Minimum value problem without derivative

Question

Each straight line with negative slope which passes through the point (3; 7) in the rectangular coordinate system Oxy, together with the x and y coordinate axes encloses right triangle. Find the smallest one among the areas of such triangles. without derivative.

Answer 1

Let $(-m)$ is a slop of the line. Hence, $m>0$.

Thus, $y-7=-m(x-3)$ is an equation of the line, which intersects the $x$-axis in the point $A$ and the $y$- axis in the point $B$.

Hence, $B(0,7+3m)$ and $A\left(\frac{3m+7}{m},0\right)$.

The area of $\Delta OAB$ it's $$\frac{(3m+7)^2}{2m}.$$ Now, by AM-GM $$\frac{(3m+7)^2}{2m}=\frac{1}{2}\left(9m+\frac{49}{m}+42\right)\geq\frac{1}{2}\left(2\sqrt{9m\cdot\frac{49}{m}}+42\right)=42.$$ The equality occurs for $9m=\frac{49}{m}$, which says that the answer is $42$.

Answer 2

HINT: the line has the equation $$y=mx+n$$ The given Point is situated on this line, then we have $$7=3m+n$$ or $$y=mx-3m+7$$ the intersection Points with the axis are $$P_1(0;7-3m)$$ $$P_2\left(\frac{3m-7}{m};0\right)$$ can you go further?

Answer 3

Let $y=ax+b$ cuts $x$-axis and $y$-axis at $(x_0,0)$ and $(0,y_0)$, respectively.

Then: $$\begin{cases} 0=ax_0+b \\ y_0=a\cdot 0+b \end{cases} \Rightarrow \begin{cases} a=-\frac{y_0}{x_0} \\ b=y_0 \end{cases} \Rightarrow y=-\frac{y_0}{x_0}x+y_0.$$

The line passes through $(3,7)$: $$7=-\frac{y_0}{x_0}\cdot 3+y_0 \Rightarrow x_0y_0=7x_0+3y_0\ge 2\sqrt{21x_0y_0} \Rightarrow \sqrt{x_0y_0}\ge 2\sqrt{21} \Rightarrow x_0y_0\ge 84 \Rightarrow \frac{x_0y_0}{2}\ge 42.$$

And equality occurs when $x_0=6, y_0=14$.