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

Let $a$ and $b$ be positive. The line $y = - a x + b$ is tangent to $y = 3 / x$ at some point exactly when $3 / x = - a x + b$ . That is, $3 = - a x^{2} + b x$ which is the same as $- a x^{2} + b x - 3$ . Since this is $0 = A x^{2} + B x + C$ with $A = - a$ and $c = - 3$ , the discriminant is $B^{2} - 4 A C = b^{2} + 12 a$ . So the discriminant is zero exactly when $b^{2} = - 12 a$ . Substituting this into $y = - a x + b$ 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 $\frac{d}{d b} (b^{2} + (12 / b)^{2}) = 0$ ( $\frac{d}{d b}$ means differentiate with respect to $b$ ).

Using the power and chain rules, gives $\frac{d}{d b} (b^{2} + (12 / b)^{2}) = 2 b + (2 (12 / b) \times - 12 / b^{2}) = 2 (b - 1 2^{2} / b^{3}) = 0$ . So $b - 1 2^{2} / b^{3} = 0$ , $b^{4} - 1 2^{2} = 0$ , $b^{4} = 1 2^{2}$ , and $b^{2} = 12$ . Substituting this value of $b$ into $\sqrt{b^{2} + (12 / b)^{2}}$ gives $\sqrt{b^{2} + (12 / b)^{2}} = \sqrt{b^{2} + 1 2^{2} / b^{2}} = \sqrt{12 + 1 2^{2} / 12} = \sqrt{12 + 12} = \sqrt{24}$ .

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