Hello, Student_richard!
Find the maximum of $\displaystyle f(x,y)\:=\:4xy$, subject to the condition $\displaystyle x^2+4y^2\:=\:1$
ive got the x values as: .$\displaystyle x \,=\,\pm\frac{\sqrt{2}}{2}$
. . and y values: .$\displaystyle y \,=\,\pm\frac{\sqrt{2}}{4}$ . . . . Right . . . good work!
So $\displaystyle \left(\frac{\sqrt{2}}{2},\:\frac{\sqrt{2}}{4}\righ t)$ and $\displaystyle \left(-\frac{\sqrt{2}}{2},\:-\frac{\sqrt{2}}{4}\right)$ are the maximums
but how do i show that they are maximums?
There must be a test for this (even with Lagrange multipliers),
. . but I'm not familiar with it.
But it is clear that $\displaystyle x$ and $\displaystyle y$ must have the same sign
. . in order for $\displaystyle f(x,y) \,=\,4xy$ to be a maximum.
That may be sufficient . . . or not.
Hello,
please check your calculations. I've got the same x-values but the y-value of the maxima is +1:
$\displaystyle M_1\left(\frac{\sqrt{2}}{2} , 1 \right)~\text{ and } ~M_2\left(-\frac{\sqrt{2}}{2} , 1 \right)$
(You only considered the condition without using the equation of the function)
To show that a certain value is indeed a maximum you have to calculate the 2nd derivative. Plug in the x-value. If f''(x) < 0 then you have a maximum.
Hello,
your results are: $\displaystyle x=\pm\frac{\sqrt{2}}{2}=\pm \sqrt{\frac{1}{2}}$ and $\displaystyle y=\pm \frac{\sqrt{2}}{4}=\pm\sqrt{\frac{1}{8}}$
Now plug in these values into the equation
$\displaystyle f(x,y) = 4xy \Longrightarrow~ f\left(\sqrt{\frac{1}{2}}, \sqrt{\frac{1}{8}} \right) = 4 \cdot \sqrt{\frac{1}{2}} \cdot \sqrt{\frac{1}{8}} = 4 \cdot \sqrt{\frac{1}{16}} =1
$
Not a Lagrange multiplier problem. The maximum of $\displaystyle f(x,y)$ is also a local maximum of $\displaystyle (f(x,y))^2$, so we want to find the extrema of $\displaystyle 4x^2y^2$ subject to $\displaystyle x^2+4y^2=1$. So we want the extrema of $\displaystyle g(y)=(1-4y^2)y^2$, which we find by setting $\displaystyle g'(y)=0$ and solving for $\displaystyle y$.
Doing this we find that $\displaystyle y=\pm \frac{\sqrt{2}}{4} $, and so the corresponding $\displaystyle x$'s are $\displaystyle x=\pm \frac{\sqrt{2}}{2} $ and $\displaystyle x=\pm \frac{\sqrt{2}}{2} $.
(note in doing this I had to assume that $\displaystyle y \ne 0$, so if the maximum turns out to be negative we will need to throw in the possibility that $\displaystyle (1,0) , (-1,0)$ give a maximum for $\displaystyle f(x,y)$ of $\displaystyle 0$)
This gives four posible points for the maximum $\displaystyle \left(\frac{\sqrt{2}}{2} ,\frac{\sqrt{2}}{4}\right) , \left(-\frac{\sqrt{2}}{2} ,\frac{\sqrt{2}}{4}\right), \left(\frac{\sqrt{2}}{2} ,-\frac{\sqrt{2}}{4}\right) ,$ $\displaystyle \left(-\frac{\sqrt{2}}{2} ,-\frac{\sqrt{2}}{4}\right)$.
Now we could do some clever things to determine which of these corresponds to the maximum, but the easiest way is just to evaluate $\displaystyle f(x,y)$ at each point and the largest will be the maximum.
RonL