Smallest value of function on a line

Question

Problem :

If the point $(\alpha, \beta)$ lies on the line $2x+3y=6$, the smallest value of $\alpha^2+\beta^2$ is

(a) $36/13$

(b) $6\sqrt{13}/13$

(c) $6$

(d) $13$

Solution : Since $\alpha,\beta$ lies on the given line $\Rightarrow 2x+3y=6 = 2\alpha + 3\beta -6=0$

$\Rightarrow \alpha = \frac{6-3\beta}{2}$

Putting the value of $\alpha$ in $\alpha^2+\beta^2$ we get $$(\frac{6-3\beta}{2})^2+\beta^2\\ \Rightarrow 13\beta^2 -36\beta +36 =0.$$

This give us parabola opening upward which has minimum value at $\dfrac{-b}{2a}$

Therefore the minimum value is $36/26$ but the answer is $36/13$ please correct thanks..

Answer 1

Hint: $|2x+3y| \leq \sqrt{(2^2+3^2)(x^2+y^2)}$

Answer 2

Another way to approach this problem is to recognize that the point $(\alpha, \beta)$ will be the point along the line that is the minimum distance to the origin. (Why?)

Answer 3

The point $(\alpha,\beta)$ with minimal $\alpha^2 + \beta^2$ is located at the intersection of the given line with a perpendicular line passing through the origin. Since the product of the slopes of two mutually perpendicular lines is $-1$, and the given line has slope $-2/3$, it follows that the perpendicular line has equation $3x = 2y$, hence $3\alpha = 2\beta$, and the rest is trivial.

Answer 4

The smallest value is obtained by the nearest point of the line to the origin so this point is the orthogonal projection of the origin onto this line. To find this point we determinate the equation of the line that pass through the origin and orthogonal to the given line

$$-3x+2y=0$$ Now the desired point is the intersection of these two lines and its coordinate is solution of this system

$$\left\{\begin{array}\\2x+3y=6\\-3x+2y=0\end{array}\right.\iff x=\frac{12}{13}=\alpha\quad;\quad y=\frac{18}{13}=\beta$$

Answer 5

The value of $\alpha^2 + \beta ^2 $ is $$\frac14(13\beta^2 - 36\beta + 36).$$ As you correctly calculated, the minimum value is acchieved at $\beta = \frac{-b}{2a}$, however, you made an incorrect step when you said that the value equals $\frac{-b}{2a}$