Find the max and min value from equation?

[Helly123]

Homework Statement

Homework Equations

f ' (x) = 0 --> to find extreme points

The Attempt at a Solution

$2x^2 + y^2 = 4$
$y = \sqrt{4 - 2x^2}$
$y ' = \frac {-2x} {\sqrt{4-2x^2}}$
$0 = \frac {-2x} {\sqrt{4-2x^2}}$
x = 0

also $4 - 2x^2 >= 0$
$4 >= 2x^2$
$\sqrt{2} >= x$
$x <= +/- \sqrt{2}$

$x= -\sqrt{2} , x = 0, x = \sqrt{2}$
$f( -\sqrt{2} ) = 0$ --> $4x + y^2 = -4\sqrt{2}$ --> min
$f( \sqrt{2} ) = 0$ --> $4x + y^2 = 4\sqrt{2}$ ---> max
$f(0) = 2$ --> $4x + y^2 = 4$

is there something wrong?

 

[BvU]

is there something wrong?

I suppose so.

f ' (x) = 0 --> to find extreme points

You manipulate things, but where does this $f(x)$ appear ?
Setting $y ' = \frac {-2x} {\sqrt{4-2x^2}}=0$ finds you extrema for y(x) at x = 0 (you recognize this is an ellipse ?) but that is not what you want!

 

[Mark44]

Homework Statement

View attachment 206531

Homework Equations

f ' (x) = 0 --> to find extreme points

The Attempt at a Solution

$2x^2 + y^2 = 4$
$y = \sqrt{4 - 2x^2}$
$y ' = \frac {-2x} {\sqrt{4-2x^2}}$
$0 = \frac {-2x} {\sqrt{4-2x^2}}$
x = 0

You aren't using the fact that the points have to lie on the given ellipse.
You don't need calculus to solve this problem. In the equation of the ellipse, solve for y, and then substitute that into the expression whose max and min values you want to find. I get a maximum value of 6.

also $4 - 2x^2 >= 0$
$4 >= 2x^2$
$\sqrt{2} >= x$
$x <= +/- \sqrt{2}$

I get what you're trying to say, but this is not how to say it. It should be
$-\sqrt 2 \le x \le \sqrt 2$

$x= -\sqrt{2} , x = 0, x = \sqrt{2}$
$f( -\sqrt{2} ) = 0$ --> $4x + y^2 = -4\sqrt{2}$ --> min
$f( \sqrt{2} ) = 0$ --> $4x + y^2 = 4\sqrt{2}$ ---> max
$f(0) = 2$ --> $4x + y^2 = 4$

is there something wrong?

 

[BvU]

I get a maximum value of 6.

So did I but I didn't want to spoil the fun.
And with calculus you can find maxima, minima and flexing points (is that the proper designation?), but on $[0,1]$ the function $f(x) = x$ has a lowest value of 0 and a maximum value of 1 and neither can be found with calculus !

 

[Mark44]

So did I but I didn't want to spoil the fun.
And with calculus you can find maxima, minima and flexing points (is that the proper designation?)

Usual term is "inflection point".

, but on $[0,1]$ the function $f(x) = x$ has a lowest value of 0 and a maximum value of 1 and neither can be found with calculus !

However, noting that the derivative is constant and positive, one can conclude that the minimum occurs at the left endpoint of the interval, and the maximum occurs at the right endpoint of the same interval.

 

[BvU]

Usual term is "inflection point".

Thanks !
Back to Helly: do you now see what needs to be done (calculus or not) ?

 

[Helly123]

$2x^2 + y^2 = 4$
$\frac{x^2}{2} + \frac{y^2}{4} = 1$ -> ellipse equation

y = 0
$x = \sqrt{2} \ or -\sqrt{2}$

x = 0
$y = 2 \ or -2$

x = 0
$y^2 = 4$
$4x + y^2 = 4$

y = 0
$x = 4\sqrt{2} \ or -4\sqrt{2}$
$4x + y^2 = 4\sqrt{2} \ or -4\sqrt{2}$

What do you mean solve for y?

