<p>A farmer uses $160$m of fencing to enclose a plot of his land in a shape that comprises an isosceles triangle and a rectangle, with the dimensions shown.</p>
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<ol class="i-list"><li><div class="list-item-paragraph">Show that the area of the plot is $\frac{320\sqrt{3}x - (3\sqrt{3} + 6)x^2}{4} m^2$.</div></li><li><div class="list-item-paragraph">Given that $x$ can vary, find the value of $x$ for which the area of the plot is stationary.</div></li><li><div class="list-item-paragraph">Explain why this value of $x$ gives the farmer the largest possible area for the plot. Find this area and give your answer correct to the nearest square metre.</div></li></ol>

A farmer uses $160$m of fencing to enclose a plot of his land in a shape that comprises an isosceles triangle and a rectangle, with the dimensions shown.

(i) Show that the area of the plot is $\frac{320\sqrt{3}x - (3\sqrt{3} + 6)x^2}{4} m^2$.

Show that the area of the plot is $\frac{320\sqrt{3}x - (3\sqrt{3} + 6)x^2}{4} m^2$.

(ii) Given that $x$ can vary, find the value of $x$ for which the area of the plot is stationary.

Given that $x$ can vary, find the value of $x$ for which the area of the plot is stationary.

(iii) Explain why this value of $x$ gives the farmer the largest possible area for the plot. Find this area and give your answer correct to the nearest square metre.

Explain why this value of $x$ gives the farmer the largest possible area for the plot. Find this area and give your answer correct to the nearest square metre.

Applications of Differentiation

A Math

A Math (IGCSE)

IP Math

2017 PLMGS P1

$y = \frac{\sqrt{3}}{2}x$, so $2y = \sqrt{3}x$

$l = \frac{1}{2}[160 - (2 + \sqrt{3})x]$

Total area: $A = \frac{1}{2}x \cdot x \sin 120° + l(\sqrt{3}x)$

$A = \frac{1}{2}x^2 \cdot \frac{\sqrt{3}}{2} + \sqrt{3}x \cdot \frac{1}{2}[160 - (2 + \sqrt{3})x]$

$= \frac{\sqrt{3}}{4}x^2 + \sqrt{3}x\left(80 - \frac{2 + \sqrt{3}}{2}x\right)$

$= \frac{\sqrt{3}}{4}x^2 + \frac{80\sqrt{3}x}{1} - \frac{\sqrt{3}(2 + \sqrt{3})}{2}x^2$

$= \frac{\sqrt{3}x^2 + 320\sqrt{3}x - 2(2\sqrt{3} + 3)x^2}{4}$

$= \frac{320\sqrt{3}x - (3\sqrt{3} + 6)x^2}{4}$ (shown)

$A = \frac{1}{4}[320\sqrt{3}x - (3\sqrt{3} + 6)x^2]$

$\frac{\mathrm{d}A}{\mathrm{d}x} = \frac{1}{4}[320\sqrt{3} - 2(3\sqrt{3} + 6)x]$

At stationary value, $\frac{\mathrm{d}A}{\mathrm{d}x} = 0$:

$x = \frac{320\sqrt{3}}{2(3\sqrt{3} + 6)}$

$\frac{\mathrm{d}^2A}{dx^2} = \frac{1}{4}(0 - 2(3\sqrt{3} + 6))$

Since $\frac{\mathrm{d}^2A}{dx^2} < 0$, by 2nd derivative test, $A$ is maximum at $x = 24.8$ m (3 s.f.). Thus, $x$ gives the largest possible area for the plot.

$A = \frac{320\sqrt{3}(24.752) - (3\sqrt{3} + 6)(24.752)^2}{4}$

2017 PLMGS P1

2017 PLMGS P1

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