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Quadratic equation to find max and min

Homework Statement
##2x^2 + y^2 = 4##
Max and min of
##4x + y^2## ?
Homework Equations
F'(x) = 0
To find critical point
Problem Statement: ##2x^2 + y^2 = 4##
Max and min of
##4x + y^2## ?
Relevant Equations: F'(x) = 0
To find critical point

X = 0 -> y = +- 2
Y =0 -> x = +- ##\sqrt 2##

I input that
Find f(xy) = 4 and f(xy) = +-##4\sqrt 2##

To find with derivative i find both x and y are zero
The answer is wrong
 
Last edited by a moderator:
32,575
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Problem Statement: ##2x^2 + y^2 = 4##
Max and min of
##4x + y^2## ?
Relevant Equations: F'(x) = 0
To find critical point

X = 0 -> y = +- 2
Y =0 -> x = +- ##\sqrt 2##

I input that
Find f(xy) = 4 and f(xy) = +-##4\sqrt 2##

To find with derivative i find both x and y are zero
The answer is wrong
There are a couple of ways to go for this problem. The goal is to find the max. and min. values of ##f(x, y) = 4x + y^2## subject to the constraint ##2x^2 + y^2 = 4##. The constraint means that you're looking for the largest/smallest values of ##f(x, y) = 4x + y^2## for points (x, y) that lie on the ellipse ##2x^2 + 4y^2 = 4##. It might be helpful to sketch the ellipse. Doing so might give you some insight into which points on the ellipse maximize or minimize the function you're working with.

One way is to solve for y in the constraint equation, and then substitute for y in f(x, y). You should then be able to use differentiation to find the max and min values.

Another way is to use Lagrange multipliers, if you know that technique.
 
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Also, you have attempted to use LaTeX markup, but what you did doesn't work on this site. Instead of $\sqrt2$, write this as ##\sqrt 2##, which renders as ##\sqrt 2##. And similar for exponents.
 
There are a couple of ways to go for this problem. The goal is to find the max. and min. values of ##f(x, y) = 4x + y^2## subject to the constraint ##2x^2 + y^2 = 4##. The constraint means that you're looking for the largest/smallest values of ##f(x, y) = 4x + y^2## for points (x, y) that lie on the ellipse ##2x^2 + 4y^2 = 4##. It might be helpful to sketch the ellipse. Doing so might give you some insight into which points on the ellipse maximize or minimize the function you're working with.

One way is to solve for y in the constraint equation, and then substitute for y in f(x, y). You should then be able to use differentiation to find the max and min values.

Another way is to use Lagrange multipliers, if you know that technique.
With Lagrange multipliers i only get one critical point. How to get the other critical point?
 
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Your first attempt involved differentiation. Extreme points can occur at points where the derivative is zero, at points in the domain where the derivative is undefined, or at endpoints of the domain. Because the constraint equation represents an ellipse, there is a domain for the x values to consider.
 
There are a couple of ways to go for this problem. The goal is to find the max. and min. values of ##f(x, y) = 4x + y^2## subject to the constraint ##2x^2 + y^2 = 4##. The constraint means that you're looking for the largest/smallest values of ##f(x, y) = 4x + y^2## for points (x, y) that lie on the ellipse ##2x^2 + 4y^2 = 4##. It might be helpful to sketch the ellipse. Doing so might give you some insight into which points on the ellipse maximize or minimize the function you're working with.

One way is to solve for y in the constraint equation, and then substitute for y in f(x, y). You should then be able to use differentiation to find the max and min values.

Another way is to use Lagrange multipliers, if you know that technique.
what "to solve for y in the constraint" means?
Why Lagrange Method only gets me 1 critical point?
 
It means to isolate y in the equation ##2x^2 + 4y^2 = 4##.
It gives you one value of x, but that corresponds to two y values.
I solve for y in the constraint and substitute it to y at f(xy). Gets me x = 4. But how to find y? I get y = undefined

About Lagrange Method, i get 2 y that is ##\sqrt{1/2}## and ##\sqrt{6/4}## but the answer it's wrong..
 
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I solve for y in the constraint and substitute it to y at f(xy). Gets me x = 4. But how to find y? I get y = undefined
x = 4 is obviously wrong. Any point (x, y) has to lie on the ellipse ##2x^2 + 4y^2 = 4##.

Please show what you did to solve for y in this equation.
Lifeforbetter said:
About Lagrange Method, i get 2 y that is ##\sqrt{1/2}## and ##\sqrt{6/4}## but the answer it's wrong..
That's wrong as well. For an example of how to use the technique of Lagrange Multipliers, see https://en.wikipedia.org/wiki/Lagrange_multiplier, examples 1a, 1b, and 2.
 
x = 4 is obviously wrong. Any point (x, y) has to lie on the ellipse ##2x^2 + 4y^2 = 4##.

