Answer
$y = -\frac{5}{3}~x+10$
Work Step by Step
Let the equation of the line be $~~y = mx+b$
Note that $b \gt 5$
Suppose the line forms a triangle in the first quadrant with the points $(0,0)$, $(0,b)$, and $(a,0)$
Then $m = -\frac{b}{a}$
Also, the line passes through the point $(3,5)$.
Then:
$y = mx+b$
$5 = 3m+b$
$5 = -\frac{3b}{a}+b$
$\frac{3b}{a} = b - 5$
$a = \frac{3b}{b-5}$
We can write an expression for the area of the triangle:
$A = \frac{1}{2}ab = \frac{3b^2}{2(b-5)}$
We can find the value of $b$ when $A'(b) = 0$:
$A(b) = \frac{3b^2}{2b-10}$
$A'(b) = \frac{6b(2b-10)-6b^2}{(2b-10)^2}= 0$
$6b^2-60b= 0$
$b^2-10b= 0$
$b(b-10)= 0$
$b = 0, 10$
When $5 \lt b \lt 10$, then $A'(b) \lt 0$
When $b \gt 10$, then $A'(b) \gt 0$
Thus, $b=10$ is the value of $b$ where the area $A$ is a minimum.
We can find the value of $a$:
$a = \frac{3b}{b-5}$
$a = \frac{3(10)}{(10)-5}$
$a = 6$
We can find the equation of the line:
$y = mx+b$
$y = -\frac{b}{a}~x+b$
$y = -\frac{10}{6}~x+10$
$y = -\frac{5}{3}~x+10$
