Question

a rancher wants to use 300 feet of fencing to enclose a rectangular area of 4400 square feet. what dimensions should the rectangle have? the rectangles dimensions should be □ feet. (simplify your answers. type the length and width, separated by commas; do not type the units.)

Explanation:

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

The perimeter formula is $2(l + w)=300$, so $l + w = 150$, which gives $l=150 - w$. The area formula is $A=lw$, and $A = 4400$. Substitute $l = 150 - w$ into the area formula: $(150 - w)w=4400$.

Step2: Expand the equation.

$150w-w^{2}=4400$. Rearrange to get a quadratic equation $w^{2}-150w + 4400=0$.

Step3: Use the quadratic formula $w=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for $aw^{2}+bw + c = 0$.

Here $a = 1$, $b=-150$, $c = 4400$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-150)^{2}-4\times1\times4400=22500 - 17600=4900$. Then $w=\frac{150\pm\sqrt{4900}}{2}=\frac{150\pm70}{2}$.

Step4: Find the two possible values of $w$.

$w_1=\frac{150 + 70}{2}=\frac{220}{2}=110$ and $w_2=\frac{150 - 70}{2}=\frac{80}{2}=40$.
If $w = 110$, then $l=150 - 110 = 40$. If $w = 40$, then $l=150 - 40=110$.

Answer:

40,110