Please show what you did to solve for y in this equation.
That's wrong as well. For an example of how to use the technique of Lagrange Multipliers, see https://en.wikipedia.org/wiki/Lagrange_multiplier, examples 1a, 1b, and 2.
The domain = {+-##\sqrt2## and +- 2}
When x = +-2 y is undefined
How is the ellipse looks like then?
 
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The domain = {+-##\sqrt2## and +- 2}
I think I know what you're trying to say, but this is not a good way to say it.
A better way would be to state the intervals along the x- and y-axes for the equation.

As you wrote above, the domain consists of four numbers, which is not correct.
Lifeforbetter said:
When x = +-2 y is undefined
How is the ellipse looks like then?
Well, do you know how to sketch the graph of an ellipse? In standard form, the equation is ##\frac {x^2}2 + \frac {y^2} 4 = 1##. The standard form provides information about the major and minor semiaxes of the ellipse.

Do you know how to isolate y in the equation ##2x^2 + y^2 = 4##?
 
I think I know what you're trying to say, but this is not a good way to say it.
A better way would be to state the intervals along the x- and y-axes for the equation.

As you wrote above, the domain consists of four numbers, which is not correct.
Well, do you know how to sketch the graph of an ellipse? In standard form, the equation is ##\frac {x^2}2 + \frac {y^2} 4 = 1##. The standard form provides information about the major and minor semiaxes of the ellipse.

Do you know how to isolate y in the equation ##2x^2 + y^2 = 4##?
To isolates y. I think i know. I get x = 1 y = ##\sqrt2##
The domain is the major and minor semiaxes right?
I realized that
Major x axis = +-##\sqrt2##
Minor y axis = +-2
Right?
So the critical point is
(0, -2) (0,2) (##-\sqrt2##,0)(##\sqrt2##,0)
So Lagrange Method is to find major x axis only? I can't find minor axis?
 
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To isolates y. I think i know. I get x = 1 y = ##\sqrt2##
No!
All you have done here is to find one point on the graph of the ellipse. To "solve an equation for y," you need to rewrite the equation so that y is on one side of the equals sign, and everything else is on the other side.

For example, solving for y in the equation 2x - y = 5 results in a new equation y = 2x - 5. Both equations are equivalent, meaning that both describe the same set of points.
Lifeforbetter said:
The domain is the major and minor semiaxes right?
No!
Considering the equation as representing a function (which in this case it is not one), the domain is the interval that contains all of the possible x-values.
Lifeforbetter said:
I realized that
Major x axis = +-##\sqrt2##
Minor y axis = +-2
Right?
No!
The major axis is 4, and runs along the y-axis. The minor axis is ##2\sqrt 2## and runs along the x-axis.
Lifeforbetter said:
So the critical point is
(0, -2) (0,2) (##-\sqrt2##,0)(##\sqrt2##,0)
No!
These are the vertices of the ellipse. They are useful in graphing the ellipse, but they are not critical points, and really have nothing to do with finding the max or min values of the function ##f(x, y) = 4x + y^2##.
Lifeforbetter said:
So Lagrange Method is to find major x axis only? I can't find minor axis?
Did you look at the link I wrote in post #9? In it there are several examples of how to use Lagrange multipliers, but you don't need to use that technique in this problem. I mentioned another way back in post #2.
 

pasmith

Homework Helper
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One can set [itex]x = \sqrt{2} \cos \theta[/itex], [itex]y = \sin \theta[/itex] for [itex]\theta \in [0, 2\pi][/itex] and the constraint [itex]2x^2 + 4y^2 = 4[/itex] is then satisfied automatically.
 
No!
All you have done here is to find one point on the graph of the ellipse. To "solve an equation for y," you need to rewrite the equation so that y is on one side of the equals sign, and everything else is on the other side.

For example, solving for y in the equation 2x - y = 5 results in a new equation y = 2x - 5. Both equations are equivalent, meaning that both describe the same set of points.
No!
Considering the equation as representing a function (which in this case it is not one), the domain is the interval that contains all of the possible x-values.
No!
The major axis is 4, and runs along the y-axis. The minor axis is ##2\sqrt 2## and runs along the x-axis.
No!
These are the vertices of the ellipse. They are useful in graphing the ellipse, but they are not critical points, and really have nothing to do with finding the max or min values of the function ##f(x, y) = 4x + y^2##.

Did you look at the link I wrote in post #9? In it there are several examples of how to use Lagrange multipliers, but you don't need to use that technique in this problem. I mentioned another way back in post #2.
Domain:{-##\sqrt2##, ##\sqrt2##}
Range:{-2,2}
Major axis is on y-axis = 4
Minor axis is on x-axis = 2##\sqrt2##
Vertices (0,-2)(0,2)
The ends point of minor axis is (-##\sqrt2##,0) (##\sqrt2##,0)

I don't have to use Lagrange because the critical points can be seen in the ellipse? That is the x of vertices and the x of end points of minor axis?
So these are the critical points to test
X1 = 1
X2 = 0
X3 = +-##\sqrt2##
 
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One can set [itex]x = \sqrt{2} \cos \theta[/itex], [itex]y = \sin \theta[/itex] for [itex]\theta \in [0, 2\pi][/itex] and the constraint [itex]2x^2 + 4y^2 = 4[/itex] is then satisfied automatically.
How to find the angle then?
 
