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

MATH110 April 2016

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Question 10
What is the shortest possible length of the line segment that is cut off by the first quadrant and is tangent to the curve $y={\frac {3}{x}}$ at some point?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Characterize the lines $y=ax+b$ that are tangent to the curve $y=3/x$. To do this, recall from the quadratic formula that there is exactly one solution to $y=ax^{2}+bx+c$ exactly when the discriminant, $b^{2}-4ac$ is zero. Moreover, note that if $f(x)$ is never negative and $x=s$ minimizes $f(x)$, then $x=s$ minimizes ${\sqrt {f(x)}}$.

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Solution

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 $\color {blue}{\sqrt {24}}$.