# Math Help - maxmium values?

1. ## maxmium values?

Find the maximum of f(x,y)=4xy, subject to the condition x2+4y2=1,

ive got the x values as

and

so and make the maximum values, but how do i show that this is the maximum?

2. Originally Posted by Student_richard
make the maximum values, but how do i show that this is the maximum?
Evaluate the functions at those points and see which one is BIGGER.

3. Hello, Student_richard!

Find the maximum of $f(x,y)\:=\:4xy$, subject to the condition $x^2+4y^2\:=\:1$

ive got the x values as: . $x \,=\,\pm\frac{\sqrt{2}}{2}$

. . and y values: . $y \,=\,\pm\frac{\sqrt{2}}{4}$ . . . . Right . . . good work!

So $\left(\frac{\sqrt{2}}{2},\:\frac{\sqrt{2}}{4}\righ t)$ and $\left(-\frac{\sqrt{2}}{2},\:-\frac{\sqrt{2}}{4}\right)$ 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 $x$ and $y$ must have the same sign
. . in order for $f(x,y) \,=\,4xy$ to be a maximum.

That may be sufficient . . . or not.

4. Originally Posted by Student_richard
Find the maximum of f(x,y)=4xy, subject to the condition x2+4y2=1,

ive got the x values as

and

so and make the maximum values, but how do i show that this is the maximum?
Hello,

please check your calculations. I've got the same x-values but the y-value of the maxima is +1:

$M_1\left(\frac{\sqrt{2}}{2} , 1 \right)~\text{ and } ~M_2\left(-\frac{\sqrt{2}}{2} , 1 \right)$

(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.

5. cheers for that, but i still dont get the y=1. are you sure about that and can anyone else verify the answer?

6. Originally Posted by Student_richard
cheers for that, but i still dont get the y=1. are you sure about that and can anyone else verify the answer?
Hello,

your results are: $x=\pm\frac{\sqrt{2}}{2}=\pm \sqrt{\frac{1}{2}}$ and $y=\pm \frac{\sqrt{2}}{4}=\pm\sqrt{\frac{1}{8}}$

Now plug in these values into the equation

$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
$

7. Originally Posted by Student_richard
Find the maximum of f(x,y)=4xy, subject to the condition x2+4y2=1,
Not a Lagrange multiplier problem. The maximum of $f(x,y)$ is also a local maximum of $(f(x,y))^2$, so we want to find the extrema of $4x^2y^2$ subject to $x^2+4y^2=1$. So we want the extrema of $g(y)=(1-4y^2)y^2$, which we find by setting $g'(y)=0$ and solving for $y$.

Doing this we find that $y=\pm \frac{\sqrt{2}}{4}$, and so the corresponding $x$'s are $x=\pm \frac{\sqrt{2}}{2}$ and $x=\pm \frac{\sqrt{2}}{2}$.

(note in doing this I had to assume that $y \ne 0$, so if the maximum turns out to be negative we will need to throw in the possibility that $(1,0) , (-1,0)$ give a maximum for $f(x,y)$ of $0$)

This gives four posible points for the maximum $\left(\frac{\sqrt{2}}{2} ,\frac{\sqrt{2}}{4}\right) , \left(-\frac{\sqrt{2}}{2} ,\frac{\sqrt{2}}{4}\right), \left(\frac{\sqrt{2}}{2} ,-\frac{\sqrt{2}}{4}\right) ,$ $\left(-\frac{\sqrt{2}}{2} ,-\frac{\sqrt{2}}{4}\right)$.

Now we could do some clever things to determine which of these corresponds to the maximum, but the easiest way is just to evaluate $f(x,y)$ at each point and the largest will be the maximum.

RonL

8. so and make the maximum values, but to test whether they are the maximum, i would find the second dereative of f(x,y) = 4xy?

is this the right idea, and how would i find the second derivative?