Lagrange Multipliers $f(x,y)=10y^2-4x^2$

Question

Find the min and max values of the function $$f(x,y)=10y^2-4x^2$$ with the constraint $$g(x,y)=x^4+y^4=1$$ I have done the following working; $$\frac{\partial f}{\partial x} = \lambda \frac{\partial g}{\partial x} \\\frac{\partial f}{\partial y} = \lambda \frac{\partial g}{\partial y}$$$$-8x= \lambda 4x^3\\20y= \lambda 4y^3$$$$x(2- \lambda x^3)=0\\y(5- \lambda y^3)=0$$$$\lambda x^3=2\\\lambda y^3=5$$ My question is how can I find the value of lambda for the constraint to hold.

Answer 1

Your system of equations is$$\left\{\begin{array}{l}-8x=4\lambda x^3\\20y=4\lambda y^3\\x^4+y^4=1.\end{array}\right.$$If $x=0$, then you can obviously take $y=\pm1$ and if $y=0$, you can obviously take $x=\pm1$; these are the only solutions in which one of the numbers $x$ or $y$ is $0$.

And there are no more solutions. In fact, if $x,y\neq0$, then it follows from the first two equations that $x^2=-\frac2\lambda$ and that $y^2=\frac5\lambda$. This is impossible, of course: if $\lambda>0$ there is no such $x$, and if $\lambda<0$ there is no such $y$.

Answer 2

There are errors in both of your equations. $$x(2+\lambda x^2)=0$$ $$y(5-\lambda y^2)=0$$ Now, if $\lambda\ge 0$, from the first equation we get $x=0$. If $\lambda\le 0$, from the second equation we get, $y=0$. Hence, at optimum points, atleast one of the coordinate is $0$. So, required optimum points are $(\pm1,0)$ and $(0,\pm1)$ on putting in the constraint.

Answer 3

$$-4=-4\sqrt{x^4+y^4}\leq10y^2-4x^2\leq10\sqrt{x^4+y^4}=10.$$ The equality occurs for $(x,y)=(1,0)$ for the left inequality and for $(x,y)=(1,0)$ for the right inequality, which says that we got the minimal value and the maximal value.

Answer 4

The solutions by Lagrange Multipliers are shown above. But if you like a traditional method, here is one. One way is to lower the "power" of the constraint, namely: $x^4+y^4 =1$. So let's put $a = x^2, b = y^2$, thus $a, b \ge 0$ and $a^2+b^2=1$.Thus we have: $f(a,b) = 10b-4a$ with $a^2+b^2=1$ and $a,b \ge 0$.From this we can use a trig substitution: $a = \cos \theta, b = \sin \theta, \theta\in [0,\frac{\pi}{2}]$, and we have: $f(\theta) = 10\sin \theta - 4\cos \theta, \theta \in [0,\frac{\pi}{2}]$. Taking derivative of $f$ we have: $f'(\theta) = 10\cos \theta + 4\sin \theta > 0 \implies f_{\text{min}} = f(0) = -4, f_{\text{max}} = f(\frac{\pi}{2}) = 10$ as claimed.

Answer 5

A different approach without using Lagrange multipliers:

Let $x^2=\cos\theta$ and $y^2=\sin\theta$ where $\displaystyle 0\le\theta\le\frac{\pi}{2}$.

$f(\theta)=10\sin\theta-4\cos\theta$

$f'(\theta)=10\cos\theta+4\sin\theta$

$f'(\theta)=0$

$\implies 10\cos\theta+4\sin\theta=0$

We note that this is impossible for any value of $\theta$ in our domain. So, the maximum and minimum occur at the end-points.

$f(0)=-4$

$\displaystyle f\left(\frac{\pi}{2}\right)=10$

So, the minimum value is $-4$ and the maximum value is $10$.