If $4x^4+9y^4=64$ then what is the maximum value of $x^2+y^2?$ using Lagrange's multiplier

Question

If $4x^4+9y^4=64$ then what is the maximum value of $x^2+y^2?$

I solved this question by Subhasish Basak method and answer is correct.But i tried Lagrange's multiplier and did not get the correct answer.I write my work here.

We have to maximize $x^2+y^2$ subject to $4x^4+9y^4-64=0$ constraint.Let $f(x,y)\equiv x^2+y^2$ and $g(x,y)\equiv 4x^4+9y^4=64 $ and let $h(x)\equiv x^2+y^2+\lambda(4x^4+9y^4-64)$

Now i found $\frac{\partial h}{\partial x},\frac{\partial h}{\partial y},\frac{\partial h}{\partial \lambda}$

$\frac{\partial h}{\partial x}=2x+16\lambda x^3=0$........(1)

$\frac{\partial h}{\partial y}=2y+36\lambda y^3=0$........(2)

$4x^4+9y^4-64=0$..............(3)

From $(1),x=0$ put $x=0$ in the third equation to get $y=\pm\frac{2\sqrt2}{\sqrt3}$,putting this value in the second equation gives $\lambda=\frac{-1}{48}$,but setting these $x=0,y=\pm\frac{2\sqrt2}{\sqrt3},\lambda=\frac{-1}{48}$ gives me $\frac{8}{3}$.Also i found $y=0$ fro second equation and put it in third equation to find $x=\pm2$ and putting this value in the first equation gives me $\lambda=\frac{-1}{32}$ but setting $x=\pm2,y=0,\lambda=\frac{-1}{32}$ gives me 4.

Neither $\frac{8}{3}$ nor $4$ are the correct maximum value $\frac{4\sqrt{13}}{3}$.Where have i gone wrong or Lagrange's multiplier is not applicable here?Please help me.

Answer 1

$(4x^4+9y^4)(1/4+1/9) \geq (\frac{1}{2}2x^2+\frac{1}{3}3y^2)^2=(x^2+y^2)^2$ by Cauchy Schwartz inequality. Now substitute the values to get the answer.

When using lagrange multiplier.

From your first two equation write $x^2$ and $y^2$ as a function of $\lambda$.(ignore those x,y=0 solutions for now).

And substitute in your third equation.

from there you can find the value of $\lambda(=\sqrt{13}/96)$ now get back to your $x^2 and y^2$ put the value of $\lambda$ there you will get the maximum value. and it will be equal to $4\sqrt{13}/3$

Answer 2

Answer:

$$y^2 = \frac{\sqrt{64-4x^4}}{3}$$

$$f(x,y) = x^2+y^2 = x^2+\frac{\sqrt{64-4x^4}}{3}$$

Find $\frac{\delta f}{\delta x}$ and set it equal to 0 and find $x^2$.

Find $y^2$,

Add $x^2+y^2 = 4.807$.

Using the other method, you get the same value.

Good Luck

Answer 3

The constraint is an equality constraint. You would need to rewrite it to = 0 and then use a function which worsens the more far off from 0 it becomes. For instance you could use the function $$-(4x^4+9y^4-64)^2$$The minus sign is because "worsening" the function value of a maximisation is always negative. The square makes sure we will always subtract something if we are off the constraint. So now try instead using the function

$$x^2+y^2 - \lambda(4x^4+9y^4-64)^2$$

Answer 4

(1) and (2) give $x(1+8\lambda x^2)=0$ and $y(1+18\lambda y^2)=0$, hence $4x^2=9y^2$. Now plug that in the constraint.

BTW: nobody cares about $\lambda$'s value.

Answer 5

Both the superellipse $ \ 4x^4 + 9y^4 \ = \ 64 \ $ and the function to be extremized, $ \ x^2 + y^2 \ \ , \ $ have "four-fold" symmetry about the origin, so we expect that extremal values will occur at symmetrically positioned points on the coordinate axes or else at four symmetrically placed points, one in each quadrant. The only problem with the result of your factoring the Lagrange equations, $$ 2x \ · \ (1 \ + \ \lambda·8x^2) \ = \ 0 \ \ \ , \ \ \ 2y \ · \ (1 \ + \ \lambda·18y^2) \ = \ 0 \ \ , \ $$ is one of interpretation.

