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The maximum value of the function $f(x, y) = xy$, and subject to condition $x^2+y^2=1$:

So do I apply Lagrange's Multiplier method to find the maximum value? I tried to find the numbers just by applying few numbers but it doesn't seem to work.

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2 Answers

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how about this inequality : $ x^2+y^2\geq 2xy$ it would simplify things.

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If you wanted to use Lagrange Multipliers, you could do the following:

$$f(x,y) = xy\\ g(x,y) = x^2+y^2 = 1$$

So

$$\nabla f(x,y) = < f_x,f_y>=<y,x>, \,\mathrm{and}\, \nabla g(x,y) = <g_x,g_y> = <2x,2y>$$

Now for the method of Lagrange Multipliers, we must find all values of $x$, $y$, $\lambda$ such that

$$\nabla f(x,y) = \lambda \nabla g(x,y)\\ x^2 + y^2 = 1$$

So $$f_x = \lambda g_x \,\,\mathrm{and}\,\, f_y=\lambda g_y$$

Thus we must solve $$ y=2\lambda x\\ x=2\lambda y \\ x^2+y^2=1 $$

If we rearrange the second equation for $y$, we get

$$ y = \frac{x}{2\lambda} $$

So

$$ 2 \lambda x = \frac{x}{2\lambda} \\ 4\lambda^2x=x\\ 4\lambda^2x-x=0\\ x(4\lambda^2-1)=0\\ $$

So $x=0$, $\lambda=\frac{1}{2}$

If $x=0$,

$y=2\lambda (0) = 0$

BUT if $x=y=0$, then the condition $x^2+y^2=1$ does not hold. So $x$,$y\neq 0$

If $\lambda = \frac{1}{2}$, $y=x$

So $$x^2+x^2 = 1\\ 2x^2=1\\ x^2=\frac{1}{2}\\ x=\pm \frac{1}{\sqrt{2}}\\ $$

Now $y=x$, so $y=\frac{1}{\sqrt{2}}$ when $x=\frac{1}{\sqrt{2}}$ and $y=-\frac{1}{\sqrt{2}}$ when $x=-\frac{1}{\sqrt{2}}$

Thus $$f(x,y) = xy = \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)\\ f(x,y) = \frac{1}{2}$$

That was a lot of work just to get the same answer as @Amire...

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