \ x \ $ (the line in violet in the graph). The vertical tangent points are then $ \left( \pm \ 8\sqrt{\frac{30}{79}} \ , \ \mp \ \frac72\sqrt{\frac{30}{79}} \right) \ \approx \ (\pm \ 4.93 \ , \ \mp 2.16) \ \ . $
The horizontal and vertical tangent lines we have found then form a box which circumscribes the given rotated ellipse.
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.