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 $\displaystyle y=1-x$ and $\displaystyle z=7x^2-4xy+9y^2$.
Now substitute the first into the second. This gives:
$\displaystyle z=7x^2-4x(1-x)+9(1-x)^2$
so
$\displaystyle z=2x^2-22x-9$
Now differentiate and equate to zero to find the turning point
$\displaystyle \frac{dz}{dx}=4x-22=0$
so
$\displaystyle x=5.5$
(you can clarify that this is a minimum by differentiating again and seeing it is positive)
Now plug this into $\displaystyle z=2x^2-22x-9$ to get $\displaystyle z=-69.5$ as the min value.
Now plug $\displaystyle x=5.5$ back into $\displaystyle y=1-x$ to get $\displaystyle y=-4.5$
so $\displaystyle 5.5,-4.5$ are the values of x and y where the minimum occurs.
HTH
Si.
Since this was listed in "Pre-Algebra and Algebra", here's how to do it without calculus:
Reduce to $\displaystyle 2x^2- 22x- 9$ as bananaxxx did and then complete the square.
$\displaystyle 2x^2- 22x- 9= 2(x^2- 11x) - 9= 2(x^2- 11x+ (11/2)^2- (11/2)^2)- 9$
$\displaystyle = 2(x^2- 11x+ 121/4)- 121/2- 9= 2(x- 11/2)^2- 139/2$
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.
$\displaystyle z = 7x^2 - 4x(1-x) + 9(1-x)^2$
$\displaystyle z = 7x^2 - 4x + 4x^2 + 9(1 - 2x + x^2)$
$\displaystyle z = 7x^2 - 4x + 4x^2 + 9 - 18x + 9x^2$
$\displaystyle z = 20x^2 - 22x + 9$
$\displaystyle x_{min} = -\frac{b}{2a} = \frac{22}{40} = \frac{11}{20}
$
$\displaystyle y = 1 - x = \frac{9}{20}$
Wilmer's calculations are correct ... he just has a sign typo for the "9".No; z = 20x^2 - 22x - 9
So x = .55, y = .45 ; min = 2.95