Science:Math Exam Resources/Courses/MATH110/April 2019/Question 08

MATH110 April 2019

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[hide] Question 08
A farmer wants to fence in an area of $10^{4}{\text{ m}}^{2}$ in a rectangular field and then divide it in half with a fence down the middle, parallel to one side. What is the shortest length of fence that the farmer can use? Justify your answer.

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

[show] Hint 1

Recall that the formula for the area of a rectangle is given by $A=l\times w$, where $l,w$ denote length and width respectively. We also have that perimeter of a rectangle is given by the formula: $P=2l+2w$. However, can you think of how the additional fence length should change this perimeter formula? You can assume that the additional fence length is parallel to the width side (so its length is given in terms of $w$). See the next hint for more information.

[show] Hint 2
Since we are told that there is an additional length of fence added parallel to the width side, the formula for the amount of fencing we need is given by: $F=P+w=2l+3w$. The problem now boils down to minimizing this function. What is something you have learned in class that lets you minimize a function? See the next hint if you need help minimizing the function $F$.

[show] Hint 3
We want to minimize the function $F=2l+3w$. However, in this form, it depends on two variables, $l,w$. Think of a way to remove the dependence on one of them, say $w$. Perhaps the area formula will come in handy? Remember: we are told that $A=10,000\,m^{2}$, and we also have the formula $A=l\times w$.

Checking a solution serves two purposes: helping you if, after having used all the hints, 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.

[show] 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. As mentioned in the hints, we know that the formula for area is given by $A=l\times w=10^{4}\,m^{2}$, and also that $F=2l+3w$ is a formula for how much fencing we will need in order to enclose our given area. There are two steps we must take before we can calculate this value. First, we need to use substitution in order to express the function $F$ in terms of $l$ alone. We could also do this in terms of $w$, but in this solution, we will do as previously mentioned. Since we are told that $A=10^{4}m^{2}$, we know that $w={\frac {10,000m^{2}}{l}}$. Notice, our units make sense, because $l,w$ are length and width, and hence are given in terms of meters. Substituting this value of $w$ into our formula for $F$, we have: $F(l)=2l+{\frac {30,000}{l}}$. Now, we are looking for a value of $l$ which will minimize the function $F(l)$. Call this minimizing value $l_{0}$. From our knowledge of the derivative, since $l_{0}$ is a local extrema, we must have $F^{'}(l_{0})=0$. We also know that: $F^{'}(l)=2-{\frac {30,000}{l^{2}}}$. Combining this information shows that: $F^{'}(l_{0})=2-{\frac {30,000}{l_{0}^{2}}}$. Solving this expression for $l_{0}$, we see that $l_{0}=(15,000)^{1/2}=10{\sqrt {150}}$. To find the final answer, we need only plug-in $l=l_{0}$ into $F$ expressed as a function of $l$ alone. That is, we have: $F(l_{0})=2l_{0}+{\frac {30,000}{l_{0}}}=20{\sqrt {150}}m+{\frac {3,000}{\sqrt {150}}}m$. Simplifying this expression, the minimum amount of fence needed to enclose this area is ${\frac {6,000}{\sqrt {150}}}$ meters, or approximately $490$meters.