A rectangular site, with one side facing a road, is to be fenced off - Junior Cycle Mathematics - Question 13 - 2014

The area $A$ of the rectangular site is given by the formula:

$A = l imes x$

Substituting the expression for $l$ from part (i), we get:

$A = (140 - 2x) imes x = 140x - 2x^2$

This simplifies to:

$A = -2x^2 + 140x$

Using the function $f(x) = -2x^2 + 140x$, we evaluate:

• For $x = 0$: $f(0) = -2(0)^2 + 140(0) = 0$
• For $x = 10$: $f(10) = -2(10)^2 + 140(10) = 1200$
• For $x = 20$: $f(20) = -2(20)^2 + 140(20) = 2400$
• For $x = 30$: $f(30) = -2(30)^2 + 140(30) = 3600$
• For $x = 40$: $f(40) = -2(40)^2 + 140(40) = 4800$
• For $x = 50$: $f(50) = -2(50)^2 + 140(50) = 6000$
• For $x = 60$: $f(60) = -2(60)^2 + 140(60) = 7200$
• For $x = 70$: $f(70) = -2(70)^2 + 140(70) = 8400$

From the graph of $f(x)$, we observe that the maximum area occurs at $x = 35$, which corresponds to an area of:

$f(35) = -2(35)^2 + 140(35) = 2450 ext{ m}^2.$

We know from part (i) that $l = 140 - 2x$. Given $l = 30$, we can set up the equation:

$30 = 140 - 2x$

Solving for $x$, we find:

ightarrow x = 55 ext{ m}.$$ Next, we substitute back into the area formula: $$A = -2(55)^2 + 140(55) = 1650 ext{ m}^2.$$