Science:Math Exam Resources/Courses/MATH104/December 2011/Question 01 (m)

MATH104 December 2011

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Question 01 (m)
Find two positive real numbers $m$ and $n$ whose product is 50 and whose sum is as small as possible.

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
A common error for this type of question is to look for solutions $m$,$n$ that are integers through trial and error. This is an optimization problem. As such, you should determine what quantity you want to optimize and what your constraint is.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We wish to minimize the quantity $S=m+n$ subject to $mn=50$. Using our constraint we rewrite $S$ as $S=m+{\frac {50}{m}}.$ To determine the minimum of this quantity $S$, we first find the critical points of $S$: ${\begin{aligned}{\frac {dS}{dm}}=1-{\frac {50}{m^{2}}}&=0\quad \rightarrow \quad m^{2}=50.\end{aligned}}$ Dropping the negative solution (since the problem requires that $m$ and $n$ be positive), we find $m={\sqrt {50}}.$ The corresponding value for $n$ is determined from the constraint $mn=50$, and is therefore equal to $n=50/{\sqrt {50}}={\sqrt {50}}.$ Is this critical point a local minimum? Apply the second derivative test: ${\frac {d^{2}S}{dm^{2}}}={\frac {100}{m^{3}}}>0$ at the critical point. Therefore, this critical point is a minimum of $S$ and the two values $m,n$ that have product equal to 50 and the smallest sum are $m={\sqrt {50}},\ n={\sqrt {50}}.$

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