Find Zi Song's minimum

Write the given expression as $(x-y-2)^{2}+(y-2)^{2}-8$ . As perfect square is non negative, the minimum value of expression is -8. And so a = -8 and hence the answer.......

Sorry.. I don't really know LaTex... I don't have the best time to learn this thing but anyways.

We take the partial derivative of the function: f(x,y) = x(x-4) - 2y(x-y) = x^2 - 4x - 2xy + 2y^2

Taking the partial derivative for x: We have partial f/ partial x = 2x - 4 - 2y = 0

Taking the partial derivative for y: We have partial f/ partial y = 4y - 2x = 0

Solving the system yields x = 4 and y = 2.

Substituting back to the original function. The minimum value is therefore -8. Hence absolute value of a = 8.

By expanding the given expression, we get $x^2-4x-2xy+2y^2$ . If we treat this as a quadratic in $x$ , we have $x^2-4x-2xy+2y^2=x^2-2x(y+2)+2y^2=[x-(y+2)]^2+[2y^2-(y+2)^2]$ . The minimum of this expression is $2y^2-(y+2)^2=y^2-4x-4$ . So we wish to minimize $y^2-4x-4$ . Completing the square, we get $(y-2)^2-8$ and the minimum of this quadratic occurs when $y=2$ and we get minimum of the original expression as $-8$ . $|-8|=\boxed{8}$ .

$x(x-4)-2y(x-y)=\frac{1}{2}(2x^2-8x+4y^2-4xy)=\frac{1}{2}[(x-4)^2+(x-2y)^2]-8$ . The expression is minimum when $x=4$ and $y=2$ . So, $a=-8$ . Answer is $|a|=\boxed{8}$

The given expression is equivalent to $(x-y-2)^2+(y-2)^2-8.$ Now since the first two terms in the sum are nonnegative and become zero when $x=4$ and $y=2$ , it follows that the minimum of the given expression is $a = -8$ . Hence $|a| = 8.$

First, we find the minimum of the expression given a constant x. Let $f(y) = x(x - 4) - 2y(x - y) = x(x - 4) - 2xy + 2y^2$ Using calculus: $f'(y) = -2x + 4y = 0, y = x/2$ We know this is the minimum and not the maximum of f(y) since $f''(y) = 4 > 0$ Now substituting into the original expression: $g(x) = x(x - 4) - x(x/2) = x^2 - 4x - x^2/2 = x^2/2 - 4x$ Since this is the equation of a parabola, the minimum is at -b/2a = -(-4)/1 = 4 and so $a = g(4) = -8, |a| = 8$

x( x- 4) - 2xy( x - y ) = $x^2$ - x( 4 + 2y ) + 2 $y^2$

From that quadratic equation, we can find the minimum value by $\frac{D}{-4a}$

So, The minimum value of that expression is :

$\frac{D}{-4a}$ = $\frac{-4y^2 + 16y +16}{-4}$ = $y^2 - 4y - 4$ = $(y-2)^2 - 8$

Because $(y-2)^2$ $\geq$ 0 , so, a = -8 , |a| = 8

F(x,y) = x^2- 4x -2xy +2y^2

It is a fn in 2 variables.

To find minima , partially diff f w.r.t. x & y resp.,

Equate both to 0 & find common solution.

. δ(f)/δx = 2x-4-2x=0

.δf/δx = -2x+4y=0

From above two eq, x=4,y=2

Therefore, a=f(4,2)=-8

|a|=8

If you let x=A, constant, then the smallest value of parabola 2yy -2yA +A(A-4) is y=A/2, ie y=x/2. Knowing that, we seek the lowest value of parabola x(x-4)-2(x/2)(x-x/2) = xx/2-4x which is at x=4. That minimum "a" is -8. So |a| is 8.

Assuming that $y$ is constant, we can find the minimum by taking the derivative. $\frac{d(x(x-4) - 2y(x-y))}{dx} = 2x - 4 - 2y = 0 \implies x = y + 2$ Thus, we can substitute $x = y + 2$ and then find the minimum of the resulting expression $\frac{d((y+2)(y-2) - 2y(2))}{dy} = 2y - 4 = 0 \implies y = 2.$ Now substituting gives a minimum of $2^2 -4\cdot 2 - 4 = -8$ . So our answer is $|a| = 8$ .

we have the following equation: x^2 - 4x - 2xy + 2y^2 = a

Multiplying both sides by 2 yields: (2y-x)^2 + (x-4)^2 = 2a + 16

The left side is always positive; hence 2a + 16 >= 0 for all x,y

=> a >= -8 => the minimum of a is -8

Thus |a| = 8

The most obvious value of x to make x(x−4)−2y(x−y) as low as possible is x=4 since if x=4, the first part, x(x−4), equals 0. To make the lowest value of x(x−4)−2y(x−y), you must maximize the value of 2y(x−y). By plugging in x=4 into 2y(x−y), you get 2y(4-y)=8y-2y^2. To make 8y-2y^2 the largest value possible y must =2. Plugging in x=4, y=2 into x(x−4)−2y(x−y), you get -8 and |-8|=8

