1. Equation of the line in intercept form

A line cutting the positive axes at

• $A(a,0)$ on the $x$-axis,
• $B(0,b)$ on the $y$-axis,

has equation $\frac{x}{a}+\frac{y}{b}=1,$ with $a>0, b>0$.

Since the line passes through $(3,5)$, $\frac{3}{a}+\frac{5}{b}=1.$

2. Area of triangle $OAB$

The triangle has base $OA=a$ and height $OB=b$, so its area is $S=\frac{1}{2}ab.$

We must minimize $ab$ subject to $\frac{3}{a}+\frac{5}{b}=1.$

3. Express one variable in terms of the other

From $\frac{3}{a}+\frac{5}{b}=1,$ we get $\frac{5}{b}=1-\frac{3}{a}=\frac{a-3}{a}.$ So, $b=\frac{5a}{a-3}, \quad a>3.$

Hence, $ab=a\cdot \frac{5a}{a-3}=\frac{5a^2}{a-3}.$ Therefore, $S=\frac{1}{2}ab=\frac{5a^2}{2(a-3)}.$

4. Minimize the function

Let $f(a)=\frac{5a^2}{2(a-3)}, \quad a>3.$

Differentiate: $f'(a)=\frac{5}{2}\cdot \frac{2a(a-3)-a^2}{(a-3)^2}$ $=\frac{5}{2}\cdot \frac{a^2-6a}{(a-3)^2}$ $=\frac{5}{2}\cdot \frac{a(a-6)}{(a-3)^2}.$

Set $f'(a)=0$: $a(a-6)=0.$ Since $a>3$, we get $a=6.$

Then $b=\frac{5a}{a-3}=\frac{5\cdot 6}{6-3}=10.$

So the minimum area is $S_{\min}=\frac{1}{2}\cdot 6\cdot 10=30.$

5. Check options

The minimum area is $30.$ So the correct option is C.

6. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

They agree.