# Math Help - Quadratic Functions

If x and y are real and x+y=1, find the minimum value of 7x^2 - 4xy + 9y^2 and the values that x and y then assume</SPAN>

Thanks.

2. Originally Posted by Lukybear
If x and y are real and x+y=1, find the minimum value of 7x^2 - 4xy + 9y^2 and the values that x and y then assume</SPAN>

Thanks.
Hi.

ok you have that $y=1-x$ and $z=7x^2-4xy+9y^2$.

Now substitute the first into the second. This gives:
$z=7x^2-4x(1-x)+9(1-x)^2$
so
$z=2x^2-22x-9$

Now differentiate and equate to zero to find the turning point
$\frac{dz}{dx}=4x-22=0$
so
$x=5.5$

(you can clarify that this is a minimum by differentiating again and seeing it is positive)

Now plug this into $z=2x^2-22x-9$ to get $z=-69.5$ as the min value.

Now plug $x=5.5$ back into $y=1-x$ to get $y=-4.5$

so $5.5,-4.5$ are the values of x and y where the minimum occurs.

HTH
Si.

3. Since this was listed in "Pre-Algebra and Algebra", here's how to do it without calculus:

Reduce to $2x^2- 22x- 9$ as bananaxxx did and then complete the square.

$2x^2- 22x- 9= 2(x^2- 11x) - 9= 2(x^2- 11x+ (11/2)^2- (11/2)^2)- 9$
$= 2(x^2- 11x+ 121/4)- 121/2- 9= 2(x- 11/2)^2- 139/2$

If x= 11/2= 5.5, x-11/2= 0 and so z= -139/2. If x is any other number, x- 11/2 is not 0 so its square is positive and z is larger than -139.2. -139.2 is the smallest possible value of z and it occurs when x= 5.5. Of course, as bananaxxx says, when x= 5.5, y= 1- 5.5= -4.5.

4. Originally Posted by bananaxxx
$z=7x^2-4x(1-x)+9(1-x)^2$
so
$z=2x^2-22x-9$
No; z = 20x^2 - 22x + 9
So x = .55, y = .45 ; min = 2.95

5. Originally Posted by Wilmer
No; z = 20x^2 - 22x - 9
So x = .55, y = .45 ; min = 2.95

6. Originally Posted by HallsofIvy
Since this was listed in "Pre-Algebra and Algebra", here's how to do it without calculus:

7. Originally Posted by Lukybear
If x and y are real and x+y=1, find the minimum value of 7x^2 - 4xy + 9y^2 and the values that x and y then assume
$z = 7x^2 - 4x(1-x) + 9(1-x)^2$

$z = 7x^2 - 4x + 4x^2 + 9(1 - 2x + x^2)$

$z = 7x^2 - 4x + 4x^2 + 9 - 18x + 9x^2$

$z = 20x^2 - 22x + 9$

$x_{min} = -\frac{b}{2a} = \frac{22}{40} = \frac{11}{20}
$

$y = 1 - x = \frac{9}{20}$

No; z = 20x^2 - 22x - 9
So x = .55, y = .45 ; min = 2.95
Wilmer's calculations are correct ... he just has a sign typo for the "9".

8. Originally Posted by skeeter
Wilmer's calculations are correct ... he just has a sign typo for the "9".
We were both wrong then

Edit: It seems the more I learn in maths...the more stupid mistakes I make. Im now a fourth year uni student and I can't multiply out brackets without a mistake! Oh dear...I blame it on the hangover.....

9. Originally Posted by bananaxxx
We were both wrong then
I disagree ... a typo is not an algebra error.

10. Originally Posted by skeeter
I disagree ... a typo is not an algebra error.
Prove it.

sigh...back to my ergodic theory...much easier than multiplying out brackets haha!

11. Originally Posted by bananaxxx
Prove it.
Wilmer's minimum value calculation is correct.

12. Originally Posted by skeeter
Wilmer's minimum value calculation is correct.
Lol i can see it is now ...just what he had written wasn't. And in a proof a typo is indistinguishable from an error!

13. Hey Bananas, if you look closely, you'll see that my -9 was simply
copying YOURS...I thought a university student would NEVER make
a typo

14. Originally Posted by Wilmer
Hey Bananas, if you look closely, you'll see that my -9 was simply
copying YOURS...I thought a university student would NEVER make
a typo
Hence why I said earlier about me making more and more simple mistakes the more maths I learn...do I sense hostility in you? And FYI university students are not machines we are human and make just as many mistakes as you retired folk.

15. Originally Posted by bananaxxx
> ...do I sense hostility in you?

You shouldn't...twas a joke.

> And FYI university students are not machines we are human...

I see you like to joke too
.

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