Answer
$$
M(x, y)=45x^{2}+40y^{2}-20xy+50
$$
(a)
$$
\begin{aligned}
M_{y}(x, y)&= 80y-20x
\end{aligned}
$$
$$
M_{y}(4, 2)= 80(2)-20(4)=80
$$
(b)
$$
\begin{aligned}
M_{x}(x, y)&= 90x-20y
\end{aligned}
$$
$$
M_{x}(3, 6)= 90 (3)-20(6)=150
$$
(c)
$$
\frac{\partial M}{\partial x}(2, 5)= 90(2)-20(5)=80
$$
(d)
$$
\frac{\partial M}{\partial y}(6, 7)= 80(7)-20(6)=440
$$
Work Step by Step
$$
M(x, y)=45x^{2}+40y^{2}-20xy+50
$$
(a)
find $M_{y}(x, y).$
differentiate $M(x, y)$ with respect to $y$ and treat $x$ constant.
$$
\begin{aligned}
M_{y}(x, y)&= \frac{\partial}{\partial y} (45x^{2}+40y^{2}-20xy+50) \\
&=80y-20x \quad\quad\quad\quad\quad\quad(1)
\end{aligned}
$$
To find $M_{y}(4, 2)$ Replacing $x$ with 4 and $y$ with 2 in equation (1) gives:
$$
M_{y}(4, 2)= 80(2)-20(4)=160-80=80
$$
(b)
find $M_{x}(x, y).$
differentiate $M(x, y)$ with respect to $x$ and treat $y$ constant.
$$
\begin{aligned}
M_{x}(x, y)&= \frac{\partial}{\partial x} (45x^{2}+40y^{2}-20xy+50) \\
&=90x-20y \quad\quad\quad\quad\quad\quad(2)
\end{aligned}
$$
To find $M_{x}(3, 6)$ Replacing $x$ with 3 and $y$ with 6 in equation (2) gives:
$$
M_{x}(3, 6)= 90 (3)-20(6)=270-120=150
$$
(c)
To find $\frac{\partial M}{\partial x}(2, 5)$ Replacing $x$ with 2 and $y$ with 5 in equation (2) gives:
$$
\frac{\partial M}{\partial x}(2, 5)=M_{x}(2, 5)= 90(2)-20(5)=180-100=80
$$
(d)
To find $\frac{\partial M}{\partial y}(6, 7)$ Replacing $x$ with 6and $y$ with 7 in equation (1) gives:
$$
\frac{\partial M}{\partial y}(6, 7)=M_{y}(6, 7)= 80(7)-20(6)=560-120=440
$$
