Find maximum value under constraints by Lagrange multipliers or another method

Question

Let $f(x,y)=(x + 2y)^2 + (3x + 4y)^2.$

Then for $(x,y)$ on the circle $x^2+y^2=1 $ what is the maximum value of $f$?

I tried to use Lagrange multiplier method with the constraint $g(x,y)= x^2 + y^2 =1.$

However, now I realize that this way is so complicated on this problem ($\lambda = 15 \pm \sqrt{221}$)

How can I solve this problem?

Answer 1

This is simple enough that you can use your constraint as a substitution, i.e., $y = \sqrt{1 - x^2}$.

Then your function is:

$$f(x) = \left(2 \sqrt{1-x^2}+x\right)^2+\left(4 \sqrt{1-x^2}+3 x\right)^2$$

Its derivative is:

$${d f(x) \over dx} = -\frac{4 \left(14 x^2+5 \sqrt{1-x^2} x-7\right)}{\sqrt{1-x^2}}$$

Set it to zero and find:

$$x = \left\{-\sqrt{\frac{1}{2}+\frac{5}{2 \sqrt{221}}},\sqrt{\frac{1}{442} \left(221-5 \sqrt{221}\right)}\right\} \approx \{ -0.817416, 0.576048 \}$$

where clearly one is a minimum, the other a maximum.

For "culture," here is the figure in three dimensions:

Answer 2

If Lagrange's is not mandatory, WLOG choose $x=\cos t, y=\sin t$ to find

$$f(x,y)=A\cos^2t+B\sin t\cos t+C\sin^2t=g(t)\text{(say)}$$

Use $\cos2t=1-2\sin^2t=2\cos^2t-1,\sin2t=2\sin t\cos t$ to find $g(t)$ to be of the form $$p\cos2t+q\sin2t+r$$

Now use $-\sqrt{p^2+q^2}\le p\cos2t+q\sin2t\le\sqrt{p^2+q^2}$

Answer 3

Exact solution with Lagrange multiplier method is: $$f_{min}=15-\sqrt{221}\approx0.134$$ if $$x=\frac{\sqrt{5 \sqrt{221}+221}}{\sqrt{442}}\approx0.817,\; y=-\frac{\sqrt{221-5 \sqrt{221}}}{\sqrt{442}}\approx-0.576$$ or $$x=-\frac{\sqrt{5 \sqrt{221}+221}}{\sqrt{442}}\approx-0.817,\; y=\frac{\sqrt{221-5 \sqrt{221}}}{\sqrt{442}}\approx0.576$$

Answer 4

It is possible to find the solution by using the polar coordinates:

$$f(\theta) = 10\cos^2\theta + 20\sin^2\theta + 28\sin\theta\cos\theta = 10 + 10\sin^2\theta + 14\sin2\theta$$

The extrema are found by deriving:

$$f'(\theta) = 20\sin\theta \cos\theta + 28\cos2\theta = 10\sin2\theta + 28\cos2\theta = 0$$

And we get $$\theta = -\frac{1}{2} \arctan\frac{14}{5} + k\frac{\pi}{2} $$ where $k$ is any integer between $0$ and $3$

We can check numerically that the maximum corresponds to $$\theta = -\frac{1}{2} \arctan\frac{14}{5} + k\pi \approx -0.6138 + k\pi$$

where $k$ is equal to $0$ or $1$

This corresponds to $x \approx 0.81754, \,y \approx -0.576$ and $x \approx -0.81754, \,y \approx 0.576$

Answer 5

The maximum value of a quadratic form on the unit circle is equal to its largest eigenvalue. $f$ simplifies to $10x^2+28xy+20y^2$ and so the eigenvalues of $f$ are the roots of $\lambda^2-30\lambda+4$.