Science:Math Exam Resources/Courses/MATH110/April 2016/Question 10/Solution 1

Let $a$ and $b$ be positive. The line $y=-ax+b$ is tangent to $y=3/x$ at some point exactly when $3/x=-ax+b$. That is, $3=-ax^{2}+bx$ which is the same as $-ax^{2}+bx-3$. Since this is $0=Ax^{2}+Bx+C$ with $A=-a$ and $c=-3$, the discriminant is $B^{2}-4AC=b^{2}+12a$. So the discriminant is zero exactly when $b^{2}=-12a$. Substituting this into $y=-ax+b$ gives $y={\frac {-b^{2}}{12}}x+b$, and such lines characterize the tangent lines of $y=3/x$. Setting $x=0$ gives $y={\frac {-b^{2}}{12}}0+b=b$, and setting $y=0$ gives $0={\frac {-b^{2}}{12}}x+b$, $-b={\frac {-b^{2}}{12}}x$, and $x=12/b$. Hence, the endpoints are $(0,b)$ and $(12/b,0)$, implying that the length of the line segment is ${\sqrt {b^{2}+(12/b)^{2}}}$.

To minimize this function, it is enough to minimize the following function $b^{2}+(12/b)^{2}$. This function is differentiable with respect to $b$, tends to positive infinity as $b\to 0$, and tends to positive infinity as $b\to \infty$. Hence, if $b$ that minimizes the above length, then it satisfies ${\frac {d}{db}}(b^{2}+(12/b)^{2})=0$ (${\frac {d}{db}}$ means differentiate with respect to $b$).

Using the power and chain rules, gives ${\frac {d}{db}}(b^{2}+(12/b)^{2})=2b+(2(12/b)\times -12/b^{2})=2(b-12^{2}/b^{3})=0$. So $b-12^{2}/b^{3}=0$, $b^{4}-12^{2}=0$, $b^{4}=12^{2}$, and $b^{2}=12$. Substituting this value of $b$ into ${\sqrt {b^{2}+(12/b)^{2}}}$ gives ${\sqrt {b^{2}+(12/b)^{2}}}={\sqrt {b^{2}+12^{2}/b^{2}}}={\sqrt {12+12^{2}/12}}={\sqrt {12+12}}={\sqrt {24}}$.

Hence, the minimum length is $\color {blue}{\sqrt {24}}$.