the perimeter of a rectangle is 60 feet. describe the possible lengths of a side if the area of the…

Explanation:

Step1: Let the length be $l$ and width be $w$.

Given perimeter $P = 2(l + w)=60$, so $l + w=30$, then $w = 30 - l$.

Step2: Set up area - inequality.

The area $A=lw$. Since $A\leq144$ and $w = 30 - l$, we have $l(30 - l)\leq144$.

Step3: Rearrange the inequality.

Expand to get $30l-l^{2}\leq144$, then $l^{2}-30l + 144\geq0$.

Step4: Solve the quadratic equation $l^{2}-30l + 144 = 0$.

Using the quadratic formula $l=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for $ax^{2}+bx + c = 0$. Here $a = 1$, $b=-30$, $c = 144$. $l=\frac{30\pm\sqrt{(-30)^{2}-4\times1\times144}}{2\times1}=\frac{30\pm\sqrt{900 - 576}}{2}=\frac{30\pm\sqrt{324}}{2}=\frac{30\pm18}{2}$. The roots are $l_1=\frac{30 + 18}{2}=24$ and $l_2=\frac{30 - 18}{2}=6$.

Step5: Determine the solution of the inequality.

The quadratic function $y=l^{2}-30l + 144$ is a parabola opening upwards. The solution of $l^{2}-30l + 144\geq0$ is $l\leq6$ or $l\geq24$. Also, since $l>0$ and $w = 30 - l>0$ (lengths are non - negative), when $l\leq6$, $w=30 - l\geq24$; when $l\geq24$, $w=30 - l\leq6$.

Answer:

$(0,6]\cup[24,30)$