 

[ehild]

$2x^2 + y^2 = 4$
$\frac{x^2}{2} + \frac{y^2}{4} = 1$ -> ellipse equation
What do you mean solve for y?

Express y as function of x.

 

[Helly123]

You aren't using the fact that the points have to lie on the given ellipse.
You don't need calculus to solve this problem. In the equation of the ellipse, solve for y, and then substitute that into the expression whose max and min values you want to find.

$y = \sqrt 4 -2x^2$ this is y as function of x
how to solve it?
$0 = \sqrt 4 -2x^2$
$4 - 2x^2 = 0$
x = -2 or 2

what am I supposed to do? using calculus?

Thanks !
Back to Helly: do you now see what needs to be done (calculus or not) ?

no...

 

[BvU]

$y = \sqrt 4 -2x^2$ ? You sure or is this a $\TeX$typo and you forgot the curly brackets { } around the argument for $\sqrt {}$ ?

this is y as function of x

Good. Now you can substitute y in the function for which you want to find the maximum and the minimum value

how to solve it?

You can not solve one equation for two unknowns -- 'solve' in the sense I seem to have let you think. The sense I meant it is as Ersbeth posted (you can accept my apologies, but I think it's better if you get used to the slight ambiguity )

To help speed up things somewhat: now that you have $y^2$ as a function of $x$, you can investigate the function $4x+y^2$ on the domain you found for $x$ -- using calculus or common sense PS:

x = -2 or 2

tut tut ! What did you find in post 7 ? (the first time y=0, that is)

 

[Helly123]

$2x^2 + y^2 = 4$
max and min of $4x + y^2?$
Answer
$y = \sqrt{4 - 2x^2}$
domain is
$4 - 2x^2 >= 0$
so we get $-\sqrt{2} < = x < = \sqrt{2}$

now, question is $4x + y^2$ max and min?
from equation $2x^2 + y^2 = 4$ we get $y^2 = 4 - 2x^2$
substitute $y^2 = 4 - 2x^2$ to $4x + y^2$
$4x + y^2 = 4x + 4 - 2x^2$
$f(x) = 4x + 4 - 2x^2$
$f ' (x) = 4 - 4x$
0 = 4 - 4x
x = 1 -> satisy domain

check x = 1, and x in domain for $4x + y^2 = 4x + 4 - 2x^2$
x = 1
$4x + y^2 = 4 + 4 - 2 = 6$

x = $-\sqrt{2}$
$4x + y^2 = -4\sqrt{2}$

x = $\sqrt{2}$
$4x + y^2 = 4\sqrt{2}$

max = 6
min = $-4\sqrt{2}$

is the concept right? I'm lack at concept... I manipulate formula to get answer I want.

 

[Helly123]

To help speed up things somewhat: now that you have $y^2$ as a function of $x$, you can investigate the function $4x+y^2$ on the domain you found for $x$ -- using calculus or common sense

oh that's substitute the y^2 to $4x+y^2$ right

 

[BvU]

Yes. Well done.

One pedantic remark: you used calculus to find the maximum. As Mark said in #3, you indeed don't really need it. Can you see how ?

I manipulate formula to get answer I want

that's what we all do !

 

[Helly123]

Yes. Well done.

One pedantic remark: you used calculus to find the maximum. As Mark said in #3, you indeed don't really need it. Can you see how ?

that's what we all do !

hmm.. I get it.
$4x + 4 -2x^2$
change it to another form
$4x + 4 -2x^2 = -2(x^2 - 2x) + 4$
$= -2(x^2 - 2x + 1) + 2 + 4$
$= -2(x-1)^2 + 6$
so $4x + y^2 = -2(x-1)^2 + 6$
max at 6, when x = 1