Question

find all values of x and y such that $p_x(x,y) = 0$ and $p_y(x,y) = 0$ simultaneously.
$p(x,y) = 3x^2 - 6xy + 7y^2 - 24x - 8y - 3$

$p_x(x,y) = 0$ and $p_y(x,y) = 0$ when $x = \square$ and $y = \square$.

Explanation:

Step1: Find \( P_x(x,y) \)

To find the partial derivative of \( P(x,y) = 3x^2 - 6xy + 7y^2 - 24x - 8y - 3 \) with respect to \( x \), we treat \( y \) as a constant.
Using the power rule and sum/difference rule of differentiation:
\( P_x(x,y)=\frac{\partial P}{\partial x}=6x - 6y - 24 \)
Set \( P_x(x,y) = 0 \), so we have the equation:
\( 6x - 6y - 24 = 0 \)
We can simplify this equation by dividing through by 6:
\( x - y - 4 = 0 \)
Which can be rewritten as:
\( x=y + 4 \) --- (1)

Step2: Find \( P_y(x,y) \)

To find the partial derivative of \( P(x,y) \) with respect to \( y \), we treat \( x \) as a constant.
Using the power rule and sum/difference rule of differentiation:
\( P_y(x,y)=\frac{\partial P}{\partial y}=- 6x+14y - 8 \)
Set \( P_y(x,y)=0 \), so we have the equation:
\( -6x + 14y-8=0 \) --- (2)

Step3: Substitute \( x=y + 4 \) into equation (2)

Substitute \( x=y + 4 \) from equation (1) into equation (2):
\( -6(y + 4)+14y-8 = 0 \)
First, expand the left - hand side:
\( -6y-24 + 14y-8=0 \)
Combine like terms:
\( (-6y + 14y)+(-24 - 8)=0 \)
\( 8y-32 = 0 \)
Add 32 to both sides:
\( 8y=32 \)
Divide both sides by 8:
\( y = 4 \)

Step4: Find the value of \( x \)

Substitute \( y = 4 \) into equation (1) (\( x=y + 4 \)):
\( x=4 + 4=8 \)

Answer:

\( x = \boxed{8} \) and \( y=\boxed{4} \)