The Lagrange method locates points on the constaint curve and a "family" of function curves $ \ f(x , y) \ = \ c \ \ $ at which the directions of the normal lines are aligned (either parallel or anti-parallel). Since the function under discussion is the "radius-squared" function, its family of curves is the set of circles centered at the origin. The alignment of normal lines occurs then at tangent points of members of the function curve family with the curve of the constraint function.

You found three "candidate" results:

$$ \mathbf{x \ = \ 0} \ \ \Rightarrow \ \ 9y^4 \ \ = \ \ 64 \ \ \Rightarrow \ \ y \ \ = \ \ \pm \left(\frac{64}{9} \right)^{1/4} \ \ \Rightarrow \ \ f \left( \ 0 \ , \ \left(\frac{64}{9} \right)^{1/4} \ \right) \ \ = \ \ \sqrt{\frac{64}{9}} \ \ = \ \ \frac83 \ \ ; $$

$$ \mathbf{y \ = \ 0} \ \ \Rightarrow \ \ 4x^4 \ \ = \ \ 64 \ \ \Rightarrow \ \ x \ \ = \ \ \pm \left(\frac{64}{4} \right)^{1/4} \ = \ \ 2 \ \ \Rightarrow \ \ f ( \ 2 \ , \ 0 \ ) \ \ = \ \ 2^2 \ \ = \ \ 4 \ \ ; $$

$$ \mathbf{\lambda \ = \ -\frac{1}{8x^2} \ = \ -\frac{1}{18y^2} } \ \ \Rightarrow \ \ y^2 \ \ = \ \ \frac49·x^2 \ \ \Rightarrow \ \ 4·x^4 \ + \ 9·\left( \frac49·x^2 \right)^2 \ \ = \ \ 64 $$ $$ \Rightarrow \ \ \frac{52}{9}·x^4 \ \ = \ \ 64 \ \ \Rightarrow \ \ x^2 \ \ = \ \ \frac{24}{\sqrt{52}} \ = \ \ \frac{12}{\sqrt{13}} \ \ \Rightarrow \ \ y^2 \ \ = \ \ \frac49·\frac{12}{\sqrt{13}} \ \ = \ \ \frac{16}{3·\sqrt{13}} $$ $$ f \left( \ \pm \frac{2 \sqrt3}{13^{1/4}} \ \approx \ \pm 1.824 \ , \ \pm \frac{4}{3^{1/2}·13^{1/4}} \ \approx \ \pm 1.216 \ \right) \ \ = \ \ \frac{52}{3·\sqrt{13}} \ \ = \ \ \frac{4·\sqrt{13}}{3} \ \ $$ [note that these four points line on the lines $ \ y^2 \ \ = \ \ \frac49·x^2 \ \Rightarrow \ y \ = \ \pm \ \frac23·x \ \ ] \ . $

The minimum value of $ \ x^2 + y^2 \ $ that you obtained corresponds to a circle [marked in red in the graph below] of radius $ \ \sqrt{\frac83} \ $ which is inscribed in the superellipse [blue curve] and is tangent to the points on its "minor axis" on the $ \ y-$axis. The intermediate value corresponds to a circle [in orange] of radius $ \ 2 \ $ which is tangent to the superellipse on its "major axis". Finally, the maximal value of $ \ \frac{4·\sqrt{13}}{3} \ $ corresponds to the circle [in violet] which circumscribes the superellipse, resulting in four tangent points on the lines $ \ y \ = \ \pm \ \frac23 x \ \ . $

So in terms of the associated geometry, there are three possible situations for tangency and the Lagrange method does find all of them. At times, considering what the method determines geometrically can be helpful in understanding the numerical results for the function extrema.