Exercise 1

Correct: We have to minimize the following function:
$$\begin{align*}z(a,b)&=[ 4- (a \cdot 1 + b)]^2 + [-1- (a \cdot 2 + b)]^2 +[ 0- (a \cdot 4 + b)]^2+[-2- (a \cdot 8 + b)]^2\end{align*}$$

Consequently, we determine the partial derivative with respect to $a$ of $z(a,b)$,
$$\begin{align*}
z'_a(a,b)&=-2(4-a-b)-4(-1-2a-b)-8(-4a-b)-16(-2-8a-b)\\
& = 28+170a+30b,
\end{align*}$$
and the partial derivative with respect to $b$ of $z(a,b)$,
$$\begin{align*}
z'_b(a,b)&=-2(4-a-b)-2(-1-2a-b)-2(-4a-b)-2(-2-8a-b)\\
&=-2+30a+8b,
\end{align*}$$
The stationary points of $z(a,b)$ are found by solving the following system of equations:
$$\begin{align*}
z'_a(a,b)&=0\\
z'_b(a,b)&=0.
\end{align*}$$
Hence, $30a=2-8b$, which gives $a=\frac{1}{15}-\frac{4}{15}b$. Therefore, $28+170(\frac{1}{15}-\frac{4}{15}b)+30b=0$ gives $b=2\frac{13}{23}$. Finally, $a=-\frac{71}{115}$.

Since $z''_{aa}(a,b)=170$, $z''_{bb}(a,b)=8$ and $z''_{ab}(a,b)=30$ it holds that $C(a,b)=170\cdot 8-30^2=460$. Hence, $C(-\frac{71}{115},2\frac{13}{23})=460>0$ and $z''_{aa}(-\frac{71}{115},2\frac{13}{23})=170>0$, which implies that $(a,b)=(-\frac{71}{115},2\frac{13}{23})$ is a minimum location.

Consequently, $y=-\frac{71}{115}x+2\frac{13}{23}$.

Go on.

Wrong: We are not looking for the minimum of the function $z(a,b)$.

See Example.

Wrong: We are looking for the equation of a regression line.

See Example.

Wrong: $39\frac{1}{3}-15\frac{1}{3}b=0$ does not give $b=\frac{23}{59}$.

Try again.