Fence it

2663 /www/nrich/html/content/id/2663/ 1 2012-12-18T10:02:26 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Imagine you had $40$ one-metre sections of fencing. <mdo:image src="FenceIt0.png"></mdo:image> What is the largest rectangular area of land you could fence off? Now imagine you could build your fence up against a wall, so you only need to use the fence for three sides of the enclosure: <mdo:image src="FenceIt1.png"></mdo:image> What is the largest area you can fence off now? Now imagine you can attach the fence to the wall shown below, at the point marked X. <mdo:image alt="L shaped enclosure " height="360" src="fenceit.png" width="500"></mdo:image> What is the largest area you can fence off now? Extension Could you enclose even greater areas if you had $40$m of flexible wire fencing that could fence off curves as well as straight lines?</mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <h3>Why do this problem?</h3> <a href="http://nrich.maths.org/public/viewer.php?obj_id=2663&part=">This problem</a> challenges students to work systematically while applying their knowledge of areas of rectangles. It offers opportunities for higher level mathematical thinking (optimising and graphing) in a context that doesn't require sophisticated mathematical content. This problem could be revisited when students are older and are able to use algebraic techniques (forming quadratic expressions, maximising by completing the square). <h3>Possible approach</h3> "Imagine you had $40$ one-metre sections of fencing, and you wanted to make a rectangular enclosure. On your whiteboards, sketch a possible rectangle you could make that uses all $40$m. Write the area of your rectangle inside." Tabulate students' responses on the board. Select the largest area that has been found so far: "Is this the largest possible area we can make with $40$m of fencing?" Once the $10$m square with area $100$m$^2$ has been found, "How can we be convinced that a larger area isn't possible?" Give students a few minutes to think about the question in pairs and develop some convincing arguments. Circulate and listen in on conversations to identify which pairs have something worth sharing. Look out for diagram representations similar to this one: <mdo:image src="FenceWork2.jpg"></mdo:image> or tables of values like this one: <mdo:image src="FenceWork1.jpg"></mdo:image> Bring the class back together and invite any pairs whose interesting insights you noticed to share their thoughts. Use this to write an agreed 'convincing argument' why a square gives the maximum area, modelling the level of rigour that you expect them to come up with in their own justifications later on. Now set the second problem: "Imagine you had a long wall that you could use as one of the sides of your enclosure, so your $40$ metres of fencing only has to go round three sides of the rectangle. Use the ideas we have shared to work with your partner and find the biggest possible enclosure now. Make sure you have a justification to convince everyone that your area is the biggest possible." <a class="pdflink" href="/content/id/2663/FenceIt.pdf">This worksheet</a> has all three parts of the <a href="/2663">problem</a> on, so you could hand this out to students who finish the second task quickly. Before the end of the lesson, allow some time for students to present their findings, focussing particularly on their justifications. <h3>Key questions</h3> What would the width of the rectangle be if the length is 1m? 2m? 3m? ... How can I be sure I have found the maximum possible area? <h3>Possible extension</h3> Some students may use algebra to represent the scenarios and use graphical methods to find/justify the optimal solution. Relaxing the straight lines constraint would allow students to consider some circle geometry: "Could you enclose even greater areas if you had 40m of flexible wire fencing that could fence off curves as well as straight lines?" <h3>Possible support</h3> Share the diagrams above to help students to represent their trial-and-improvement strategies in a clear way that shows how the area changes. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">Choose a length for your rectangle. What must the width be if the fence is $40$m long? What's the area? What happens to the area when you change your chosen length? </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> The solutions below all refer to this diagram: <mdo:image alt="" height="360" src="fenceit3.png" width="347"></mdo:image> Many solutions got very close to the optimum solution : Pauline and Amy from Moorfield Junior School, Mollie from St.Michael's C.of.E Primary School, and Akalya and Vanessa from Devonshire Primary School all arrived at the same conclusion. This is what Akalya and Vanessa wrote: For this problem you have to do a trial and error method. My answers were: $x = 14$m $y = 18$m, and the third side is $8$m. The maximum area was $192$m². Fred and Matt from Albion Heights Junior Middle School noticed that there were two solutions that gave an area of 192m²: I (Matt) began to use paper, but soon realised it could be solved mentally. We realised that on the left side, ten metres of fencing would be required for every possible shape. So we would have to make a shape with $30$m. Using a trial and error method, we came upon two maximums: $8\times 14$ (where $x =14$) and $7\times16$ (where $x=16$). Both rectangles have an area of $112$m². To add up the amount of fencing, add two times the width to the length. ($16+14=30$, $14+16=30$) To get the maximum area, we add the $10\times8$ rectangle to the $8\times14$ (or to the $7\times16$) rectangle to get $192$m². Amy from Mason Middle School spotted that if you didn't stick to whole numbers you could improve on $192$m²: If $x = 15$, $y = 17.5$, and the other side $= 7.5$, the total area will be $192.5$m². Tom from De La Salle explained how he reached this optimum solution: I have found that to maximize the area of the pen: $y= 17.5$ and therefore $x = 15$. I found this out by listing all the possible values of $y$ ($11$ to $25$) and the corresponding values of $x$ ($28$ to $0$). I then calculated the area for each iteration. However, I found that the area was equal when y was either $17$ or $18$. This led me to believe that the optimal solution lay half way between $17$ and $18$. Therefore, for the above reasons, $y$ should be $17.5$ to maximize the pen. Well done Amy and Tom. </mdoxml> 2 3 0 0 1 0 0 Fence it If you have only 40 metres of fencing available, what is the maximum area of land you can fence off? 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