I have been given a function $f(x,y)= x^2 + y^2$ whose maximum and minimum values have been sought (if existent) under the constraint that $3x^2 + 4xy +16y^2=140$.
This looks pretty much to be a problem, wherein the technique of Lagrange Multipliers can be exploited. So, I generate the Lagrangian and set the partial derivatives to 0; which gives the following set of equations:-
(i) $$2(1+3λ)x + 4λy = 0$$
(ii) $$4λx + 2(1+16λ)y = 0$$
(iii) The constraint itself; $$3x^2 + 4xy +16y^2=140$$.
But, solving this system is proving to be quite quite tedious.
Solving (i) and (ii); I retrieve $44λ^2+19λ+1=0$ whose solutions are not much-elegant (I know it's subjective!)
I then proceed to write $x$ in terms of $λ$ and $y$ from (i) and thereafter, substitute a value of $λ$ from above, getting $y$in terms of $x$.
Now, I substitute the above value of $y$ into (iii) which generates an equation in $x$; pending which the value of $x$ can be ascertained. Thereafter; the value of $y$ can be ascertained, too. And, then the substitution of these $x$ and $y$ into the parent equation of $f(x,y)= x^2 + y^2$ to get the desired result.
Now, I have to repeat the above two steps for the other value of $λ$, which just adds to the complexity.
So, my questions are:-
1) Does there exist any more efficient method for solving the system? I need to do this stuff at exam-hall where average time is around 8 mins and i assume that there might be a compulsion to use Lagrange Multipliers.
2) Can this sum be done in a more easy manner than using the concept of Lagrange multipliers; which may be employed in absence of the above compulsion?
Best Answer
Here's one way: Substitute $x=r\cos \theta$ and $y=r\sin\theta$. Then, the constraint becomes $$ 3\cos^2\theta +4\sin\theta\cos\theta +16\sin^2\theta = \frac{140}{r^2}. $$Using the double angle formula and $\sin^2 \theta+\cos^2 \theta= 1$, this is equivalent to $$ 2\sin2\theta -\frac{13}{2}\cos2\theta = \frac{140}{r^2}-\frac{19}{2}. $$ The LHS takes values in $[-\sqrt{2^2+(\frac{13}{2})^2},\sqrt{2^2+(\frac{13}{2})^2}]$ so the minimum and maximum of $r^2$ can be obtained by solving $$ \frac{140}{r^2}-\frac{19}{2}=\pm\frac{\sqrt{185}}{2}. $$ This gives $$ r^2=\frac{35(19\pm\sqrt{185})}{22}. $$