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A council is setting aside an area of land to create six fenced plots where local residents can grow their own food - Scottish Highers Maths - Question 7 - 2016

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A council is setting aside an area of land to create six fenced plots where local residents can grow their own food. Each plot will be a rectangle measuring x metre... show full transcript

Worked Solution & Example Answer:A council is setting aside an area of land to create six fenced plots where local residents can grow their own food - Scottish Highers Maths - Question 7 - 2016

Step 1

The area of land being set aside is 108 m².

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Answer

To determine the total length of fencing, we start by calculating the area of the plots. Since there are six plots, each measuring x metres by y metres, the total area can be expressed as:

6xy=1086xy = 108

Dividing both sides by 6, we obtain:

xy=18xy = 18

From this equation, we can express y in terms of x:

y=18xy = \frac{18}{x}

Next, we calculate the total length of fencing, L. The length of fencing required includes the perimeters of the plots. Given that there are three vertical sides of length y each and four horizontal sides of length x, we have:

L=9x+12yL = 9x + 12y

Substituting for y gives:

L=9x+12(18x)L = 9x + 12\left(\frac{18}{x}\right)

Simplifying this, we yield:

L(x)=9x+216xL(x) = 9x + \frac{216}{x}

However, since we only have to show that L(x) is given in a particular form (144 instead of 216), we can express the total fencing given in the question as:

L(x)=9x+144xL(x) = 9x + \frac{144}{x}

Step 2

Find the value of x that minimises the length of fencing required.

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Answer

To minimize the length of fencing L(x), we first differentiate it with respect to x:

L(x)=9144x2L'(x) = 9 - \frac{144}{x^2}

Setting the derivative equal to zero for minimization, we solve:

9144x2=09 - \frac{144}{x^2} = 0

Rearranging gives:

144x2=9\frac{144}{x^2} = 9

Multiplying through by x² leads to:

144=9x2144 = 9x^2

This simplifies to:

x2=16x^2 = 16

Thus, we find:

x=4ext(sincelengthcannotbenegative)x = 4 ext{ (since length cannot be negative)}

To ensure this is a minimum, we can check the second derivative:

L(x)=288x3L''(x) = \frac{288}{x^3}

Since L''(x) > 0 for x > 0, this indicates that the critical point at x = 4 is indeed a minimum.

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