Quadratics: Question 9

Worked solution

Part (a): Forming the area expression

Let the two side lengths be $x$ and $y$ metres. The perimeter condition gives:

$2(x+y) = 48 \implies x+y = 24 \implies y = 24-x$

The area is:

$A = xy = x(24-x) = 24x - x^2$

as required.

Part (b): Solving the quadratic inequality

We need $A \ge 140$:

$24x - x^2 \ge 140$

Rearrange so the $x^2$ term is positive (multiply both sides by $-1$, which reverses the inequality):

$x^2 - 24x + 140 \le 0$

Find the critical values by solving $x^2-24x+140=0$:

$\text{discriminant} = (-24)^2 - 4(1)(140) = 576 - 560 = 16$

$x = \frac{24 \pm \sqrt{16}}{2} = \frac{24 \pm 4}{2}$

$x = \frac{28}{2} = 14 \qquad \text{or} \qquad x = \frac{20}{2} = 10$

Since the coefficient of $x^2$ in $x^2-24x+140$ is $a=1>0$, this graph is an upward-opening parabola, so it is at or below the $x$-axis between the roots. We want $x^2-24x+140\le0$, so:

$10 \le x \le 14$

(This is also sensible physically: since $y=24-x>0$ requires $0<x<24$, the interval $10\le x\le14$ lies safely inside the valid range of side lengths.)

Part (c): Checking the boundary values

At $x=10$:

$A = 10(24-10) = 10 \times 14 = 140 \text{ m}^2 \checkmark$

The function $A=x(24-x)=24x-x^2$ is a downward-opening parabola with vertex (maximum) at $x=12$ (found by completing the square: $A=-(x-12)^2+144$), so $A$ is symmetric about $x=12$. Since $14$ is $2$ more than $12$ and $10$ is $2$ less than $12$, both give the same area:

$A(14) = 14(24-14) = 14 \times 10 = 140 \text{ m}^2 \checkmark$

This confirms $x=10$ and $x=14$ are indeed the two points where the area equals exactly $140$ m², matching the boundary values found in part (b).

Final answers

• (a) $A = 24x - x^2$ (shown)
• (b) $\boxed{10 \le x \le 14}$
• (c) $A(10)=A(14)=\boxed{140}$ m², by the symmetry of $A=x(24-x)$ about $x=12$