Using Differentiation:

Let z = x(x−4)−2y(x−y), let's differentiate z with respect to x (we will consider y constant).

i.e. We are trying to answer the following question: For a certain value of y,what is the value of x that yields the minimum value of z?

z = x^2 - 4x - 2yx + 2y^2

dz/dx = 2x - 4 - 2y + 0

when dz/dx = 0

2x - 4 - 2y = 0

x = 2 + y

d(dz/dx)/dx = 2 (> 0 which ensures z is minimum when x = y+2 for a constant y)

(this means for any y value x = y+2 let z be minimum)

let's reformulate z in terms of y only:

z = y^2 - 4y - 4

Let's find the value of y that let z be minimum:

dz/dy = 2y-4

when 2y-4 = 0

y = 2

z = 8

d(dz/dy)/dy = 2 (> 0 which ensures z is minimum at 2 NOT maximum)

We have x(x-4)-2y(x-y)=a Let x=4 so that the first term will be equal to 0 and the equation will be;

-2y(4-y)=a. To get the minimum value of a, y must be a positive integer less than 4 for us to have the term -2y be negative and for the term 4-y be positive so that when you multiply the two, the result is negative. Therefore y=2 and a=-2(2)(4-2)=-8 and its absolute value is 8.

To get the smallest a : x ( x -4) needs to be small and 2 _y_(_x_-_y_) needs to be high. Both _x_ and _y_ must be positive. To have the first multiplication smaller or equal to zero, _x_ should be less than four. To have the second multiplication bigger than 0, _y_ must be smaller than _x_. After checking all the possibilities we see that the smallest _a_ we get with _x_=4 and _y_=2. 4(4-4)-2 2(4-2)=-8. |-8|=8

2y(x-y)=-(2y²-2yx), which is a quadratic function whose roots are 0 and x, so the maximum value that this function can assume is when y=x/2, 2(x/2)²-2(x/2)x=-x²/2. Now we have that the minimum value of x(x-4)-2y(x-y) is x²-4x-x²/2=x²/2-4x which also is a quadratic function whose roots are 0 and 8, so the the minimum value is reached when x=4, so a=(4)²/2 -4(4)=-8, and |a|=8

initially, consider y as a constant. Now the expression is a quadratic in a single variable x . Now differentiate the expression and on equating the differential to zero we get a relation between x and y as...... x=2+y . Now substitute this equation in original expression so that it transforms into an expression in a single variable . now we can get the least value of the final expression as (-8) . And thus the answer is 8.

Let f ( x , y ) = x ( x - 4) - 2 y ( x - y ) = x ^2 - 4 x - 2 xy + 2 y ^2 . To find the minimum of f ( x , y ) , we will differentiate f ( x , y ) partially , first , w.r.t. x treating y as a constant and then w.r.t. y treating x as a constant and equate both equations to zero to find the values of x and y which on substituting in f ( x , y ) will give us its minimum value .Therefore ,

\frac{\partial f ( x , y )}{\partial x } = 2 x - 4 - 2 y = 0

\Rightarrow x - y = 2 ........eq-1

Similarly , \frac{\partial f ( x , y )}{\partial y } = -2 x + 4 y = 0

\Rightarrow x = 2 y . Substituting the value of x in eq-1 , we get

2 y - y = 2 \Rightarrow y = 2 . Therefore , x = 2 y = 4 . Thus , f ( x , y ) is minimum when x = 4 and y = 2 . Hence , the minimum value of f ( x , y ) ,i.e. a = 4 \times (4 - 4) - 2 \times 2 \times (4 - 2) = -8 .Therefore , | a | = 8 .

Expanding the statement yields $x^2-4x-2yx+2y^2$ . Completing the square for x yields $x^2-4x+4-2yx+2y^2-4$ , or $(x-2)^2+2y^2-2yx-4$ . Slightly changing this to $(x-2)^2+2(y^2-yx)-4$ and completing the square for y yields $(x-2)^2+2(y^2-yx+\frac{x^2}{4})-4-\frac{x^2}{2}$ . Factoring yields $(x-2)^2+2(y-\frac{x}{2})^2-4-\frac{x^2}{2}$ . Since y only appears once, and since the statement $2(y-\frac{x}{2})^2$ is always $\geq 0$ , we can just set $y=\frac{x}{2}$ , making that term 0, and effectively get rid of it. Now we're left with the task of minimizing $(x-2)^2-4-\frac{x^2}{2}$ . Expanding and simplifying yields $\frac{x^2}{2}-4x$ . Differentiating this yields $x-4$ . The minimum is achieved when this is equal to 0, so $x=4$ . Substituting 4 for x in $\frac{x^2}{2}-4x$ yields -8, so |-8|=8.