Single Variable Calculus, Chapter 4, 4.7, Section 4.7, Problem 50

Determine the equation of the line through the point $(3,5)$ that cuts off the least area of the first quadrant.



By using Point Slope Form, we can determine the equation of the line that pass through the point $(3,5)$.

$
\begin{equation}
\begin{aligned}
y - 5 &= m(x-3)\\
\\
y &= mx - 3m + 5
\end{aligned}
\end{equation}
$


So the $y$-intercept when $x=$ will be,
$ y = m(0) - 3m + 5$

And the $x$-intercept when $y = 0$ will be,

$
\begin{equation}
\begin{aligned}
0 &= mx - 3m + 5\\
\\
x &= 3 - \frac{5}{m}
\end{aligned}
\end{equation}
$







$
\begin{equation}
\begin{aligned}
\text{So, the area of the triangle will be }A &= \frac{1}{2} \left( 3-\frac{5}{m}\right) \left( 5-3m\right)\\
\\
A &= 15 - \frac{25}{2m} - \frac{9m}{2}
\end{aligned}
\end{equation}
$


If we take the derivative of $A$, we get...
$\displaystyle A' = \frac{25}{2m^2} - \frac{9}{2}$

when $A'=0$,

$
\begin{equation}
\begin{aligned}
\frac{25}{2m^2} &= \frac{9}{2}\\
\\
m^2 &= \frac{25}{9}\\
\\
m &= -\frac{5}{3} \text{ and } m = \frac{5}{3}
\end{aligned}
\end{equation}
$


$\displaystyle m = \frac{-5}{3}$ respects a minimum since $\displaystyle A''\left( \frac{-5}{3} \right) > 0$

$
\begin{equation}
\begin{aligned}
\text{Therefore, the equation of the line is } y &= mx - 3m + 5 = - \frac{5}{3} x - 3 \left( \frac{-5}{3}\right)+5\\
\\
y &= \frac{-5}{3}x + 10
\end{aligned}
\end{equation}
$