There are a couple of ways to go for this problem. The goal is to find the max. and min. values of ##f(x, y) = 4x + y^2## subject to the constraint ##2x^2 + y^2 = 4##. The constraint means that you're looking for the largest/smallest values of ##f(x, y) = 4x + y^2## for points (x, y) that lie on the ellipse ##2x^2 + 4y^2 = 4##. It might be helpful to sketch the ellipse. Doing so might give you some insight into which points on the ellipse maximize or minimize the function you're working with.

One way is to solve for y in the constraint equation, and then substitute for y in f(x, y). You should then be able to use differentiation to find the max and min values.

Another way is to use Lagrange multipliers, if you know that technique.
So. The 3 method mentioned. At least 2 method applied to get all the x.
Because using differentiation only gets me x=1
And using ellipse graph only gets me x=+-##\sqrt2##
By Lagrange only gets me x = +-##\sqrt2##
But by the Lagrange doesn't give me x=1?
 
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Domain:{-##\sqrt2##, ##\sqrt2##}
Range:{-2,2}
Major axis is on y-axis = 4
Minor axis is on x-axis = 2##\sqrt2##
Vertices (0,-2)(0,2)
The ends point of minor axis is (-##\sqrt2##,0) (##\sqrt2##,0)
These are all correct, but when you talk about domain and range, you are talking about a function, and the equation##2x^2 + y^2 = 4## doesn't represent a function. Each x value in the interval ##[-\sqrt 2, \sqrt 2]## pairs with two y values.
Lifeforbetter said:
I don't have to use Lagrange because the critical points can be seen in the ellipse?
You don't have to use Lagrange multipliers because there is at least one other method you can use. Obviously and critical points have to be on the ellipse, but I don't think you can get them merely by inspection.
Lifeforbetter said:
That is the x of vertices and the x of end points of minor axis?
So these are the critical points to test
X1 = 1
X2 = 0
X3 = +-##\sqrt2##
x = 1 and x = ##\pm \sqrt 2## are critical points, but x = 0 is not.
 
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So. The 3 method mentioned. At least 2 method applied to get all the x.
Because using differentiation only gets me x=1
And using ellipse graph only gets me x=+-##\sqrt2##
By Lagrange only gets me x = +-##\sqrt2##
But by the Lagrange doesn't give me x=1?
If you use Lagrange multipliers correctly, it does give you x = 1.
 
These are all correct, but when you talk about domain and range, you are talking about a function, and the equation##2x^2 + y^2 = 4## doesn't represent a function. Each x value in the interval ##[-\sqrt 2, \sqrt 2]## pairs with two y values.
You don't have to use Lagrange multipliers because there is at least one other method you can use. Obviously and critical points have to be on the ellipse, but I don't think you can get them merely by inspection.

x = 1 and x = ##\pm \sqrt 2## are critical points, but x = 0 is not.
What is the other method besides Lagrange and without inspection? Since i have to find max and min.
Because Lagrange gives me x=1
Differentiation gives me x=1
 
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From post #2:
One way is to solve for y in the constraint equation, and then substitute for y in f(x, y). You should then be able to use differentiation to find the max and min values.
Do you know how to get y isolated in the equation ##2x^2 + y^2 = 4##? It's not clear to me that you are able to do this. If you are able to do so, then replace y in ##f(x, y) = 4x + y^2##, which will give you a function with only one variable. Differentiating that equation gives you a critical point at x = 1, and you can check the two endpoints of the x-interval ##[-\sqrt 2, \sqrt 2]##.
 
From post #2:Do you know how to get y isolated in the equation ##2x^2 + y^2 = 4##? It's not clear to me that you are able to do this. If you are able to do so, then replace y in ##f(x, y) = 4x + y^2##, which will give you a function with only one variable. Differentiating that equation gives you a critical point at x = 1, and you can check the two endpoints of the x-interval ##[-\sqrt 2, \sqrt 2]##.
##2x^2 + y^2 = 4##
##y^2 = 4 - 2x^2##
##f(x, y) = 4x + y^2##
##f(x) = 4x + 4 - 2x^2##
f'(x) = 0
x = 1
 
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##2x^2 + y^2 = 4##
##y^2 = 4 - 2x^2##
##f(x, y) = 4x + y^2##
##f(x) = 4x + 4 - 2x^2##
f'(x) = 0
x = 1
Yes.
There are two points on the ellipse associated with x = 1. One of them gives the maximum value for the function of interest. The other point of interest occurs at one of the endpoints of the interval on the x-axis.
 

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