Maximum and Minimum problem

[sparklemetink]

Find the smallest possible value of the quantity x^2+ y^2 under the restriction that 2x+3y=6.

I am getting closer to what the answer in the book is, but not quite there. Maybe you can tell me what I’m doing wrong. Here is what I have…
2x+3y=6
3y=6-2x
y=2-2/3x

x^2+(2-2/3x)^2=6
x^2+(2-2/3x)^2-6=0
x^2+4-8/3x+4/(9x^2 )-6
13/(9x^2 )-8/3x-2
Then, I did the x=-b/2a and got…
x=-(-8/3)/2(13/9) =24/26=12/13
The answer in the book is 36/13 .

 

[mmm4444bot]

Find the smallest possible value of the quantity x^2+ y^2 This is what you need to find.

x^2+(2-2/3x)^2=6 You've already decided that the smallest possible value is 6 ?

x^2+4-8/3x+4/(9x^2 )-6 What happened to the equal sign, and why are you dividing by x ?

Time to abandon this ship!

Your strategy is good.

You determined the correct expression for y, in terms of x.

You substituted this expression properly, into the expression x^2 + y^2.

So, multiply out everything, and combine like-terms, to end up with a quadratic expression (that is, a polynomial, not an equation; I'll write the equation below: Y = quadratic polynomial).

You find the smallest value of this quadratic expression (that is, the smallest Y) by realizing that the graph of the following equation is a parabola that opens upward (because the leading coefficient is positive), followed by realizing that the Y-coordinate of the vertex point is the smallest value of Y, yes?

Y = x^2 + (2 - 2x/3)^2

Note that Y is not the same as the y in your exercise.

Y = x^2 + y^2

 

[mmm4444bot]

x=-(-8/3)/2(13/9) =24/26=12/13

The answer in the book is 36/13

I just took another look at your post, and read it to the end (after goofing around with graphs on a related thought).

I guess that I was misled by all of your sloppy typing; I see your mistake now. It's a mistake in logic.

You found x.

But they're asking for Y (as defined in my first response), when x = 12/13.

In other words, you're not done, yet.

You have the x-coordinate of the vertex point. Now calculate the Y-coordinate of the vertex point.

I mean, substitute x = 12/13 into the quadratic polynomial 13x^2/9 - 8x/3 + 4, and evaluate.

Cheers

 

[sparklemetink]

Sorry about the slopping typing. I can't get it to look neat; wish I could.

I think when I got to the 12/13 I didn't know where to plug it into. Thank you, I now see that.

 

[mmm4444bot]

I can't get [my typing] to look neat; wish I could.

I don't care so much about "unneat" typing; I care about inaccurate statements. In other words, if the numbers, symbols, and operations show what you're actually doing, I don't care what it looks like.

When I call your typing "sloppy", I mean "careless" because your expressions/equations are not correct as typed. It's very misleading, when you don't tell people the truth.

If you did not set the expression x^2 + y^2 equal to six, why would you type that ?

You better find a way to communicate clearly (other than wishing), when discussing mathematics.

I mean, you might begin to receive less and less help, once people realize that you consistently waste time with false statements.

You can, of course, simply take your chances in life. Seems like most people do.

Click HERE to learn about standard conventions for typing mathematics using a keyboard.

On the freemathhelp boards, begin using the [Preview] button (it's right next to the [Submit] button) to proofread your posts before submitting.

 

[sparklemetink]

I'm sorry if I offended you. I did in fact write on my paper that x^2 + (2-2/3x)^2 = 6. Sorry. I was asking for help to see where I went wrong. After my last comment, I did look at ways to type more clearly mathmatically. I'm trying the best I can. I typed it neatly in Microsoft Word and tried to upload it as an attachment, but ".docx" files are not accepted here. Forgive me.

 

[mmm4444bot]

I'm sorry if I offended you.

There is no way for you to offend me.

I was asking for help to see where I went wrong.

And how do you expect us to do that, when we have no confidence that what you post is what you're doing?

I initially abandoned ship because I had no idea what was going on. I came back only to verify the line equation, since I was using it in an unrelated thought experiment. That's when I noticed, "Hey! x = 12/13 is correct".

I'm puzzled how you managed to get the correct x-coordinate, after the initial goof, and then dividing by x^2, and doing that other stuff that's wrong.

Or, maybe you didn't do all of that stuff. Who knows what you did? (Right now, only you do, and that's my point.) There's no way for anybody else to know what your mistakes are.


I typed it neatly in Microsoft Word and tried to upload it

Well, that's an okay idea, as far as showing what you've actually done.

Upload a screenshot of the Word document, instead, saved in .JPG format. That's easy, if you've got any version of Windows. Let me know, if you need instructions.

Otherwise, I suppose, in future posts, that you could ask for somebody to format your posted work in LaTex, so that you could verify that your typing accurately represents what you intend. (There might not always be people with time to do that, however.)

x^2+4-8/3x+4/(9x^2 )-6

\(\displaystyle x^2 \;+\; 4 \;-\; \frac{8}{3} x \;+\; \frac{4}{9x^2} \;-\; 6 \;=\; 0 \;\text{?}\)